Mixing
Calculate the MSE(T)
$begingroup$
Suppose that each measurement $X_i$ can be seen as $X_i = μ + U_i$, where the $U_i$ are i.i.d. $N(0, σ^2)$ random variables.
Consider the estimator $T=frac{1}{20}sum_{i=1}^{20}U_i^2$ for $σ^2$.
Calculate the MSE(T).
Hint: use that $E[U_i^4]=3σ^4$.
Solve:
$MSE(T)=Var(T)+(E[T]-sigma^2)^2$
$E[T]=frac{1}{20}cdot 20 cdot E[U_i^2]=E[U_i^2]=sigma^2$
$Var(T)=frac{1}{20^2} cdot 20 cdot Var(U_i^2)= frac{Var(U_i^2)}{20}$
$Var(U_i^2)=E[U_i^4]-(E[U_i^2])^2=3σ^4-sigma^4=2σ^4$
So $Var(T)=frac{σ^4}{10}$ and $MSE(T)=frac{σ^4}{10}+(sigma^2-sigma^2)^2=frac{σ^4}{10}$
Is this the correct way to solve it? My only doubt is when I compute $E[T]$, I don't know if I have to assume it equal to zero since the distribution is $N=(0,sigma^2)$ or maybe just assume that $T$ is un unbiased estimator for $sigma^2$ (what I did)?
probability statistics
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
$endgroup$
|
show 1 more comment
$begingroup$
Suppose that each measurement $X_i$ can be seen as $X_i = μ + U_i$, where the $U_i$ are i.i.d. $N(0, σ^2)$ random variables.
Consider the estimator $T=frac{1}{20}sum_{i=1}^{20}U_i^2$ for $σ^2$.
Calculate the MSE(T).
Hint: use that $E[U_i^4]=3σ^4$.
Solve:
$MSE(T)=Var(T)+(E[T]-sigma^2)^2$
$E[T]=frac{1}{20}cdot 20 cdot E[U_i^2]=E[U_i^2]=sigma^2$
$Var(T)=frac{1}{20^2} cdot 20 cdot Var(U_i^2)= frac{Var(U_i^2)}{20}$
$Var(U_i^2)=E[U_i^4]-(E[U_i^2])^2=3σ^4-sigma^4=2σ^4$
So $Var(T)=frac{σ^4}{10}$ and $MSE(T)=frac{σ^4}{10}+(sigma^2-sigma^2)^2=frac{σ^4}{10}$
Is this the correct way to solve it? My only doubt is when I compute $E[T]$, I don't know if I have to assume it equal to zero since the distribution is $N=(0,sigma^2)$ or maybe just assume that $T$ is un unbiased estimator for $sigma^2$ (what I did)?
probability statistics
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
$endgroup$
1
$begingroup$
Same question was asked sometime back (now deleted) and surprisingly with the same formatting as far as I remember.
$endgroup$
– StubbornAtom
Jan 27 at 16:19
$begingroup$
Yes, it was a friend of mine (we are doing the same course and we are struggling with the same question) and since it was posted in a wrong moment (just a few views and no answer) we decided to repost it with my account and see if someone can help us! :)
$endgroup$
– Mark Jacon
Jan 27 at 16:22
2
$begingroup$
We don't draw attention to our questions by deleting previous copies and posting the same thing again (within a few hours in this case). Why not wait for someone to answer?
$endgroup$
– StubbornAtom
Jan 27 at 16:28
$begingroup$
Ah ok sorry, now I know it!
$endgroup$
– Mark Jacon
Jan 27 at 16:29
1
$begingroup$
@MarkJacon Do you have two accounts?
$endgroup$
– callculus
Jan 27 at 16:32
|
show 1 more comment
$begingroup$
Suppose that each measurement $X_i$ can be seen as $X_i = μ + U_i$, where the $U_i$ are i.i.d. $N(0, σ^2)$ random variables.
Consider the estimator $T=frac{1}{20}sum_{i=1}^{20}U_i^2$ for $σ^2$.
Calculate the MSE(T).
Hint: use that $E[U_i^4]=3σ^4$.
