Computing the expectation of the number of balls in a box The 2019 Stack Overflow Developer Survey Results Are InThere is two boxes with one with 8 balls and one with 4 ballsdrawing balls from box without replacemntRandom distribution of colored balls into boxes.Optimal Number of White BallsCompute possible outcomes when get balls from a boxPoisson Approximation Problem involving putting balls into boxesCompute expected received balls from boxesput n balls into n boxesA question of probability regarding expectation and variance of a random variable.Distributing 5 distinct balls into 3 distinct boxes
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Computing the expectation of the number of balls in a box
The 2019 Stack Overflow Developer Survey Results Are InThere is two boxes with one with 8 balls and one with 4 ballsdrawing balls from box without replacemntRandom distribution of colored balls into boxes.Optimal Number of White BallsCompute possible outcomes when get balls from a boxPoisson Approximation Problem involving putting balls into boxesCompute expected received balls from boxesput n balls into n boxesA question of probability regarding expectation and variance of a random variable.Distributing 5 distinct balls into 3 distinct boxes
$begingroup$
- There are $r$ boxes and $n$ balls.
- Each ball is placed in a box with equal probability, independently of the other balls.
- Let $X_i$ be the number of balls in box $i$,
$1 leq i leq r$. - Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.
I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.
probability-theory
$endgroup$
add a comment |
$begingroup$
- There are $r$ boxes and $n$ balls.
- Each ball is placed in a box with equal probability, independently of the other balls.
- Let $X_i$ be the number of balls in box $i$,
$1 leq i leq r$. - Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.
I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.
probability-theory
$endgroup$
$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
6 hours ago
$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
6 hours ago
$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
5 hours ago
$begingroup$
Have you studied covariance matrices, or vector-valued random variables, at all? That would seem to me to provide the most compact notation for solving this problem.
$endgroup$
– Daniel Schepler
21 mins ago
add a comment |
$begingroup$
- There are $r$ boxes and $n$ balls.
- Each ball is placed in a box with equal probability, independently of the other balls.
- Let $X_i$ be the number of balls in box $i$,
$1 leq i leq r$. - Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.
I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.
probability-theory
$endgroup$
- There are $r$ boxes and $n$ balls.
- Each ball is placed in a box with equal probability, independently of the other balls.
- Let $X_i$ be the number of balls in box $i$,
$1 leq i leq r$. - Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.
I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.
probability-theory
probability-theory
edited 6 hours ago
Felix Marin
68.9k7110147
68.9k7110147
asked 6 hours ago
631631
585
585
$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
6 hours ago
$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
6 hours ago
$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
5 hours ago
$begingroup$
Have you studied covariance matrices, or vector-valued random variables, at all? That would seem to me to provide the most compact notation for solving this problem.
$endgroup$
– Daniel Schepler
21 mins ago
add a comment |
$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
6 hours ago
$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
6 hours ago
$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
5 hours ago
$begingroup$
Have you studied covariance matrices, or vector-valued random variables, at all? That would seem to me to provide the most compact notation for solving this problem.
$endgroup$
– Daniel Schepler
21 mins ago
$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
6 hours ago
$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
6 hours ago
$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
6 hours ago
$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
6 hours ago
$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
5 hours ago
$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
5 hours ago
$begingroup$
Have you studied covariance matrices, or vector-valued random variables, at all? That would seem to me to provide the most compact notation for solving this problem.
$endgroup$
– Daniel Schepler
21 mins ago
$begingroup$
Have you studied covariance matrices, or vector-valued random variables, at all? That would seem to me to provide the most compact notation for solving this problem.
$endgroup$
– Daniel Schepler
21 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:
$$ mathbbE[X_i] = fracnr $$
Now, we would like to know what is $mathbbE[X_i X_j] $.
We begin by making the following observation:
$$X_i = n - sum_jneq iX_j $$
Which gives us:
$$ X_isum_jneq iX_j = nX_i - X_i^2$$
Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:
beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
&= frac1r mathbbE[nX_i] \
&= fracn^2r^2
endalign
$endgroup$
1
$begingroup$
If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
$endgroup$
– Daniel Schepler
1 hour ago
$begingroup$
Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
$endgroup$
– Sean Lee
1 hour ago
1
$begingroup$
I've now expanded VHarisop's answer with my calculations for part two of the question.
