Let $X$ be the number of heads showing, and let $Y$ be the number of tails showing. Compute $Cov(X, Y)$ and...
$begingroup$
Suppose you flip four fair coins. Let $X$ be the number of heads showing, and
let $Y$ be the number of tails showing. Compute $Cov(X, Y)$ and $Corr(X, Y)$.
Although it is not stated, it is clear that $X,Y$ are Binomial to $n=4, theta=1/2$.
Thus we know that variance of a binom dist is $ntheta(1-theta)=1$ in our case.
We also know that the expectation of a binom dist is $ntheta=2$ in our case.
Thus, our formula becomes:
$cov(X,Y)=E(XY)-E(X)E(Y)=E(XY)-(2)(2)$
How do I calculate $E(X)E(Y)$?
Can I treat $X$ and $Y$ as independent?
probability statistics covariance variance expected-value
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
$endgroup$
add a comment |
$begingroup$
Suppose you flip four fair coins. Let $X$ be the number of heads showing, and
let $Y$ be the number of tails showing. Compute $Cov(X, Y)$ and $Corr(X, Y)$.
Although it is not stated, it is clear that $X,Y$ are Binomial to $n=4, theta=1/2$.
Thus we know that variance of a binom dist is $ntheta(1-theta)=1$ in our case.
We also know that the expectation of a binom dist is $ntheta=2$ in our case.
Thus, our formula becomes:
$cov(X,Y)=E(XY)-E(X)E(Y)=E(XY)-(2)(2)$
How do I calculate $E(X)E(Y)$?
Can I treat $X$ and $Y$ as independent?
probability statistics covariance variance expected-value
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
$endgroup$
$begingroup$
I suspect that $X$ and $Y$ are binomial. This may be what you intended
$endgroup$
– Henry
Nov 26 '18 at 21:34
1
$begingroup$
They are not independent note that $Y=4-X$.
$endgroup$
– karakfa
Nov 26 '18 at 21:35
$begingroup$
Note that X and $X$ are regarded as "different" symbols sometimes. Mathematicians use different fonts of the same symbol all the time within the literature to mean different things. It is good practice, then, to stay consistent with one's notation throughout any given text.
$endgroup$
– Shaun
Nov 26 '18 at 21:39
add a comment |
$begingroup$
Suppose you flip four fair coins. Let $X$ be the number of heads showing, and
let $Y$ be the number of tails showing. Compute $Cov(X, Y)$ and $Corr(X, Y)$.
Although it is not stated, it is clear that $X,Y$ are Binomial to $n=4, theta=1/2$.
Thus we know that variance of a binom dist is $ntheta(1-theta)=1$ in our case.
We also know that the expectation of a binom dist is $ntheta=2$ in our case.
Thus, our formula becomes:
$cov(X,Y)=E(XY)-E(X)E(Y)=E(XY)-(2)(2)$
How do I calculate $E(X)E(Y)$?
Can I treat $X$ and $Y$ as independent?
probability statistics covariance variance expected-value
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
$endgroup$
Suppose you flip four fair coins. Let $X$ be the number of heads showing, and
let $Y$ be the number of tails showing. Compute $Cov(X, Y)$ and $Corr(X, Y)$.
Although it is not stated, it is clear that $X,Y$ are Binomial to $n=4, theta=1/2$.
Thus we know that variance of a binom dist is $ntheta(1-theta)=1$ in our case.
We also know that the expectation of a binom dist is $ntheta=2$ in our case.
Thus, our formula becomes:
$cov(X,Y)=E(XY)-E(X)E(Y)=E(XY)-(2)(2)$
How do I calculate $E(X)E(Y)$?
Can I treat $X$ and $Y$ as independent?
probability statistics covariance variance expected-value
probability statistics covariance variance expected-value
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
edited Nov 26 '18 at 21:42
K Split X
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
asked Nov 26 '18 at 21:28
K Split XK Split X
4,19611131
4,19611131
$begingroup$
I suspect that $X$ and $Y$ are binomial. This may be what you intended
$endgroup$
– Henry
Nov 26 '18 at 21:34
1
$begingroup$
They are not independent note that $Y=4-X$.
