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Transformation of random variables and joint distributions


Plotting confidence intervalsWhat is the PDF of a variable where a parameter is itself a random variable?NProbability not reliability analysis?Mathematica function to calculate equivalent NormalDistribution from a WeibullDistributionPDF for square of Rician random variable?Convolve discrete random variables efficientlyDistribution of Function of Random Sum of Random VariablesSketching Normal Probability Distributions GraphsConstruct Distribution Histogram From Random VariableNormal distribution plot construction













3












$begingroup$


Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation



$y_i$ ~ NormalDistribution[0,$σ_y$ ]



I want to obtain with Mathematica:



  1. The distribution of:
    $x = bary = frac sum_i=1^ny_in$


  2. The joint distribution of $ (x,y_i )$


Thank you for your helpful comments










share|improve this question









New contributor




Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 4




    $begingroup$
    What have you tried? For example, have you seen the documentation on TransformedDistribution and ProbabilityDistribution?
    $endgroup$
    – JimB
    6 hours ago










  • $begingroup$
    @JimB . I tried this TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]. The result is NormalDistribution[0, [Sigma]y]. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
    $endgroup$
    – Andrea2810
    5 hours ago











  • $begingroup$
    You need to "index" the variable y or else Mathematica thinks it is a single variable.
    $endgroup$
    – JimB
    1 hour ago















3












$begingroup$


Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation



$y_i$ ~ NormalDistribution[0,$σ_y$ ]



I want to obtain with Mathematica:



  1. The distribution of:
    $x = bary = frac sum_i=1^ny_in$


  2. The joint distribution of $ (x,y_i )$


Thank you for your helpful comments










share|improve this question









New contributor




Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 4




    $begingroup$
    What have you tried? For example, have you seen the documentation on TransformedDistribution and ProbabilityDistribution?
    $endgroup$
    – JimB
    6 hours ago










  • $begingroup$
    @JimB . I tried this TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]. The result is NormalDistribution[0, [Sigma]y]. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
    $endgroup$
    – Andrea2810
    5 hours ago











  • $begingroup$
    You need to "index" the variable y or else Mathematica thinks it is a single variable.
    $endgroup$
    – JimB
    1 hour ago













3












3








3





$begingroup$


Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation



$y_i$ ~ NormalDistribution[0,$σ_y$ ]



I want to obtain with Mathematica:



  1. The distribution of:
    $x = bary = frac sum_i=1^ny_in$


  2. The joint distribution of $ (x,y_i )$


Thank you for your helpful comments










share|improve this question









New contributor




Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Given a variable $y_i$, normally distributed with 0 mean and $σ_y$ standard deviation



$y_i$ ~ NormalDistribution[0,$σ_y$ ]



I want to obtain with Mathematica:



  1. The distribution of:
    $x = bary = frac sum_i=1^ny_in$


  2. The joint distribution of $ (x,y_i )$


Thank you for your helpful comments







probability-or-statistics






share|improve this question









New contributor




Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 2 hours ago









mjw

9679




9679






New contributor




Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 6 hours ago









Andrea2810Andrea2810

162




162




New contributor




Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Andrea2810 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 4




    $begingroup$
    What have you tried? For example, have you seen the documentation on TransformedDistribution and ProbabilityDistribution?
    $endgroup$
    – JimB
    6 hours ago










  • $begingroup$
    @JimB . I tried this TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]. The result is NormalDistribution[0, [Sigma]y]. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
    $endgroup$
    – Andrea2810
    5 hours ago











  • $begingroup$
    You need to "index" the variable y or else Mathematica thinks it is a single variable.
    $endgroup$
    – JimB
    1 hour ago












  • 4




    $begingroup$
    What have you tried? For example, have you seen the documentation on TransformedDistribution and ProbabilityDistribution?
    $endgroup$
    – JimB
    6 hours ago










  • $begingroup$
    @JimB . I tried this TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]. The result is NormalDistribution[0, [Sigma]y]. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
    $endgroup$
    – Andrea2810
    5 hours ago











  • $begingroup$
    You need to "index" the variable y or else Mathematica thinks it is a single variable.
    $endgroup$
    – JimB
    1 hour ago







4




4




$begingroup$
What have you tried? For example, have you seen the documentation on TransformedDistribution and ProbabilityDistribution?
$endgroup$
– JimB
6 hours ago




$begingroup$
What have you tried? For example, have you seen the documentation on TransformedDistribution and ProbabilityDistribution?
$endgroup$
– JimB
6 hours ago












$begingroup$
@JimB . I tried this TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]. The result is NormalDistribution[0, [Sigma]y]. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago





$begingroup$
@JimB . I tried this TransformedDistribution[Sum[y, i, n]/n, y [Distributed] NormalDistribution[0, [Sigma]y]]. The result is NormalDistribution[0, [Sigma]y]. However, the correct result should be NormalDistribution[0, [Sigma]y / Sqrt[n]]
$endgroup$
– Andrea2810
5 hours ago













$begingroup$
You need to "index" the variable y or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
1 hour ago




$begingroup$
You need to "index" the variable y or else Mathematica thinks it is a single variable.
$endgroup$
– JimB
1 hour ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.