$endgroup$
– Daniel Schepler
49 mins ago
add a comment |
$begingroup$
For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
Specifically, you know that for a fixed box, the probability of putting a ball in it
is $frac1r$. Let
$$
Y_k^(i) = begincases
1 &, text if ball $k$ was placed in box $i$ \
0 &, text otherwise
endcases,
$$
which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
Then you can write
$$
X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
$$
For the second part, you can proceed similarly: $X_i = sum_k=1^n Y_k^(i)$ and $X_j = sum_ell=1^n Y_ell^(j)$, so:
$$
X_i X_j = sum_k=1^n sum_ell=1^n Y_k^(i) Y_ell^(j) implies
mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n mathbbE(Y_k^(i) Y_ell^(j)).
$$
We will first treat the case where $i ne j$. Then, for each term in the sum such that $k = ell$, we must have $Y_k^(i) Y_ell^(j) = Y_k^(i) Y_k^(j) = 0$ since it impossible for ball $k$ to be placed both in box $i$ and in box $j$. On the other hand, if $k ne ell$, then the events corresponding to $Y_k^(i)$ and $Y_ell^(j)$ are independent since the placement of balls $k$ and $ell$ are independent, which implies that $Y_k^(i)$ and $Y_ell^(j)$ are independent random variables. Therefore, in this case,
$$mathbbE(Y_k^(i) Y_ell^(j)) = mathbbE(Y_k^(i)) mathbbE(Y_ell^(j)) = frac1r cdot frac1r.$$
In summary, if $i ne j$, then
$$mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n delta_k ne ell cdot frac1r^2 = fracn(n-1)r^2$$
where $delta_k ne ell$ represents the indicator value which is 1 when $k ne ell$ and 0 when $k = ell$.
For the case $i = j$, I will leave the similar computation of $mathbbE(X_i^2)$ to you, with just the hint that the difference is in the expected value of $mathbbE(Y_k^(i) Y_ell^(j))$ for the case $k = ell$.
$endgroup$
1
$begingroup$
I decided to add my solution to part two, using your notation, to your answer to avoid having an answer split between your part and a part I would post separately. Feel free to edit it more to your liking, or even revert the addition if you prefer.
$endgroup$
– Daniel Schepler
51 mins ago
add a comment |
$begingroup$
Think of placing the ball in box "$i$" as success and not placing it as a failure.
This situation can be represented using the Hypergeometric Distribution.
$$
P(X=k) = fracK choose k N- Kchoose n - kN choose n.
$$
$N$ is the population size (number of boxes $r$)
$K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)
$n$ is the number of draws (the number of balls $n$).
$k$ is the number of observed successes (the number of balls in box "$i$").
The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
$$E[X_i]=nfrac1r=fracnr$$
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:
$$ mathbbE[X_i] = fracnr $$
Now, we would like to know what is $mathbbE[X_i X_j] $.
We begin by making the following observation:
$$X_i = n - sum_jneq iX_j $$
Which gives us:
$$ X_isum_jneq iX_j = nX_i - X_i^2$$
Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:
beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
&= frac1r mathbbE[nX_i] \
&= fracn^2r^2
endalign
$endgroup$
1
$begingroup$
If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
$endgroup$
– Daniel Schepler
1 hour ago
$begingroup$
Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
$endgroup$
– Sean Lee
1 hour ago
1
$begingroup$
I've now expanded VHarisop's answer with my calculations for part two of the question.
$endgroup$
– Daniel Schepler
49 mins ago
add a comment |
$begingroup$
Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:
$$ mathbbE[X_i] = fracnr $$
Now, we would like to know what is $mathbbE[X_i X_j] $.
We begin by making the following observation:
$$X_i = n - sum_jneq iX_j $$
Which gives us:
$$ X_isum_jneq iX_j = nX_i - X_i^2$$
Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:
beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
&= frac1r mathbbE[nX_i] \
&= fracn^2r^2
endalign
$endgroup$
1
$begingroup$
If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
$endgroup$
– Daniel Schepler
1 hour ago
$begingroup$
Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
$endgroup$
– Sean Lee
1 hour ago
1
$begingroup$
I've now expanded VHarisop's answer with my calculations for part two of the question.
$endgroup$
– Daniel Schepler
49 mins ago
add a comment |
$begingroup$
Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:
$$ mathbbE[X_i] = fracnr $$
Now, we would like to know what is $mathbbE[X_i X_j] $.
We begin by making the following observation:
$$X_i = n - sum_jneq iX_j $$
Which gives us:
$$ X_isum_jneq iX_j = nX_i - X_i^2$$
Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:
beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
&= frac1r mathbbE[nX_i] \
&= fracn^2r^2
endalign
$endgroup$
Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:
$$ mathbbE[X_i] = fracnr $$
Now, we would like to know what is $mathbbE[X_i X_j] $.