Solve:
$MSE(T)=Var(T)+(E[T]-sigma^2)^2$
$E[T]=frac{1}{20}cdot 20 cdot E[U_i^2]=E[U_i^2]=sigma^2$
$Var(T)=frac{1}{20^2} cdot 20 cdot Var(U_i^2)= frac{Var(U_i^2)}{20}$
$Var(U_i^2)=E[U_i^4]-(E[U_i^2])^2=3σ^4-sigma^4=2σ^4$
So $Var(T)=frac{σ^4}{10}$ and $MSE(T)=frac{σ^4}{10}+(sigma^2-sigma^2)^2=frac{σ^4}{10}$
Is this the correct way to solve it? My only doubt is when I compute $E[T]$, I don't know if I have to assume it equal to zero since the distribution is $N=(0,sigma^2)$ or maybe just assume that $T$ is un unbiased estimator for $sigma^2$ (what I did)?
probability statistics
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
$endgroup$
Suppose that each measurement $X_i$ can be seen as $X_i = μ + U_i$, where the $U_i$ are i.i.d. $N(0, σ^2)$ random variables.
Consider the estimator $T=frac{1}{20}sum_{i=1}^{20}U_i^2$ for $σ^2$.
Calculate the MSE(T).
Hint: use that $E[U_i^4]=3σ^4$.
Solve:
$MSE(T)=Var(T)+(E[T]-sigma^2)^2$
$E[T]=frac{1}{20}cdot 20 cdot E[U_i^2]=E[U_i^2]=sigma^2$
$Var(T)=frac{1}{20^2} cdot 20 cdot Var(U_i^2)= frac{Var(U_i^2)}{20}$
$Var(U_i^2)=E[U_i^4]-(E[U_i^2])^2=3σ^4-sigma^4=2σ^4$
So $Var(T)=frac{σ^4}{10}$ and $MSE(T)=frac{σ^4}{10}+(sigma^2-sigma^2)^2=frac{σ^4}{10}$
Is this the correct way to solve it? My only doubt is when I compute $E[T]$, I don't know if I have to assume it equal to zero since the distribution is $N=(0,sigma^2)$ or maybe just assume that $T$ is un unbiased estimator for $sigma^2$ (what I did)?
probability statistics
probability statistics
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
edited Feb 1 at 22:39
Lee David Chung Lin
4,42841242
4,42841242
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
asked Jan 27 at 16:09
Mark JaconMark Jacon
1127
1127
1
$begingroup$
Same question was asked sometime back (now deleted) and surprisingly with the same formatting as far as I remember.
$endgroup$
– StubbornAtom
Jan 27 at 16:19
$begingroup$
Yes, it was a friend of mine (we are doing the same course and we are struggling with the same question) and since it was posted in a wrong moment (just a few views and no answer) we decided to repost it with my account and see if someone can help us! :)
$endgroup$
– Mark Jacon
Jan 27 at 16:22
2
$begingroup$
We don't draw attention to our questions by deleting previous copies and posting the same thing again (within a few hours in this case). Why not wait for someone to answer?
$endgroup$
– StubbornAtom
Jan 27 at 16:28
$begingroup$
Ah ok sorry, now I know it!
$endgroup$
– Mark Jacon
Jan 27 at 16:29
1
$begingroup$
@MarkJacon Do you have two accounts?
$endgroup$
– callculus
Jan 27 at 16:32
|
show 1 more comment
1
$begingroup$
Same question was asked sometime back (now deleted) and surprisingly with the same formatting as far as I remember.
$endgroup$
– StubbornAtom
Jan 27 at 16:19
$begingroup$
Yes, it was a friend of mine (we are doing the same course and we are struggling with the same question) and since it was posted in a wrong moment (just a few views and no answer) we decided to repost it with my account and see if someone can help us! :)
$endgroup$
– Mark Jacon
Jan 27 at 16:22
2
$begingroup$
We don't draw attention to our questions by deleting previous copies and posting the same thing again (within a few hours in this case). Why not wait for someone to answer?
$endgroup$
– StubbornAtom
Jan 27 at 16:28
$begingroup$
Ah ok sorry, now I know it!
$endgroup$
– Mark Jacon
Jan 27 at 16:29
1
$begingroup$
@MarkJacon Do you have two accounts?
$endgroup$
– callculus
Jan 27 at 16:32
1
1
$begingroup$
Same question was asked sometime back (now deleted) and surprisingly with the same formatting as far as I remember.