First the distribution of the mean:



marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]


$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$



So we see that the marginal distribution of $bary$ is



NormalDistribution[0, σ/Sqrt[n]]


The joint distribution of $bary$ and, say, $y_1$ is given by



jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm


$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$



So the general distribution is a multivariate normal



MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]


The general form of the joint density function can then be found with



FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]


$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$






share|improve this answer











$endgroup$












  • $begingroup$
    Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
    $endgroup$
    – mjw
    56 mins ago











  • $begingroup$
    @mjw Good. Answers should always be scrutinized and challenged if desired.
    $endgroup$
    – JimB
    54 mins ago



















0












$begingroup$

Here is the distribution of $x=overliney$ (Part I of your question):



n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]


The result is



NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]


UPDATE



Okay, here is how to do it with $n$ a variable:



a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n]; 
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];


Now



x [Distributed] p[5] (* n=5, for example *)


Again, the result is



x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]





share|improve this answer











$endgroup$












  • $begingroup$
    I am not sure, but shouldn't be n instead of 5 here TransformedDistribution[Sum[y[k]/n, k, 5], a] ? And what if I want to leave n, without assigning a value to n? Thanks @mjw
    $endgroup$
    – Andrea2810
    2 hours ago











  • $begingroup$
    Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
    $endgroup$
    – mjw
    2 hours ago











  • $begingroup$
    Let's go with five because it is clearer. The result is NormalDistribution[0,[Sigma]/Sqrt[5]]. Not sure why Mathematica puts Abs[] around $sigma$. Obviously, $sigma>0$.
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
    $endgroup$
    – Andrea2810
    2 hours ago










  • $begingroup$
    a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
    $endgroup$
    – mjw
    2 hours ago



















0












$begingroup$

just modified @mjw's answer,



n = 100;(*for example*)ClearAll[y]; 
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]


JointDistribution can be composed by ProductDistribution,
if these random variables are independent.

if not,you have to use Copula



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D


enter image description here



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]


=>



1., -0.00256777, -0.00256777, 1.


enter image description here

I'm not sure about correlation,but it's okay.






share|improve this answer











$endgroup$












  • $begingroup$
    I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Exactly, the two variables are not independent unfortunately
    $endgroup$
    – Andrea2810
    2 hours ago










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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.



First the distribution of the mean:



marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]


$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$



So we see that the marginal distribution of $bary$ is



NormalDistribution[0, σ/Sqrt[n]]


The joint distribution of $bary$ and, say, $y_1$ is given by



jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm


$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$



So the general distribution is a multivariate normal



MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]


The general form of the joint density function can then be found with



FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]


$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$






share|improve this answer











$endgroup$












  • $begingroup$
    Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
    $endgroup$
    – mjw
    56 mins ago











  • $begingroup$
    @mjw Good. Answers should always be scrutinized and challenged if desired.
    $endgroup$
    – JimB
    54 mins ago
















2












$begingroup$

I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.



First the distribution of the mean:



marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]


$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$



So we see that the marginal distribution of $bary$ is



NormalDistribution[0, σ/Sqrt[n]]


The joint distribution of $bary$ and, say, $y_1$ is given by



jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm


$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$



So the general distribution is a multivariate normal



MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]


The general form of the joint density function can then be found with



FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]


$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$






share|improve this answer











$endgroup$












  • $begingroup$
    Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
    $endgroup$
    – mjw
    56 mins ago











  • $begingroup$
    @mjw Good. Answers should always be scrutinized and challenged if desired.
    $endgroup$
    – JimB
    54 mins ago














2












2








2





$begingroup$

I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.



First the distribution of the mean:



marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]


$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$



So we see that the marginal distribution of $bary$ is



NormalDistribution[0, σ/Sqrt[n]]


The joint distribution of $bary$ and, say, $y_1$ is given by



jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm


$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$



So the general distribution is a multivariate normal



MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]


The general form of the joint density function can then be found with



FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]


$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$






share|improve this answer











$endgroup$



I don't know how to get Mathematica to get the joint distribution explicitly for a general value of $n$ but here is how one can easily see the pattern to figure out the general solution.



