Computing $mathbb{E}[X_{1} mid X_{1} + X_{2} + cdots + X_{n} = x]$ if $X_{1}, ldots X_{N}$ are i.i.d
$begingroup$
I would like to compute $mathbb{E}[X_{1} mid X_{1} + cdots X_{n} = x]$ if $X_{1}, ldots, X_{n}$ are i.i.d random variables.
Intuitively, I think that the answer should be $x/n$ because there are $n$ random variables with the same distribution. If they each contribute the same amount, then we would have $x/n cdot n = x$, which is what we want the sum to equal.
How can I show this, though?
probability probability-theory expected-value
asked Dec 4 '18 at 18:54
josephjoseph
510111
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|
show 2 more comments
$begingroup$
I would like to compute $mathbb{E}[X_{1} mid X_{1} + cdots X_{n} = x]$ if $X_{1}, ldots, X_{n}$ are i.i.d random variables.
Intuitively, I think that the answer should be $x/n$ because there are $n$ random variables with the same distribution. If they each contribute the same amount, then we would have $x/n cdot n = x$, which is what we want the sum to equal.
How can I show this, though?
probability probability-theory expected-value
asked Dec 4 '18 at 18:54
josephjoseph
510111
$endgroup$
$begingroup$
Presumably $E[sum_k X_k | sum_k X_k = x] = x$, assuming the event is not a null event.
$endgroup$
– copper.hat
Dec 4 '18 at 18:58
$begingroup$
Yes, that is correct. @copper.hat
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– joseph
Dec 4 '18 at 18:58
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Is $p[sum_k X_k=x] > 0$?
$endgroup$
– copper.hat
Dec 4 '18 at 18:59
$begingroup$
I don't know. I'm not given any information other than $X_{1}, ldots X_{n}$ being i.i.d.
$endgroup$
– joseph
Dec 4 '18 at 19:00
$begingroup$
The keyword here is linear. Expectation is linear, even when it is conditional. That is how you "show this". Either you already have a theorem which says as much, or the point of this exercise is to prove a special case of that theorem.
$endgroup$
– Arthur
Dec 4 '18 at 19:05
|
show 2 more comments
$begingroup$
I would like to compute $mathbb{E}[X_{1} mid X_{1} + cdots X_{n} = x]$ if $X_{1}, ldots, X_{n}$ are i.i.d random variables.
Intuitively, I think that the answer should be $x/n$ because there are $n$ random variables with the same distribution. If they each contribute the same amount, then we would have $x/n cdot n = x$, which is what we want the sum to equal.
How can I show this, though?
probability probability-theory expected-value
asked Dec 4 '18 at 18:54
josephjoseph
510111
$endgroup$
I would like to compute $mathbb{E}[X_{1} mid X_{1} + cdots X_{n} = x]$ if $X_{1}, ldots, X_{n}$ are i.i.d random variables.
Intuitively, I think that the answer should be $x/n$ because there are $n$ random variables with the same distribution. If they each contribute the same amount, then we would have $x/n cdot n = x$, which is what we want the sum to equal.
How can I show this, though?
probability probability-theory expected-value
probability probability-theory expected-value
asked Dec 4 '18 at 18:54
josephjoseph
510111
asked Dec 4 '18 at 18:54
josephjoseph
510111
asked Dec 4 '18 at 18:54
josephjoseph
510111
asked Dec 4 '18 at 18:54
josephjoseph
510111
asked Dec 4 '18 at 18:54
josephjoseph
510111
510111
$begingroup$
Presumably $E[sum_k X_k | sum_k X_k = x] = x$, assuming the event is not a null event.
$endgroup$
– copper.hat
Dec 4 '18 at 18:58
$begingroup$
Yes, that is correct. @copper.hat
$endgroup$
– joseph
Dec 4 '18 at 18:58
$begingroup$
Is $p[sum_k X_k=x] > 0$?
$endgroup$
– copper.hat
Dec 4 '18 at 18:59
$begingroup$
I don't know. I'm not given any information other than $X_{1}, ldots X_{n}$ being i.i.d.
$endgroup$
– joseph
Dec 4 '18 at 19:00
$begingroup$
The keyword here is linear. Expectation is linear, even when it is conditional. That is how you "show this". Either you already have a theorem which says as much, or the point of this exercise is to prove a special case of that theorem.
$endgroup$
– Arthur
Dec 4 '18 at 19:05
|
show 2 more comments
$begingroup$
Presumably $E[sum_k X_k | sum_k X_k = x] = x$, assuming the event is not a null event.
$endgroup$
– copper.hat
Dec 4 '18 at 18:58
$begingroup$
Yes, that is correct. @copper.hat
$endgroup$
– joseph
Dec 4 '18 at 18:58
$begingroup$
Is $p[sum_k X_k=x] > 0$?
$endgroup$
– copper.hat
Dec 4 '18 at 18:59
$begingroup$
I don't know. I'm not given any information other than $X_{1}, ldots X_{n}$ being i.i.d.