$endgroup$
– karakfa
Nov 26 '18 at 21:35
$begingroup$
Note that X and $X$ are regarded as "different" symbols sometimes. Mathematicians use different fonts of the same symbol all the time within the literature to mean different things. It is good practice, then, to stay consistent with one's notation throughout any given text.
$endgroup$
– Shaun
Nov 26 '18 at 21:39
add a comment |
$begingroup$
I suspect that $X$ and $Y$ are binomial. This may be what you intended
$endgroup$
– Henry
Nov 26 '18 at 21:34
1
$begingroup$
They are not independent note that $Y=4-X$.
$endgroup$
– karakfa
Nov 26 '18 at 21:35
$begingroup$
Note that X and $X$ are regarded as "different" symbols sometimes. Mathematicians use different fonts of the same symbol all the time within the literature to mean different things. It is good practice, then, to stay consistent with one's notation throughout any given text.
$endgroup$
– Shaun
Nov 26 '18 at 21:39
$begingroup$
I suspect that $X$ and $Y$ are binomial. This may be what you intended
$endgroup$
– Henry
Nov 26 '18 at 21:34
$begingroup$
I suspect that $X$ and $Y$ are binomial. This may be what you intended
$endgroup$
– Henry
Nov 26 '18 at 21:34
1
1
$begingroup$
They are not independent note that $Y=4-X$.
$endgroup$
– karakfa
Nov 26 '18 at 21:35
$begingroup$
They are not independent note that $Y=4-X$.
$endgroup$
– karakfa
Nov 26 '18 at 21:35
$begingroup$
Note that X and $X$ are regarded as "different" symbols sometimes. Mathematicians use different fonts of the same symbol all the time within the literature to mean different things. It is good practice, then, to stay consistent with one's notation throughout any given text.
$endgroup$
– Shaun
Nov 26 '18 at 21:39
$begingroup$
Note that X and $X$ are regarded as "different" symbols sometimes. Mathematicians use different fonts of the same symbol all the time within the literature to mean different things. It is good practice, then, to stay consistent with one's notation throughout any given text.
$endgroup$
– Shaun
Nov 26 '18 at 21:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$E(X)=E(Y)=2$ as you computed. $E(X^2)=E(Y^2)=5$, so $Var(X)=Var(Y)=1$ and $E(XY)=E(X(4-X))=4E(X)-E(X^2)=3$, so $Cov(X,Y)=-1$ and $Cor(X,Y)=-1$.
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
$endgroup$
add a comment |
$begingroup$
How do I calculate $E(X)E(Y)$?
You just did. You likely mean $E(XY)$.
Can I treat $X$ and $Y$ as independent?
No, they are clearly very dependent, and indeed $Y=4-X$ is a linear relation, so you immediately know the correlation coefficent.
You might use $mathsf E(XY)=mathsf E(4X-X^2)$, so $mathsf {Cov}(X,Y)=mathsf E(4X-X^2)-mathsf E(X)(mathsf E(4-X)$
In short: $$mathsf Cov(X,Y)~{=mathsf Cov(X,4-X) \= -mathsf Var(X)}$$
And you know the variance of a binomially distributed random variable.
From this you can calculate the correlation coefficient, if feel the you need to.
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
$endgroup$
$begingroup$
If I define $Y=4-X$, then how can I calculate the variance of $Y$? It would be $var(4)-var(X)=0-1=-1$? How is this allowed.
$endgroup$
– K Split X
Nov 27 '18 at 0:27
1
$begingroup$
@KSplitX - no. The variance of $a+bX$ is $b^2$ times the variance of $X$. Here $a=4$ and $b=-1$ so $b^2=1$ and the variance of $Y$ is equal to the variance of $X$
$endgroup$
– Henry
Nov 27 '18 at 1:07
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%2f3014949%2flet-x-be-the-number-of-heads-showing-and-let-y-be-the-number-of-tails-showi%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$E(X)=E(Y)=2$ as you computed. $E(X^2)=E(Y^2)=5$, so $Var(X)=Var(Y)=1$ and $E(XY)=E(X(4-X))=4E(X)-E(X^2)=3$, so $Cov(X,Y)=-1$ and $Cor(X,Y)=-1$.