First the distribution of the mean:



marginalDistribution = TransformedDistribution[Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n],
Assumptions -> [Sigma] > 0]
#, marginalDistribution/.n-># &/@Range[2,10]


$$
beginarraycc
2 & textNormalDistributionleft[0,fracsigma sqrt2right] \
3 & textNormalDistributionleft[0,fracsigma sqrt3right] \
4 & textNormalDistributionleft[0,fracsigma 2right] \
5 & textNormalDistributionleft[0,fracsigma sqrt5right] \
6 & textNormalDistributionleft[0,fracsigma sqrt6right] \
7 & textNormalDistributionleft[0,fracsigma sqrt7right] \
8 & textNormalDistributionleft[0,fracsigma 2 sqrt2right] \
9 & textNormalDistributionleft[0,fracsigma 3right] \
10 & textNormalDistributionleft[0,fracsigma sqrt10right] \
endarray
$$



So we see that the marginal distribution of $bary$ is



NormalDistribution[0, σ/Sqrt[n]]


The joint distribution of $bary$ and, say, $y_1$ is given by



jointDistribution = TransformedDistribution[y[1], Sum[y[i], i, n]/n, 
Table[y[i] [Distributed] NormalDistribution[0, [Sigma]], i, n]]
#, jointDistribution /. n -> # & /@ Range[2, 10] // TableForm


$$
beginarraycc
2 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^22 \
fracsigma ^22 & fracsigma ^22 \
endarray
right)right] \
3 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^23 \
fracsigma ^23 & fracsigma ^23 \
endarray
right)right] \
4 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^24 \
fracsigma ^24 & fracsigma ^24 \
endarray
right)right] \
5 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^25 \
fracsigma ^25 & fracsigma ^25 \
endarray
right)right] \
6 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^26 \
fracsigma ^26 & fracsigma ^26 \
endarray
right)right] \
7 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^27 \
fracsigma ^27 & fracsigma ^27 \
endarray
right)right] \
8 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^28 \
fracsigma ^28 & fracsigma ^28 \
endarray
right)right] \
9 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^29 \
fracsigma ^29 & fracsigma ^29 \
endarray
right)right] \
10 & textMultinormalDistributionleft[0,0,left(
beginarraycc
sigma ^2 & fracsigma ^210 \
fracsigma ^210 & fracsigma ^210 \
endarray
right)right] \
endarray
$$



So the general distribution is a multivariate normal



MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n]


The general form of the joint density function can then be found with



FullSimplify[PDF[MultinormalDistribution[0, 0, σ^2, σ^2/n, σ^2/n, σ^2/n], y, ybar],
Assumptions -> σ > 0, n > 1]


$$fracn e^-fracn left(n textybar^2+y^2-2 y textybarright)2 (n-1) sigma ^22 pi sqrtn-1 sigma ^2$$







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 1 hour ago









JimBJimB

18k12863




18k12863











  • $begingroup$
    Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
    $endgroup$
    – mjw
    56 mins ago











  • $begingroup$
    @mjw Good. Answers should always be scrutinized and challenged if desired.
    $endgroup$
    – JimB
    54 mins ago

















  • $begingroup$
    Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
    $endgroup$
    – mjw
    56 mins ago











  • $begingroup$
    @mjw Good. Answers should always be scrutinized and challenged if desired.
    $endgroup$
    – JimB
    54 mins ago
















$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
56 mins ago





$begingroup$
Anyway, I like your answer! I'll have to look at it to understand (not obvious (to me) that this would be the solution).
$endgroup$
– mjw
56 mins ago













$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
54 mins ago





$begingroup$
@mjw Good. Answers should always be scrutinized and challenged if desired.
$endgroup$
– JimB
54 mins ago












0












$begingroup$

Here is the distribution of $x=overliney$ (Part I of your question):



n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]


The result is



NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]


UPDATE



Okay, here is how to do it with $n$ a variable:



a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n]; 
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];


Now



x [Distributed] p[5] (* n=5, for example *)


Again, the result is



x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]





share|improve this answer











$endgroup$












  • $begingroup$
    I am not sure, but shouldn't be n instead of 5 here TransformedDistribution[Sum[y[k]/n, k, 5], a] ? And what if I want to leave n, without assigning a value to n? Thanks @mjw
    $endgroup$
    – Andrea2810
    2 hours ago