We begin by making the following observation:
$$X_i = n - sum_jneq iX_j $$
Which gives us:
$$ X_isum_jneq iX_j = nX_i - X_i^2$$
Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:
beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
&= frac1r mathbbE[nX_i] \
&= fracn^2r^2
endalign
edited 5 hours ago
answered 6 hours ago
Sean LeeSean Lee
801214
801214
1
$begingroup$
If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
$endgroup$
– Daniel Schepler
1 hour ago
$begingroup$
Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
$endgroup$
– Sean Lee
1 hour ago
1
$begingroup$
I've now expanded VHarisop's answer with my calculations for part two of the question.
$endgroup$
– Daniel Schepler
49 mins ago
add a comment |
1
$begingroup$
If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
$endgroup$
– Daniel Schepler
1 hour ago
$begingroup$
Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
$endgroup$
– Sean Lee
1 hour ago
1
$begingroup$
I've now expanded VHarisop's answer with my calculations for part two of the question.
$endgroup$
– Daniel Schepler
49 mins ago
1
1
$begingroup$
If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
$endgroup$
– Daniel Schepler
1 hour ago
$begingroup$
If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
$endgroup$
– Daniel Schepler
1 hour ago
$begingroup$
Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
$endgroup$
– Sean Lee
1 hour ago
$begingroup$
Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
$endgroup$
– Sean Lee
1 hour ago
1
1
$begingroup$
I've now expanded VHarisop's answer with my calculations for part two of the question.
$endgroup$
– Daniel Schepler
49 mins ago
$begingroup$
I've now expanded VHarisop's answer with my calculations for part two of the question.
$endgroup$
– Daniel Schepler
49 mins ago
add a comment |
$begingroup$
For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
Specifically, you know that for a fixed box, the probability of putting a ball in it
is $frac1r$. Let
$$
Y_k^(i) = begincases
1 &, text if ball $k$ was placed in box $i$ \
0 &, text otherwise
endcases,
$$
which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
Then you can write
$$
X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
$$
For the second part, you can proceed similarly: $X_i = sum_k=1^n Y_k^(i)$ and $X_j = sum_ell=1^n Y_ell^(j)$, so:
$$
X_i X_j = sum_k=1^n sum_ell=1^n Y_k^(i) Y_ell^(j) implies
mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n mathbbE(Y_k^(i) Y_ell^(j)).
$$
We will first treat the case where $i ne j$. Then, for each term in the sum such that $k = ell$, we must have $Y_k^(i) Y_ell^(j) = Y_k^(i) Y_k^(j) = 0$ since it impossible for ball $k$ to be placed both in box $i$ and in box $j$. On the other hand, if $k ne ell$, then the events corresponding to $Y_k^(i)$ and $Y_ell^(j)$ are independent since the placement of balls $k$ and $ell$ are independent, which implies that $Y_k^(i)$ and $Y_ell^(j)$ are independent random variables. Therefore, in this case,
$$mathbbE(Y_k^(i) Y_ell^(j)) = mathbbE(Y_k^(i)) mathbbE(Y_ell^(j)) = frac1r cdot frac1r.$$
In summary, if $i ne j$, then
$$mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n delta_k ne ell cdot frac1r^2 = fracn(n-1)r^2$$
where $delta_k ne ell$ represents the indicator value which is 1 when $k ne ell$ and 0 when $k = ell$.
For the case $i = j$, I will leave the similar computation of $mathbbE(X_i^2)$ to you, with just the hint that the difference is in the expected value of $mathbbE(Y_k^(i) Y_ell^(j))$ for the case $k = ell$.
$endgroup$
1
$begingroup$
I decided to add my solution to part two, using your notation, to your answer to avoid having an answer split between your part and a part I would post separately. Feel free to edit it more to your liking, or even revert the addition if you prefer.
