Mixing
Expected values of IID random variables
$begingroup$
I'm trying to make sure I am understanding the following properties:
For $X_1, ldots, X_n$ IID random variables with mean $mu$ and standard deviation $sigma$,
$E(c) = c$, where $c$ is any constant
$E(X_i) = mu$,
$E(mu) = mu$ (from the first equality)
$E(X_i - mu) = E(X_i)-E(mu) = mu-mu = 0$
Are these correct? Additionally, I'd like to think about
$E([X-mu]^{2})$, and
($sum_{i=1}^{k} [X-mu])^{3}$
Where $X$ is any of the $X_i$'s (since they are identically distributed, I can say $X_i = X$, for all $i$, correct?)
Self-studying these concepts, any guidance very much appreciated.
probability
asked Jan 23 at 4:34
mXdXmXdX
848
$endgroup$
add a comment |
$begingroup$
I'm trying to make sure I am understanding the following properties:
For $X_1, ldots, X_n$ IID random variables with mean $mu$ and standard deviation $sigma$,
$E(c) = c$, where $c$ is any constant
$E(X_i) = mu$,
$E(mu) = mu$ (from the first equality)
$E(X_i - mu) = E(X_i)-E(mu) = mu-mu = 0$
Are these correct? Additionally, I'd like to think about
$E([X-mu]^{2})$, and
($sum_{i=1}^{k} [X-mu])^{3}$
Where $X$ is any of the $X_i$'s (since they are identically distributed, I can say $X_i = X$, for all $i$, correct?)
Self-studying these concepts, any guidance very much appreciated.
probability
asked Jan 23 at 4:34
mXdXmXdX
848
$endgroup$
add a comment |
$begingroup$
I'm trying to make sure I am understanding the following properties:
For $X_1, ldots, X_n$ IID random variables with mean $mu$ and standard deviation $sigma$,
$E(c) = c$, where $c$ is any constant
$E(X_i) = mu$,
$E(mu) = mu$ (from the first equality)
$E(X_i - mu) = E(X_i)-E(mu) = mu-mu = 0$
Are these correct? Additionally, I'd like to think about
$E([X-mu]^{2})$, and
($sum_{i=1}^{k} [X-mu])^{3}$
Where $X$ is any of the $X_i$'s (since they are identically distributed, I can say $X_i = X$, for all $i$, correct?)
Self-studying these concepts, any guidance very much appreciated.
probability
asked Jan 23 at 4:34
mXdXmXdX
848
$endgroup$
I'm trying to make sure I am understanding the following properties:
For $X_1, ldots, X_n$ IID random variables with mean $mu$ and standard deviation $sigma$,
$E(c) = c$, where $c$ is any constant
$E(X_i) = mu$,
$E(mu) = mu$ (from the first equality)
$E(X_i - mu) = E(X_i)-E(mu) = mu-mu = 0$
Are these correct? Additionally, I'd like to think about
$E([X-mu]^{2})$, and
($sum_{i=1}^{k} [X-mu])^{3}$
Where $X$ is any of the $X_i$'s (since they are identically distributed, I can say $X_i = X$, for all $i$, correct?)
Self-studying these concepts, any guidance very much appreciated.
probability
probability
asked Jan 23 at 4:34
mXdXmXdX
848
asked Jan 23 at 4:34
mXdXmXdX
848
asked Jan 23 at 4:34
mXdXmXdX
848
asked Jan 23 at 4:34
mXdXmXdX
848
asked Jan 23 at 4:34
mXdXmXdX
848
848
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What you have done is correct. $E(X-mu)^{2}=E(X^{2}-2mu X+mu^{2})=EX^{2} -2mu EX+mu^{2} =(sigma^{2}+mu^{2})-2mu^{2} +mu^{2}=sigma^{2}$. In general $E(X-mu)^{2}$ is always the variance of $X$. T0 calculate $E( sumlimits_{i=1}^{k}(X_i-mu))^{3}$ you have to expand $(sumlimits_{i=1}^{k}X_i-mu)^{3}$ as $sumlimits_{i=1}^{k}sumlimits_{j=1}^{k}sumlimits_{l=1}^{k} (X_i -mu)(X_j-mu)(X_l-mu)$. [ I have used the following: $sigma^{2}=EX^{2}-mu^{2}$ by definition . Hence $EX^{2}=mu^{2}+sigma^{2}$]. How do we compute $E(X_i -mu)(X_j-mu)(X_l-mu)$?. The value is $0$ if any two of $i,j,l$ are distinct and it is $E(X-mu)^{3}$ if $i=j=l$. You can calculate $E(X-mu)^{3}$ by expanding the cube (using Binomial Theorem).