  • $begingroup$
    Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
    $endgroup$
    – mjw
    2 hours ago











  • $begingroup$
    Let's go with five because it is clearer. The result is NormalDistribution[0,[Sigma]/Sqrt[5]]. Not sure why Mathematica puts Abs[] around $sigma$. Obviously, $sigma>0$.
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
    $endgroup$
    – Andrea2810
    2 hours ago










  • $begingroup$
    a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
    $endgroup$
    – mjw
    2 hours ago
















0












$begingroup$

Here is the distribution of $x=overliney$ (Part I of your question):



n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]


The result is



NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]


UPDATE



Okay, here is how to do it with $n$ a variable:



a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n]; 
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];


Now



x [Distributed] p[5] (* n=5, for example *)


Again, the result is



x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]





share|improve this answer











$endgroup$












  • $begingroup$
    I am not sure, but shouldn't be n instead of 5 here TransformedDistribution[Sum[y[k]/n, k, 5], a] ? And what if I want to leave n, without assigning a value to n? Thanks @mjw
    $endgroup$
    – Andrea2810
    2 hours ago











  • $begingroup$
    Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
    $endgroup$
    – mjw
    2 hours ago











  • $begingroup$
    Let's go with five because it is clearer. The result is NormalDistribution[0,[Sigma]/Sqrt[5]]. Not sure why Mathematica puts Abs[] around $sigma$. Obviously, $sigma>0$.
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
    $endgroup$
    – Andrea2810
    2 hours ago










  • $begingroup$
    a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
    $endgroup$
    – mjw
    2 hours ago














0












0








0





$begingroup$

Here is the distribution of $x=overliney$ (Part I of your question):



n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]


The result is



NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]


UPDATE



Okay, here is how to do it with $n$ a variable:



a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n]; 
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];


Now



x [Distributed] p[5] (* n=5, for example *)


Again, the result is



x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]





share|improve this answer











$endgroup$



Here is the distribution of $x=overliney$ (Part I of your question):



n = 5; (* for example *)
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
TransformedDistribution[Sum[y[k]/n, k, 5], a]


The result is



NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]


UPDATE



Okay, here is how to do it with $n$ a variable:



a[n_] := Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n]; 
p[n_] := TransformedDistribution[Sum[y[k]/n, k, n], a[n]];


Now



x [Distributed] p[5] (* n=5, for example *)


Again, the result is



x [Distributed] NormalDistribution[0, Abs[[Sigma]]/Sqrt[5]]






share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 3 hours ago









mjwmjw

9679




9679











  • $begingroup$
    I am not sure, but shouldn't be n instead of 5 here TransformedDistribution[Sum[y[k]/n, k, 5], a] ? And what if I want to leave n, without assigning a value to n? Thanks @mjw
    $endgroup$
    – Andrea2810
    2 hours ago











  • $begingroup$
    Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
    $endgroup$
    – mjw
    2 hours ago











  • $begingroup$
    Let's go with five because it is clearer. The result is NormalDistribution[0,[Sigma]/Sqrt[5]]. Not sure why Mathematica puts Abs[] around $sigma$. Obviously, $sigma>0$.
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
    $endgroup$
    – Andrea2810
    2 hours ago










  • $begingroup$
    a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
    $endgroup$
    – mjw
    2 hours ago

















  • $begingroup$
    I am not sure, but shouldn't be n instead of 5 here TransformedDistribution[Sum[y[k]/n, k, 5], a] ? And what if I want to leave n, without assigning a value to n? Thanks @mjw
    $endgroup$
    – Andrea2810
    2 hours ago











  • $begingroup$
    Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
    $endgroup$
    – mjw
    2 hours ago











  • $begingroup$
    Let's go with five because it is clearer. The result is NormalDistribution[0,[Sigma]/Sqrt[5]]. Not sure why Mathematica puts Abs[] around $sigma$. Obviously, $sigma>0$.
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
    $endgroup$
    – Andrea2810
    2 hours ago










  • $begingroup$
    a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
    $endgroup$
    – mjw
    2 hours ago
















$begingroup$
I am not sure, but shouldn't be n instead of 5 here TransformedDistribution[Sum[y[k]/n, k, 5], a] ? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 hours ago





$begingroup$
I am not sure, but shouldn't be n instead of 5 here TransformedDistribution[Sum[y[k]/n, k, 5], a] ? And what if I want to leave n, without assigning a value to n? Thanks @mjw
$endgroup$
– Andrea2810
2 hours ago