$endgroup$
– joseph
Dec 4 '18 at 19:00
$begingroup$
The keyword here is linear. Expectation is linear, even when it is conditional. That is how you "show this". Either you already have a theorem which says as much, or the point of this exercise is to prove a special case of that theorem.
$endgroup$
– Arthur
Dec 4 '18 at 19:05
$begingroup$
Presumably $E[sum_k X_k | sum_k X_k = x] = x$, assuming the event is not a null event.
$endgroup$
– copper.hat
Dec 4 '18 at 18:58
$begingroup$
Presumably $E[sum_k X_k | sum_k X_k = x] = x$, assuming the event is not a null event.
$endgroup$
– copper.hat
Dec 4 '18 at 18:58
$begingroup$
Yes, that is correct. @copper.hat
$endgroup$
– joseph
Dec 4 '18 at 18:58
$begingroup$
Yes, that is correct. @copper.hat
$endgroup$
– joseph
Dec 4 '18 at 18:58
$begingroup$
Is $p[sum_k X_k=x] > 0$?
$endgroup$
– copper.hat
Dec 4 '18 at 18:59
$begingroup$
Is $p[sum_k X_k=x] > 0$?
$endgroup$
– copper.hat
Dec 4 '18 at 18:59
$begingroup$
I don't know. I'm not given any information other than $X_{1}, ldots X_{n}$ being i.i.d.
$endgroup$
– joseph
Dec 4 '18 at 19:00
$begingroup$
I don't know. I'm not given any information other than $X_{1}, ldots X_{n}$ being i.i.d.
$endgroup$
– joseph
Dec 4 '18 at 19:00
$begingroup$
The keyword here is linear. Expectation is linear, even when it is conditional. That is how you "show this". Either you already have a theorem which says as much, or the point of this exercise is to prove a special case of that theorem.
$endgroup$
– Arthur
Dec 4 '18 at 19:05
$begingroup$
The keyword here is linear. Expectation is linear, even when it is conditional. That is how you "show this". Either you already have a theorem which says as much, or the point of this exercise is to prove a special case of that theorem.
$endgroup$
– Arthur
Dec 4 '18 at 19:05
|
show 2 more comments
1 Answer
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$begingroup$
Your idea is exactly right. To express it rigorously, let $S_n=X_1+...+X_n$. The key observation is the following
Claim: $E(X_1|S_n)=E(X_i|S_n)$ for any $i$.
To see this, let $sigma$ be any permutation of the indices $1,dots, n$. Let $tilde{X_i}=X_{sigma(i)}$ and let $tilde{S_n}=sum_i tilde{X}_i$. Note that the random vector $(X_1,dots, X_n)$ has the same distribution as the random vector $(tilde{X_1}, dots, tilde{X_n})$ (by iid-ness). By applying the measurable function $(x_1, dots,x_n)to (x_1, sum_ix_i)$ to each of these random vectors,we conclude that $(X_1,S_n)=^d(tilde{X}_1,tilde{S_n})$. In particular, $E(X_1|S_n)=E(tilde{X}_1|tilde{S_n})$. On the other hand, $S_n=tilde{S_n}$ and $tilde{X}_1=X_{sigma(1)}$. So $E(X_1|S_n)=E(X_{sigma(1)}|S_n)$. But $sigma$ is arbitrary, so $sigma(1)$ could be any index $i$. This establishes the claim.
The rest of your argument goes through essentially unchanged. Noting that$sum_i E(X_i|S_n)=E(S_n|S_n)=S_n$,and applying the Claim, we get $nE(X_1|S_n)=S_n$ as desired.
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your idea is exactly right. To express it rigorously, let $S_n=X_1+...+X_n$. The key observation is the following
Claim: $E(X_1|S_n)=E(X_i|S_n)$ for any $i$.
To see this, let $sigma$ be any permutation of the indices $1,dots, n$. Let $tilde{X_i}=X_{sigma(i)}$ and let $tilde{S_n}=sum_i tilde{X}_i$. Note that the random vector $(X_1,dots, X_n)$ has the same distribution as the random vector $(tilde{X_1}, dots, tilde{X_n})$ (by iid-ness). By applying the measurable function $(x_1, dots,x_n)to (x_1, sum_ix_i)$ to each of these random vectors,we conclude that $(X_1,S_n)=^d(tilde{X}_1,tilde{S_n})$. In particular, $E(X_1|S_n)=E(tilde{X}_1|tilde{S_n})$. On the other hand, $S_n=tilde{S_n}$ and $tilde{X}_1=X_{sigma(1)}$. So $E(X_1|S_n)=E(X_{sigma(1)}|S_n)$. But $sigma$ is arbitrary, so $sigma(1)$ could be any index $i$. This establishes the claim.
The rest of your argument goes through essentially unchanged. Noting that$sum_i E(X_i|S_n)=E(S_n|S_n)=S_n$,and applying the Claim, we get $nE(X_1|S_n)=S_n$ as desired.
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
$endgroup$
add a comment |
$begingroup$
Your idea is exactly right. To express it rigorously, let $S_n=X_1+...+X_n$. The key observation is the following
Claim: $E(X_1|S_n)=E(X_i|S_n)$ for any $i$.