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
$endgroup$
add a comment |
$begingroup$
$E(X)=E(Y)=2$ as you computed. $E(X^2)=E(Y^2)=5$, so $Var(X)=Var(Y)=1$ and $E(XY)=E(X(4-X))=4E(X)-E(X^2)=3$, so $Cov(X,Y)=-1$ and $Cor(X,Y)=-1$.
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
$endgroup$
add a comment |
$begingroup$
$E(X)=E(Y)=2$ as you computed. $E(X^2)=E(Y^2)=5$, so $Var(X)=Var(Y)=1$ and $E(XY)=E(X(4-X))=4E(X)-E(X^2)=3$, so $Cov(X,Y)=-1$ and $Cor(X,Y)=-1$.
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
$endgroup$
$E(X)=E(Y)=2$ as you computed. $E(X^2)=E(Y^2)=5$, so $Var(X)=Var(Y)=1$ and $E(XY)=E(X(4-X))=4E(X)-E(X^2)=3$, so $Cov(X,Y)=-1$ and $Cor(X,Y)=-1$.
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
edited Nov 26 '18 at 22:35
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
answered Nov 26 '18 at 22:20
herb steinbergherb steinberg
2,5582310
2,5582310
add a comment |
add a comment |
$begingroup$
How do I calculate $E(X)E(Y)$?
You just did. You likely mean $E(XY)$.
Can I treat $X$ and $Y$ as independent?
No, they are clearly very dependent, and indeed $Y=4-X$ is a linear relation, so you immediately know the correlation coefficent.
You might use $mathsf E(XY)=mathsf E(4X-X^2)$, so $mathsf {Cov}(X,Y)=mathsf E(4X-X^2)-mathsf E(X)(mathsf E(4-X)$
In short: $$mathsf Cov(X,Y)~{=mathsf Cov(X,4-X) \= -mathsf Var(X)}$$
And you know the variance of a binomially distributed random variable.
From this you can calculate the correlation coefficient, if feel the you need to.
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
$endgroup$
$begingroup$
If I define $Y=4-X$, then how can I calculate the variance of $Y$? It would be $var(4)-var(X)=0-1=-1$? How is this allowed.
$endgroup$
– K Split X
Nov 27 '18 at 0:27
1
$begingroup$
@KSplitX - no. The variance of $a+bX$ is $b^2$ times the variance of $X$. Here $a=4$ and $b=-1$ so $b^2=1$ and the variance of $Y$ is equal to the variance of $X$
$endgroup$
– Henry
Nov 27 '18 at 1:07
add a comment |
$begingroup$
How do I calculate $E(X)E(Y)$?
You just did. You likely mean $E(XY)$.
Can I treat $X$ and $Y$ as independent?
No, they are clearly very dependent, and indeed $Y=4-X$ is a linear relation, so you immediately know the correlation coefficent.
You might use $mathsf E(XY)=mathsf E(4X-X^2)$, so $mathsf {Cov}(X,Y)=mathsf E(4X-X^2)-mathsf E(X)(mathsf E(4-X)$
In short: $$mathsf Cov(X,Y)~{=mathsf Cov(X,4-X) \= -mathsf Var(X)}$$
And you know the variance of a binomially distributed random variable.
From this you can calculate the correlation coefficient, if feel the you need to.
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
$endgroup$
$begingroup$
If I define $Y=4-X$, then how can I calculate the variance of $Y$? It would be $var(4)-var(X)=0-1=-1$? How is this allowed.
$endgroup$
– K Split X
Nov 27 '18 at 0:27
1
$begingroup$
@KSplitX - no. The variance of $a+bX$ is $b^2$ times the variance of $X$. Here $a=4$ and $b=-1$ so $b^2=1$ and the variance of $Y$ is equal to the variance of $X$
$endgroup$
– Henry
Nov 27 '18 at 1:07
add a comment |
$begingroup$
How do I calculate $E(X)E(Y)$?
You just did. You likely mean $E(XY)$.
Can I treat $X$ and $Y$ as independent?
No, they are clearly very dependent, and indeed $Y=4-X$ is a linear relation, so you immediately know the correlation coefficent.