$endgroup$
– StubbornAtom
Jan 27 at 16:19
$begingroup$
Same question was asked sometime back (now deleted) and surprisingly with the same formatting as far as I remember.
$endgroup$
– StubbornAtom
Jan 27 at 16:19
$begingroup$
Yes, it was a friend of mine (we are doing the same course and we are struggling with the same question) and since it was posted in a wrong moment (just a few views and no answer) we decided to repost it with my account and see if someone can help us! :)
$endgroup$
– Mark Jacon
Jan 27 at 16:22
$begingroup$
Yes, it was a friend of mine (we are doing the same course and we are struggling with the same question) and since it was posted in a wrong moment (just a few views and no answer) we decided to repost it with my account and see if someone can help us! :)
$endgroup$
– Mark Jacon
Jan 27 at 16:22
2
2
$begingroup$
We don't draw attention to our questions by deleting previous copies and posting the same thing again (within a few hours in this case). Why not wait for someone to answer?
$endgroup$
– StubbornAtom
Jan 27 at 16:28
$begingroup$
We don't draw attention to our questions by deleting previous copies and posting the same thing again (within a few hours in this case). Why not wait for someone to answer?
$endgroup$
– StubbornAtom
Jan 27 at 16:28
$begingroup$
Ah ok sorry, now I know it!
$endgroup$
– Mark Jacon
Jan 27 at 16:29
$begingroup$
Ah ok sorry, now I know it!
$endgroup$
– Mark Jacon
Jan 27 at 16:29
1
1
$begingroup$
@MarkJacon Do you have two accounts?
$endgroup$
– callculus
Jan 27 at 16:32
$begingroup$
@MarkJacon Do you have two accounts?
$endgroup$
– callculus
Jan 27 at 16:32
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The calculation is correct. It is a fact that $T$ is an unbiased estimator of $sigma^2$ and not something you need to assume.
The situation often seen where one has to divide by $n-1$ to have an unbiased estimator for the variance is when the data is first used to estimate the unknown population mean. However, that is not the case here.
Here it is given that $U_i$ has mean zero. Namely, $U_i = sigma Z_i$ where $Z_i$ are iid standard Normal such that each $Z_i^2$ is a Chi-square distribution with $1$ degree of freedom and $sum_{i=1}^n Z_i$ is a Chi-sq with $n$ degree of freedom (instead of the reduced $n-1$).
The mean of a standard Chi-sq is $1$ as in $E[Z_i] = 1$. Therefore, $T = frac1n sum U_i^2$ has mean
$$E[T] = frac1n sum_{i=1}^nleft( sigma^2 E[Z_i^2] right) = frac1n sigma^2 sum_{i=1}^n 1 = sigma^2$$
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
$endgroup$
$begingroup$
If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you.
$endgroup$
– Lee David Chung Lin
Feb 7 at 8:16
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089751%2fcalculate-the-mset%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The calculation is correct. It is a fact that $T$ is an unbiased estimator of $sigma^2$ and not something you need to assume.
The situation often seen where one has to divide by $n-1$ to have an unbiased estimator for the variance is when the data is first used to estimate the unknown population mean. However, that is not the case here.
Here it is given that $U_i$ has mean zero. Namely, $U_i = sigma Z_i$ where $Z_i$ are iid standard Normal such that each $Z_i^2$ is a Chi-square distribution with $1$ degree of freedom and $sum_{i=1}^n Z_i$ is a Chi-sq with $n$ degree of freedom (instead of the reduced $n-1$).
The mean of a standard Chi-sq is $1$ as in $E[Z_i] = 1$. Therefore, $T = frac1n sum U_i^2$ has mean
$$E[T] = frac1n sum_{i=1}^nleft( sigma^2 E[Z_i^2] right) = frac1n sigma^2 sum_{i=1}^n 1 = sigma^2$$
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
$endgroup$
$begingroup$
If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you.
$endgroup$
– Lee David Chung Lin
Feb 7 at 8:16
add a comment |
$begingroup$
The calculation is correct. It is a fact that $T$ is an unbiased estimator of $sigma^2$ and not something you need to assume.
The situation often seen where one has to divide by $n-1$ to have an unbiased estimator for the variance is when the data is first used to estimate the unknown population mean. However, that is not the case here.