To see this, let $sigma$ be any permutation of the indices $1,dots, n$. Let $tilde{X_i}=X_{sigma(i)}$ and let $tilde{S_n}=sum_i tilde{X}_i$. Note that the random vector $(X_1,dots, X_n)$ has the same distribution as the random vector $(tilde{X_1}, dots, tilde{X_n})$ (by iid-ness). By applying the measurable function $(x_1, dots,x_n)to (x_1, sum_ix_i)$ to each of these random vectors,we conclude that $(X_1,S_n)=^d(tilde{X}_1,tilde{S_n})$. In particular, $E(X_1|S_n)=E(tilde{X}_1|tilde{S_n})$. On the other hand, $S_n=tilde{S_n}$ and $tilde{X}_1=X_{sigma(1)}$. So $E(X_1|S_n)=E(X_{sigma(1)}|S_n)$. But $sigma$ is arbitrary, so $sigma(1)$ could be any index $i$. This establishes the claim.
The rest of your argument goes through essentially unchanged. Noting that$sum_i E(X_i|S_n)=E(S_n|S_n)=S_n$,and applying the Claim, we get $nE(X_1|S_n)=S_n$ as desired.
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
$endgroup$
add a comment |
$begingroup$
Your idea is exactly right. To express it rigorously, let $S_n=X_1+...+X_n$. The key observation is the following
Claim: $E(X_1|S_n)=E(X_i|S_n)$ for any $i$.
To see this, let $sigma$ be any permutation of the indices $1,dots, n$. Let $tilde{X_i}=X_{sigma(i)}$ and let $tilde{S_n}=sum_i tilde{X}_i$. Note that the random vector $(X_1,dots, X_n)$ has the same distribution as the random vector $(tilde{X_1}, dots, tilde{X_n})$ (by iid-ness). By applying the measurable function $(x_1, dots,x_n)to (x_1, sum_ix_i)$ to each of these random vectors,we conclude that $(X_1,S_n)=^d(tilde{X}_1,tilde{S_n})$. In particular, $E(X_1|S_n)=E(tilde{X}_1|tilde{S_n})$. On the other hand, $S_n=tilde{S_n}$ and $tilde{X}_1=X_{sigma(1)}$. So $E(X_1|S_n)=E(X_{sigma(1)}|S_n)$. But $sigma$ is arbitrary, so $sigma(1)$ could be any index $i$. This establishes the claim.
The rest of your argument goes through essentially unchanged. Noting that$sum_i E(X_i|S_n)=E(S_n|S_n)=S_n$,and applying the Claim, we get $nE(X_1|S_n)=S_n$ as desired.
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
$endgroup$
Your idea is exactly right. To express it rigorously, let $S_n=X_1+...+X_n$. The key observation is the following
Claim: $E(X_1|S_n)=E(X_i|S_n)$ for any $i$.
To see this, let $sigma$ be any permutation of the indices $1,dots, n$. Let $tilde{X_i}=X_{sigma(i)}$ and let $tilde{S_n}=sum_i tilde{X}_i$. Note that the random vector $(X_1,dots, X_n)$ has the same distribution as the random vector $(tilde{X_1}, dots, tilde{X_n})$ (by iid-ness). By applying the measurable function $(x_1, dots,x_n)to (x_1, sum_ix_i)$ to each of these random vectors,we conclude that $(X_1,S_n)=^d(tilde{X}_1,tilde{S_n})$. In particular, $E(X_1|S_n)=E(tilde{X}_1|tilde{S_n})$. On the other hand, $S_n=tilde{S_n}$ and $tilde{X}_1=X_{sigma(1)}$. So $E(X_1|S_n)=E(X_{sigma(1)}|S_n)$. But $sigma$ is arbitrary, so $sigma(1)$ could be any index $i$. This establishes the claim.
The rest of your argument goes through essentially unchanged. Noting that$sum_i E(X_i|S_n)=E(S_n|S_n)=S_n$,and applying the Claim, we get $nE(X_1|S_n)=S_n$ as desired.
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
answered Dec 4 '18 at 19:15
Mike HawkMike Hawk
1,500110
1,500110
add a comment |
add a comment |
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$begingroup$
Presumably $E[sum_k X_k | sum_k X_k = x] = x$, assuming the event is not a null event.
$endgroup$
– copper.hat
Dec 4 '18 at 18:58
$begingroup$
Yes, that is correct. @copper.hat
$endgroup$
– joseph
Dec 4 '18 at 18:58
$begingroup$
Is $p[sum_k X_k=x] > 0$?
$endgroup$
– copper.hat
Dec 4 '18 at 18:59
$begingroup$
I don't know. I'm not given any information other than $X_{1}, ldots X_{n}$ being i.i.d.
$endgroup$
– joseph
Dec 4 '18 at 19:00
$begingroup$
The keyword here is linear. Expectation is linear, even when it is conditional. That is how you "show this". Either you already have a theorem which says as much, or the point of this exercise is to prove a special case of that theorem.
$endgroup$
– Arthur
Dec 4 '18 at 19:05