Conditional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d. Am I correct?
$begingroup$
Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.
Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$
Am I correct in thinking this?
Thanks!
(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.
asymptotics statistical-inference
edited Dec 8 '18 at 14:31
Bernard
119k740113
asked Dec 8 '18 at 14:14
L200123L200123
855
$endgroup$
add a comment |
$begingroup$
Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.
Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$
Am I correct in thinking this?
Thanks!
(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.
asymptotics statistical-inference
edited Dec 8 '18 at 14:31
Bernard
119k740113
asked Dec 8 '18 at 14:14
L200123L200123
855
$endgroup$
1
$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59
add a comment |
$begingroup$
Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.
Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$
Am I correct in thinking this?
Thanks!
(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.
asymptotics statistical-inference
edited Dec 8 '18 at 14:31
Bernard
119k740113
asked Dec 8 '18 at 14:14
L200123L200123
855
$endgroup$
Contitional expectation $E(X_1 mid overline{X}_n)$ if $X_1,dots,X_n$ are i.i.d.
Since $X_1,dots,X_n$ are i.i.d, then $E(X_1 mid overline{X}_n) = E(X_1)=overline{X}_n)$
Am I correct in thinking this?
Thanks!
(Just to overclearify $overline{X}_n$ is the sample mean)
Question is from Van Der Vaart: Asymptotic Statistics.
asymptotics statistical-inference
asymptotics statistical-inference
edited Dec 8 '18 at 14:31
Bernard
119k740113
asked Dec 8 '18 at 14:14
L200123L200123
855
edited Dec 8 '18 at 14:31
Bernard
119k740113
asked Dec 8 '18 at 14:14
L200123L200123
855
edited Dec 8 '18 at 14:31
Bernard
119k740113
edited Dec 8 '18 at 14:31
Bernard
119k740113
edited Dec 8 '18 at 14:31
Bernard
119k740113
119k740113
asked Dec 8 '18 at 14:14
L200123L200123
855
asked Dec 8 '18 at 14:14
L200123L200123
855
asked Dec 8 '18 at 14:14
L200123L200123
855
855
1
$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59
add a comment |
1
$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59
1
1
$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59
$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$
Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$
$$nE(X_1|bar{X}_n)=nbar{X}_n$$
Hence,
$$E(X_1|bar{X}_n)=bar{X}_n$$
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
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1 Answer
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$begingroup$
We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$
Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$
$$nE(X_1|bar{X}_n)=nbar{X}_n$$
Hence,
$$E(X_1|bar{X}_n)=bar{X}_n$$
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$
Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$
$$nE(X_1|bar{X}_n)=nbar{X}_n$$
Hence,
$$E(X_1|bar{X}_n)=bar{X}_n$$
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
add a comment |
$begingroup$
We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$
Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$
$$nE(X_1|bar{X}_n)=nbar{X}_n$$
Hence,
$$E(X_1|bar{X}_n)=bar{X}_n$$
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
$endgroup$
We have $$E(X_i |bar{X}_n)= E(X_j |bar{X}_n)$$
Summing them up $$sum_{i=1}^n E(X_i|bar{X}_n)=E(sum_{i=1}^n X_i|bar{X}_n)=E(nbar{X}_n|bar{X}_n)=nbar{X}_n$$
$$nE(X_1|bar{X}_n)=nbar{X}_n$$
Hence,
$$E(X_1|bar{X}_n)=bar{X}_n$$
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
answered Dec 8 '18 at 14:18
Siong Thye GohSiong Thye Goh
100k1466117
100k1466117
add a comment |
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$begingroup$
The answer from @SiongThyeGoh is right. Nevertheless, is not true that $bar X_n=E(X_1)$ (and then is also not true that $E(X_1|bar X_n)=E(X_1)$ ). Check that $E(X_1|bar X_n)$ and $bar X_n$ are random variables (the same r.v., indeed), but $E(X_1)$ is a constant.
$endgroup$
– Alejandro Nasif Salum
Dec 8 '18 at 14:59