You might use $mathsf E(XY)=mathsf E(4X-X^2)$, so $mathsf {Cov}(X,Y)=mathsf E(4X-X^2)-mathsf E(X)(mathsf E(4-X)$
In short: $$mathsf Cov(X,Y)~{=mathsf Cov(X,4-X) \= -mathsf Var(X)}$$
And you know the variance of a binomially distributed random variable.
From this you can calculate the correlation coefficient, if feel the you need to.
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
$endgroup$
How do I calculate $E(X)E(Y)$?
You just did. You likely mean $E(XY)$.
Can I treat $X$ and $Y$ as independent?
No, they are clearly very dependent, and indeed $Y=4-X$ is a linear relation, so you immediately know the correlation coefficent.
You might use $mathsf E(XY)=mathsf E(4X-X^2)$, so $mathsf {Cov}(X,Y)=mathsf E(4X-X^2)-mathsf E(X)(mathsf E(4-X)$
In short: $$mathsf Cov(X,Y)~{=mathsf Cov(X,4-X) \= -mathsf Var(X)}$$
And you know the variance of a binomially distributed random variable.
From this you can calculate the correlation coefficient, if feel the you need to.
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
answered Nov 26 '18 at 22:47
Graham KempGraham Kemp
85.1k43378
85.1k43378
$begingroup$
If I define $Y=4-X$, then how can I calculate the variance of $Y$? It would be $var(4)-var(X)=0-1=-1$? How is this allowed.
$endgroup$
– K Split X
Nov 27 '18 at 0:27
1
$begingroup$
@KSplitX - no. The variance of $a+bX$ is $b^2$ times the variance of $X$. Here $a=4$ and $b=-1$ so $b^2=1$ and the variance of $Y$ is equal to the variance of $X$
$endgroup$
– Henry
Nov 27 '18 at 1:07
add a comment |
$begingroup$
If I define $Y=4-X$, then how can I calculate the variance of $Y$? It would be $var(4)-var(X)=0-1=-1$? How is this allowed.
$endgroup$
– K Split X
Nov 27 '18 at 0:27
1
$begingroup$
@KSplitX - no. The variance of $a+bX$ is $b^2$ times the variance of $X$. Here $a=4$ and $b=-1$ so $b^2=1$ and the variance of $Y$ is equal to the variance of $X$
$endgroup$
– Henry
Nov 27 '18 at 1:07
$begingroup$
If I define $Y=4-X$, then how can I calculate the variance of $Y$? It would be $var(4)-var(X)=0-1=-1$? How is this allowed.
$endgroup$
– K Split X
Nov 27 '18 at 0:27
$begingroup$
If I define $Y=4-X$, then how can I calculate the variance of $Y$? It would be $var(4)-var(X)=0-1=-1$? How is this allowed.
$endgroup$
– K Split X
Nov 27 '18 at 0:27
1
1
$begingroup$
@KSplitX - no. The variance of $a+bX$ is $b^2$ times the variance of $X$. Here $a=4$ and $b=-1$ so $b^2=1$ and the variance of $Y$ is equal to the variance of $X$
$endgroup$
– Henry
Nov 27 '18 at 1:07
$begingroup$
@KSplitX - no. The variance of $a+bX$ is $b^2$ times the variance of $X$. Here $a=4$ and $b=-1$ so $b^2=1$ and the variance of $Y$ is equal to the variance of $X$
$endgroup$
– Henry
Nov 27 '18 at 1:07
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%2f3014949%2flet-x-be-the-number-of-heads-showing-and-let-y-be-the-number-of-tails-showi%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
$begingroup$
I suspect that $X$ and $Y$ are binomial. This may be what you intended
$endgroup$
– Henry
Nov 26 '18 at 21:34
1
$begingroup$
They are not independent note that $Y=4-X$.
$endgroup$
– karakfa
Nov 26 '18 at 21:35
$begingroup$
Note that X and $X$ are regarded as "different" symbols sometimes. Mathematicians use different fonts of the same symbol all the time within the literature to mean different things. It is good practice, then, to stay consistent with one's notation throughout any given text.
$endgroup$
– Shaun
Nov 26 '18 at 21:39