$endgroup$
– Daniel Schepler
51 mins ago
add a comment |
$begingroup$
For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
Specifically, you know that for a fixed box, the probability of putting a ball in it
is $frac1r$. Let
$$
Y_k^(i) = begincases
1 &, text if ball $k$ was placed in box $i$ \
0 &, text otherwise
endcases,
$$
which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
Then you can write
$$
X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
$$
For the second part, you can proceed similarly: $X_i = sum_k=1^n Y_k^(i)$ and $X_j = sum_ell=1^n Y_ell^(j)$, so:
$$
X_i X_j = sum_k=1^n sum_ell=1^n Y_k^(i) Y_ell^(j) implies
mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n mathbbE(Y_k^(i) Y_ell^(j)).
$$
We will first treat the case where $i ne j$. Then, for each term in the sum such that $k = ell$, we must have $Y_k^(i) Y_ell^(j) = Y_k^(i) Y_k^(j) = 0$ since it impossible for ball $k$ to be placed both in box $i$ and in box $j$. On the other hand, if $k ne ell$, then the events corresponding to $Y_k^(i)$ and $Y_ell^(j)$ are independent since the placement of balls $k$ and $ell$ are independent, which implies that $Y_k^(i)$ and $Y_ell^(j)$ are independent random variables. Therefore, in this case,
$$mathbbE(Y_k^(i) Y_ell^(j)) = mathbbE(Y_k^(i)) mathbbE(Y_ell^(j)) = frac1r cdot frac1r.$$
In summary, if $i ne j$, then
$$mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n delta_k ne ell cdot frac1r^2 = fracn(n-1)r^2$$
where $delta_k ne ell$ represents the indicator value which is 1 when $k ne ell$ and 0 when $k = ell$.
For the case $i = j$, I will leave the similar computation of $mathbbE(X_i^2)$ to you, with just the hint that the difference is in the expected value of $mathbbE(Y_k^(i) Y_ell^(j))$ for the case $k = ell$.
$endgroup$
1
$begingroup$
I decided to add my solution to part two, using your notation, to your answer to avoid having an answer split between your part and a part I would post separately. Feel free to edit it more to your liking, or even revert the addition if you prefer.
$endgroup$
– Daniel Schepler
51 mins ago
add a comment |
$begingroup$
For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
Specifically, you know that for a fixed box, the probability of putting a ball in it
is $frac1r$. Let
$$
Y_k^(i) = begincases
1 &, text if ball $k$ was placed in box $i$ \
0 &, text otherwise
endcases,
$$
which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
Then you can write
$$
X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
$$
For the second part, you can proceed similarly: $X_i = sum_k=1^n Y_k^(i)$ and $X_j = sum_ell=1^n Y_ell^(j)$, so:
$$
X_i X_j = sum_k=1^n sum_ell=1^n Y_k^(i) Y_ell^(j) implies
mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n mathbbE(Y_k^(i) Y_ell^(j)).
$$
We will first treat the case where $i ne j$. Then, for each term in the sum such that $k = ell$, we must have $Y_k^(i) Y_ell^(j) = Y_k^(i) Y_k^(j) = 0$ since it impossible for ball $k$ to be placed both in box $i$ and in box $j$. On the other hand, if $k ne ell$, then the events corresponding to $Y_k^(i)$ and $Y_ell^(j)$ are independent since the placement of balls $k$ and $ell$ are independent, which implies that $Y_k^(i)$ and $Y_ell^(j)$ are independent random variables. Therefore, in this case,
$$mathbbE(Y_k^(i) Y_ell^(j)) = mathbbE(Y_k^(i)) mathbbE(Y_ell^(j)) = frac1r cdot frac1r.$$
In summary, if $i ne j$, then
$$mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n delta_k ne ell cdot frac1r^2 = fracn(n-1)r^2$$
where $delta_k ne ell$ represents the indicator value which is 1 when $k ne ell$ and 0 when $k = ell$.
For the case $i = j$, I will leave the similar computation of $mathbbE(X_i^2)$ to you, with just the hint that the difference is in the expected value of $mathbbE(Y_k^(i) Y_ell^(j))$ for the case $k = ell$.
$endgroup$
For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
Specifically, you know that for a fixed box, the probability of putting a ball in it
is $frac1r$. Let
$$
Y_k^(i) = begincases
1 &, text if ball $k$ was placed in box $i$ \
0 &, text otherwise
endcases,
$$
which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
Then you can write
$$
X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
$$
For the second part, you can proceed similarly: $X_i = sum_k=1^n Y_k^(i)$ and $X_j = sum_ell=1^n Y_ell^(j)$, so:
$$
X_i X_j = sum_k=1^n sum_ell=1^n Y_k^(i) Y_ell^(j) implies
mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n mathbbE(Y_k^(i) Y_ell^(j)).