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
$begingroup$
Thanks. Having trouble understanding why $E[X^{2}] = (sigma^{2} + mu^{2})$, though.
$endgroup$
– mXdX
Jan 23 at 5:33
$begingroup$
Also, that last part is the cube of the entire sum, not just the summand.
$endgroup$
– mXdX
Jan 23 at 5:34
1
$begingroup$
I have added another line to my answer. @mXdX
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:35
1
$begingroup$
@mXdX Sorry that I ddin't read the last part correctly. I have revised my answer.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:38
1
$begingroup$
@mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-mu)^{3}$. For the fourth power you can use a similar argument.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:23
|
show 3 more comments
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%2f3084086%2fexpected-values-of-iid-random-variables%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
$begingroup$
What you have done is correct. $E(X-mu)^{2}=E(X^{2}-2mu X+mu^{2})=EX^{2} -2mu EX+mu^{2} =(sigma^{2}+mu^{2})-2mu^{2} +mu^{2}=sigma^{2}$. In general $E(X-mu)^{2}$ is always the variance of $X$. T0 calculate $E( sumlimits_{i=1}^{k}(X_i-mu))^{3}$ you have to expand $(sumlimits_{i=1}^{k}X_i-mu)^{3}$ as $sumlimits_{i=1}^{k}sumlimits_{j=1}^{k}sumlimits_{l=1}^{k} (X_i -mu)(X_j-mu)(X_l-mu)$. [ I have used the following: $sigma^{2}=EX^{2}-mu^{2}$ by definition . Hence $EX^{2}=mu^{2}+sigma^{2}$]. How do we compute $E(X_i -mu)(X_j-mu)(X_l-mu)$?. The value is $0$ if any two of $i,j,l$ are distinct and it is $E(X-mu)^{3}$ if $i=j=l$. You can calculate $E(X-mu)^{3}$ by expanding the cube (using Binomial Theorem).
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
$begingroup$
Thanks. Having trouble understanding why $E[X^{2}] = (sigma^{2} + mu^{2})$, though.
$endgroup$
– mXdX
Jan 23 at 5:33
$begingroup$
Also, that last part is the cube of the entire sum, not just the summand.
$endgroup$
– mXdX
Jan 23 at 5:34
1
$begingroup$
I have added another line to my answer. @mXdX
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:35
1
$begingroup$
@mXdX Sorry that I ddin't read the last part correctly. I have revised my answer.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:38
1
$begingroup$
@mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-mu)^{3}$. For the fourth power you can use a similar argument.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:23
|
show 3 more comments
$begingroup$
What you have done is correct. $E(X-mu)^{2}=E(X^{2}-2mu X+mu^{2})=EX^{2} -2mu EX+mu^{2} =(sigma^{2}+mu^{2})-2mu^{2} +mu^{2}=sigma^{2}$. In general $E(X-mu)^{2}$ is always the variance of $X$. T0 calculate $E( sumlimits_{i=1}^{k}(X_i-mu))^{3}$ you have to expand $(sumlimits_{i=1}^{k}X_i-mu)^{3}$ as $sumlimits_{i=1}^{k}sumlimits_{j=1}^{k}sumlimits_{l=1}^{k} (X_i -mu)(X_j-mu)(X_l-mu)$. [ I have used the following: $sigma^{2}=EX^{2}-mu^{2}$ by definition . Hence $EX^{2}=mu^{2}+sigma^{2}$]. How do we compute $E(X_i -mu)(X_j-mu)(X_l-mu)$?. The value is $0$ if any two of $i,j,l$ are distinct and it is $E(X-mu)^{3}$ if $i=j=l$. You can calculate $E(X-mu)^{3}$ by expanding the cube (using Binomial Theorem).
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
$begingroup$
Thanks. Having trouble understanding why $E[X^{2}] = (sigma^{2} + mu^{2})$, though.
$endgroup$
– mXdX
Jan 23 at 5:33
$begingroup$
Also, that last part is the cube of the entire sum, not just the summand.
$endgroup$
– mXdX
Jan 23 at 5:34
1
$begingroup$
I have added another line to my answer. @mXdX
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:35
1
$begingroup$
@mXdX Sorry that I ddin't read the last part correctly. I have revised my answer.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:38
1
$begingroup$
@mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-mu)^{3}$. For the fourth power you can use a similar argument.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:23
|
show 3 more comments
$begingroup$
What you have done is correct. $E(X-mu)^{2}=E(X^{2}-2mu X+mu^{2})=EX^{2} -2mu EX+mu^{2} =(sigma^{2}+mu^{2})-2mu^{2} +mu^{2}=sigma^{2}$. In general $E(X-mu)^{2}$ is always the variance of $X$. T0 calculate $E( sumlimits_{i=1}^{k}(X_i-mu))^{3}$ you have to expand $(sumlimits_{i=1}^{k}X_i-mu)^{3}$ as $sumlimits_{i=1}^{k}sumlimits_{j=1}^{k}sumlimits_{l=1}^{k} (X_i -mu)(X_j-mu)(X_l-mu)$. [ I have used the following: $sigma^{2}=EX^{2}-mu^{2}$ by definition . Hence $EX^{2}=mu^{2}+sigma^{2}$]. How do we compute $E(X_i -mu)(X_j-mu)(X_l-mu)$?. The value is $0$ if any two of $i,j,l$ are distinct and it is $E(X-mu)^{3}$ if $i=j=l$. You can calculate $E(X-mu)^{3}$ by expanding the cube (using Binomial Theorem).