$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago





$begingroup$
Oh yes, you are right! I started with 10 and changed to five as I was trying it out. I'll fix it ... Thanks!
$endgroup$
– mjw
2 hours ago













$begingroup$
Let's go with five because it is clearer. The result is NormalDistribution[0,[Sigma]/Sqrt[5]]. Not sure why Mathematica puts Abs[] around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 hours ago




$begingroup$
Let's go with five because it is clearer. The result is NormalDistribution[0,[Sigma]/Sqrt[5]]. Not sure why Mathematica puts Abs[] around $sigma$. Obviously, $sigma>0$.
$endgroup$
– mjw
2 hours ago












$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 hours ago




$begingroup$
Yes, sure it is clearer. Do you have any idea of how can I use n as a parameter, without assigning a value to n?
$endgroup$
– Andrea2810
2 hours ago












$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
2 hours ago





$begingroup$
a[n_] = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
$endgroup$
– mjw
2 hours ago












0












$begingroup$

just modified @mjw's answer,



n = 100;(*for example*)ClearAll[y]; 
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]


JointDistribution can be composed by ProductDistribution,
if these random variables are independent.

if not,you have to use Copula



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D


enter image description here



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]


=>



1., -0.00256777, -0.00256777, 1.


enter image description here

I'm not sure about correlation,but it's okay.






share|improve this answer











$endgroup$












  • $begingroup$
    I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Exactly, the two variables are not independent unfortunately
    $endgroup$
    – Andrea2810
    2 hours ago















0












$begingroup$

just modified @mjw's answer,



n = 100;(*for example*)ClearAll[y]; 
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]


JointDistribution can be composed by ProductDistribution,
if these random variables are independent.

if not,you have to use Copula



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D


enter image description here



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]


=>



1., -0.00256777, -0.00256777, 1.


enter image description here

I'm not sure about correlation,but it's okay.






share|improve this answer











$endgroup$












  • $begingroup$
    I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Exactly, the two variables are not independent unfortunately
    $endgroup$
    – Andrea2810
    2 hours ago













0












0








0





$begingroup$

just modified @mjw's answer,



n = 100;(*for example*)ClearAll[y]; 
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]


JointDistribution can be composed by ProductDistribution,
if these random variables are independent.

if not,you have to use Copula



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D


enter image description here



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]


=>



1., -0.00256777, -0.00256777, 1.


enter image description here

I'm not sure about correlation,but it's okay.






share|improve this answer











$endgroup$



just modified @mjw's answer,



n = 100;(*for example*)ClearAll[y]; 
a = Table[y[k] [Distributed] NormalDistribution[0, [Sigma]], k, 1, n];
meanDist = TransformedDistribution[Sum[y[k]/100, k, 100], a]


JointDistribution can be composed by ProductDistribution,
if these random variables are independent.

if not,you have to use Copula



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
RandomVariate[joint, 100] // Histogram3D


enter image description here



joint = ProductDistribution[meanDist, 
Last@*List @@ Part[a, 1]] /. [Sigma] -> 1;
m1 = RandomVariate[meanDist /. [Sigma] -> 1, 100000];
m2 = RandomVariate[
Last@*List @@ Part[a, 1] /. [Sigma] -> 1, 100000];
Correlation[Thread[List[m1, m2]]]
ListPlot[Thread[List[m1, m2]]]


=>



1., -0.00256777, -0.00256777, 1.


enter image description here

I'm not sure about correlation,but it's okay.







share|improve this answer














share|improve this answer



share|improve this answer








edited 1 hour ago

























answered 2 hours ago









XminerXminer

19918




19918











  • $begingroup$
    I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Exactly, the two variables are not independent unfortunately
    $endgroup$
    – Andrea2810
    2 hours ago
















  • $begingroup$
    I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
    $endgroup$
    – mjw
    2 hours ago










  • $begingroup$
    Exactly, the two variables are not independent unfortunately
    $endgroup$
    – Andrea2810
    2 hours ago















$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago




$begingroup$
I believe that the distributions are not independent. Since $overlinex$ is computed from $y_i$ and other $y_j$'s, it would seem to be dependent. We could compute whether or not the distributions are dependent ...
$endgroup$
– mjw
2 hours ago












$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago




$begingroup$
I would also recommend using 10^6 rather than 100, you'll get a sharper plot!
$endgroup$
– mjw
2 hours ago












$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago




$begingroup$
Exactly, the two variables are not independent unfortunately
$endgroup$
– Andrea2810
2 hours ago










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