$$
We will first treat the case where $i ne j$. Then, for each term in the sum such that $k = ell$, we must have $Y_k^(i) Y_ell^(j) = Y_k^(i) Y_k^(j) = 0$ since it impossible for ball $k$ to be placed both in box $i$ and in box $j$. On the other hand, if $k ne ell$, then the events corresponding to $Y_k^(i)$ and $Y_ell^(j)$ are independent since the placement of balls $k$ and $ell$ are independent, which implies that $Y_k^(i)$ and $Y_ell^(j)$ are independent random variables. Therefore, in this case,
$$mathbbE(Y_k^(i) Y_ell^(j)) = mathbbE(Y_k^(i)) mathbbE(Y_ell^(j)) = frac1r cdot frac1r.$$
In summary, if $i ne j$, then
$$mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n delta_k ne ell cdot frac1r^2 = fracn(n-1)r^2$$
where $delta_k ne ell$ represents the indicator value which is 1 when $k ne ell$ and 0 when $k = ell$.
For the case $i = j$, I will leave the similar computation of $mathbbE(X_i^2)$ to you, with just the hint that the difference is in the expected value of $mathbbE(Y_k^(i) Y_ell^(j))$ for the case $k = ell$.
edited 53 mins ago
Daniel Schepler
9,3341821
9,3341821
answered 6 hours ago
VHarisopVHarisop
1,228421
1,228421
1
$begingroup$
I decided to add my solution to part two, using your notation, to your answer to avoid having an answer split between your part and a part I would post separately. Feel free to edit it more to your liking, or even revert the addition if you prefer.
$endgroup$
– Daniel Schepler
51 mins ago
add a comment |
1
$begingroup$
I decided to add my solution to part two, using your notation, to your answer to avoid having an answer split between your part and a part I would post separately. Feel free to edit it more to your liking, or even revert the addition if you prefer.
$endgroup$
– Daniel Schepler
51 mins ago
1
1
$begingroup$
I decided to add my solution to part two, using your notation, to your answer to avoid having an answer split between your part and a part I would post separately. Feel free to edit it more to your liking, or even revert the addition if you prefer.
$endgroup$
– Daniel Schepler
51 mins ago
$begingroup$
I decided to add my solution to part two, using your notation, to your answer to avoid having an answer split between your part and a part I would post separately. Feel free to edit it more to your liking, or even revert the addition if you prefer.
$endgroup$
– Daniel Schepler
51 mins ago
add a comment |
$begingroup$
Think of placing the ball in box "$i$" as success and not placing it as a failure.
This situation can be represented using the Hypergeometric Distribution.
$$
P(X=k) = fracK choose k N- Kchoose n - kN choose n.
$$
$N$ is the population size (number of boxes $r$)
$K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)
$n$ is the number of draws (the number of balls $n$).
$k$ is the number of observed successes (the number of balls in box "$i$").
The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
$$E[X_i]=nfrac1r=fracnr$$
$endgroup$
add a comment |
$begingroup$
Think of placing the ball in box "$i$" as success and not placing it as a failure.
This situation can be represented using the Hypergeometric Distribution.
$$
P(X=k) = fracK choose k N- Kchoose n - kN choose n.
$$
$N$ is the population size (number of boxes $r$)
$K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)
$n$ is the number of draws (the number of balls $n$).
$k$ is the number of observed successes (the number of balls in box "$i$").
The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
$$E[X_i]=nfrac1r=fracnr$$
$endgroup$
add a comment |
$begingroup$
Think of placing the ball in box "$i$" as success and not placing it as a failure.
This situation can be represented using the Hypergeometric Distribution.
$$
P(X=k) = fracK choose k N- Kchoose n - kN choose n.
$$
$N$ is the population size (number of boxes $r$)
$K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)
$n$ is the number of draws (the number of balls $n$).
$k$ is the number of observed successes (the number of balls in box "$i$").
The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
$$E[X_i]=nfrac1r=fracnr$$
$endgroup$
Think of placing the ball in box "$i$" as success and not placing it as a failure.
This situation can be represented using the Hypergeometric Distribution.
$$
P(X=k) = fracK choose k N- Kchoose n - kN choose n.
$$
$N$ is the population size (number of boxes $r$)
$K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)
$n$ is the number of draws (the number of balls $n$).
$k$ is the number of observed successes (the number of balls in box "$i$").
The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
$$E[X_i]=nfrac1r=fracnr$$
answered 6 hours ago
RScrlliRScrlli
761114
761114
add a comment |
add a comment |
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$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
6 hours ago
$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
6 hours ago
$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
5 hours ago
$begingroup$
Have you studied covariance matrices, or vector-valued random variables, at all? That would seem to me to provide the most compact notation for solving this problem.
$endgroup$
– Daniel Schepler
21 mins ago