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
$endgroup$
What you have done is correct. $E(X-mu)^{2}=E(X^{2}-2mu X+mu^{2})=EX^{2} -2mu EX+mu^{2} =(sigma^{2}+mu^{2})-2mu^{2} +mu^{2}=sigma^{2}$. In general $E(X-mu)^{2}$ is always the variance of $X$. T0 calculate $E( sumlimits_{i=1}^{k}(X_i-mu))^{3}$ you have to expand $(sumlimits_{i=1}^{k}X_i-mu)^{3}$ as $sumlimits_{i=1}^{k}sumlimits_{j=1}^{k}sumlimits_{l=1}^{k} (X_i -mu)(X_j-mu)(X_l-mu)$. [ I have used the following: $sigma^{2}=EX^{2}-mu^{2}$ by definition . Hence $EX^{2}=mu^{2}+sigma^{2}$]. How do we compute $E(X_i -mu)(X_j-mu)(X_l-mu)$?. The value is $0$ if any two of $i,j,l$ are distinct and it is $E(X-mu)^{3}$ if $i=j=l$. You can calculate $E(X-mu)^{3}$ by expanding the cube (using Binomial Theorem).
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
edited Jan 23 at 5:47
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
answered Jan 23 at 5:25
Kavi Rama MurthyKavi Rama Murthy
66.6k53067
66.6k53067
$begingroup$
Thanks. Having trouble understanding why $E[X^{2}] = (sigma^{2} + mu^{2})$, though.
$endgroup$
– mXdX
Jan 23 at 5:33
$begingroup$
Also, that last part is the cube of the entire sum, not just the summand.
$endgroup$
– mXdX
Jan 23 at 5:34
1
$begingroup$
I have added another line to my answer. @mXdX
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:35
1
$begingroup$
@mXdX Sorry that I ddin't read the last part correctly. I have revised my answer.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:38
1
$begingroup$
@mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-mu)^{3}$. For the fourth power you can use a similar argument.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:23
|
show 3 more comments
$begingroup$
Thanks. Having trouble understanding why $E[X^{2}] = (sigma^{2} + mu^{2})$, though.
$endgroup$
– mXdX
Jan 23 at 5:33
$begingroup$
Also, that last part is the cube of the entire sum, not just the summand.
$endgroup$
– mXdX
Jan 23 at 5:34
1
$begingroup$
I have added another line to my answer. @mXdX
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:35
1
$begingroup$
@mXdX Sorry that I ddin't read the last part correctly. I have revised my answer.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:38
1
$begingroup$
@mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-mu)^{3}$. For the fourth power you can use a similar argument.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:23
$begingroup$
Thanks. Having trouble understanding why $E[X^{2}] = (sigma^{2} + mu^{2})$, though.
$endgroup$
– mXdX
Jan 23 at 5:33
$begingroup$
Thanks. Having trouble understanding why $E[X^{2}] = (sigma^{2} + mu^{2})$, though.
$endgroup$
– mXdX
Jan 23 at 5:33
$begingroup$
Also, that last part is the cube of the entire sum, not just the summand.
$endgroup$
– mXdX
Jan 23 at 5:34
$begingroup$
Also, that last part is the cube of the entire sum, not just the summand.
$endgroup$
– mXdX
Jan 23 at 5:34
1
1
$begingroup$
I have added another line to my answer. @mXdX
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:35
$begingroup$
I have added another line to my answer. @mXdX
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:35
1
1
$begingroup$
@mXdX Sorry that I ddin't read the last part correctly. I have revised my answer.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:38
$begingroup$
@mXdX Sorry that I ddin't read the last part correctly. I have revised my answer.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:38
1
1
$begingroup$
@mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-mu)^{3}$. For the fourth power you can use a similar argument.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:23
$begingroup$
@mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-mu)^{3}$. For the fourth power you can use a similar argument.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 7:23
|
show 3 more comments
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%2f3084086%2fexpected-values-of-iid-random-variables%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