Here it is given that $U_i$ has mean zero. Namely, $U_i = sigma Z_i$ where $Z_i$ are iid standard Normal such that each $Z_i^2$ is a Chi-square distribution with $1$ degree of freedom and $sum_{i=1}^n Z_i$ is a Chi-sq with $n$ degree of freedom (instead of the reduced $n-1$).
The mean of a standard Chi-sq is $1$ as in $E[Z_i] = 1$. Therefore, $T = frac1n sum U_i^2$ has mean
$$E[T] = frac1n sum_{i=1}^nleft( sigma^2 E[Z_i^2] right) = frac1n sigma^2 sum_{i=1}^n 1 = sigma^2$$
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
$endgroup$
$begingroup$
If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you.
$endgroup$
– Lee David Chung Lin
Feb 7 at 8:16
add a comment |
$begingroup$
The calculation is correct. It is a fact that $T$ is an unbiased estimator of $sigma^2$ and not something you need to assume.
The situation often seen where one has to divide by $n-1$ to have an unbiased estimator for the variance is when the data is first used to estimate the unknown population mean. However, that is not the case here.
Here it is given that $U_i$ has mean zero. Namely, $U_i = sigma Z_i$ where $Z_i$ are iid standard Normal such that each $Z_i^2$ is a Chi-square distribution with $1$ degree of freedom and $sum_{i=1}^n Z_i$ is a Chi-sq with $n$ degree of freedom (instead of the reduced $n-1$).
The mean of a standard Chi-sq is $1$ as in $E[Z_i] = 1$. Therefore, $T = frac1n sum U_i^2$ has mean
$$E[T] = frac1n sum_{i=1}^nleft( sigma^2 E[Z_i^2] right) = frac1n sigma^2 sum_{i=1}^n 1 = sigma^2$$
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
$endgroup$
The calculation is correct. It is a fact that $T$ is an unbiased estimator of $sigma^2$ and not something you need to assume.
The situation often seen where one has to divide by $n-1$ to have an unbiased estimator for the variance is when the data is first used to estimate the unknown population mean. However, that is not the case here.
Here it is given that $U_i$ has mean zero. Namely, $U_i = sigma Z_i$ where $Z_i$ are iid standard Normal such that each $Z_i^2$ is a Chi-square distribution with $1$ degree of freedom and $sum_{i=1}^n Z_i$ is a Chi-sq with $n$ degree of freedom (instead of the reduced $n-1$).
The mean of a standard Chi-sq is $1$ as in $E[Z_i] = 1$. Therefore, $T = frac1n sum U_i^2$ has mean
$$E[T] = frac1n sum_{i=1}^nleft( sigma^2 E[Z_i^2] right) = frac1n sigma^2 sum_{i=1}^n 1 = sigma^2$$
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
answered Feb 1 at 22:50
Lee David Chung LinLee David Chung Lin
4,42841242
4,42841242
$begingroup$
If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you.
$endgroup$
– Lee David Chung Lin
Feb 7 at 8:16
add a comment |
$begingroup$
If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you.
$endgroup$
– Lee David Chung Lin
Feb 7 at 8:16
$begingroup$
If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you.
$endgroup$
– Lee David Chung Lin
Feb 7 at 8:16
$begingroup$
If you're no longer interested in this question, or if you find this answer helpful, would you mind clicking on the green check (accept) to "conclude" this post? Otherwise it stays in the wrong queue and clog up the system. Thank you.
$endgroup$
– Lee David Chung Lin
Feb 7 at 8:16
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089751%2fcalculate-the-mset%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Same question was asked sometime back (now deleted) and surprisingly with the same formatting as far as I remember.
$endgroup$
– StubbornAtom
Jan 27 at 16:19
$begingroup$
Yes, it was a friend of mine (we are doing the same course and we are struggling with the same question) and since it was posted in a wrong moment (just a few views and no answer) we decided to repost it with my account and see if someone can help us! :)
$endgroup$
– Mark Jacon
Jan 27 at 16:22
2
$begingroup$
We don't draw attention to our questions by deleting previous copies and posting the same thing again (within a few hours in this case). Why not wait for someone to answer?
$endgroup$
– StubbornAtom
Jan 27 at 16:28
$begingroup$
Ah ok sorry, now I know it!
$endgroup$
– Mark Jacon
Jan 27 at 16:29
1
$begingroup$
@MarkJacon Do you have two accounts?
$endgroup$
– callculus
Jan 27 at 16:32