Mixing
Conditional Expectation of One Member of a Multinomial Distribution, Conditioned on a Second Member
I have a homework problem, which asks:
Let X = $(X_1, X_2, X_3, X_4, X_5) sim text{Mult}_5(n, p)$ with p = $(p_1, p_2, p_3, p_4, p_5)$.
(a) Find $E(X_1 | X_2) text{ and } Var(X_1 | X_2)$.
(b) Find $E(X_1 | X_2 + X_3)$.
So here is my approach to part (a):
To find $E(X_1|X_2 = x_2)$, I think of $X_2$ being set at a fixed value and remove it from the distribution. So first I have to normalize p, so say that:
$$ p_1^{'} = frac{p_1}{1-p_2}, p_3^{'} = frac{p_3}{1-p_2}, p_4^{'} = frac{p_4}{1-p_2}, p_5^{'} = frac{p_5}{1-p_2} $$
then let $p^{'} = (p_1^{'}, p_3^{'}, p_4^{'}, p_5^{'})$
So with $X_2$ fixed, I have a new distribution: $text{Mult}_4(n-X_2, p^{'})$. And to find the probability of $(X_1|X_2)$, I can treat it like any other marginal of a multinomial distribution, and say that: $(X_1|X_2 = x_2) sim text{Bin}(n-X_2, p_1^{'})$
I know that the expected value for a binomial distribution is $n cdot p$ and the variance is $n cdot p(1-p)$, so I can say that:
$$ E(X_1|X_2) = p_1^{'}(n-X_2)$$
$$ Var(X_1|X_2) = p_1^{'}(n-X_2)(1-p_1^{'}) $$
So I have two questions.
Is this approach OK?
If so, can I simply take the same steps for (b) but adding $X_2$ and $X_3$ together?
probability statistics conditional-expectation expected-value
asked Nov 20 '18 at 4:24
JStorm
255
add a comment |
I have a homework problem, which asks:
Let X = $(X_1, X_2, X_3, X_4, X_5) sim text{Mult}_5(n, p)$ with p = $(p_1, p_2, p_3, p_4, p_5)$.
(a) Find $E(X_1 | X_2) text{ and } Var(X_1 | X_2)$.
(b) Find $E(X_1 | X_2 + X_3)$.
So here is my approach to part (a):
To find $E(X_1|X_2 = x_2)$, I think of $X_2$ being set at a fixed value and remove it from the distribution. So first I have to normalize p, so say that:
$$ p_1^{'} = frac{p_1}{1-p_2}, p_3^{'} = frac{p_3}{1-p_2}, p_4^{'} = frac{p_4}{1-p_2}, p_5^{'} = frac{p_5}{1-p_2} $$
then let $p^{'} = (p_1^{'}, p_3^{'}, p_4^{'}, p_5^{'})$
So with $X_2$ fixed, I have a new distribution: $text{Mult}_4(n-X_2, p^{'})$. And to find the probability of $(X_1|X_2)$, I can treat it like any other marginal of a multinomial distribution, and say that: $(X_1|X_2 = x_2) sim text{Bin}(n-X_2, p_1^{'})$
I know that the expected value for a binomial distribution is $n cdot p$ and the variance is $n cdot p(1-p)$, so I can say that:
$$ E(X_1|X_2) = p_1^{'}(n-X_2)$$
$$ Var(X_1|X_2) = p_1^{'}(n-X_2)(1-p_1^{'}) $$
So I have two questions.
Is this approach OK?
If so, can I simply take the same steps for (b) but adding $X_2$ and $X_3$ together?
probability statistics conditional-expectation expected-value
asked Nov 20 '18 at 4:24
JStorm
255
It looks good to me. You may adopt a similar approach in part b).
– BGM
Nov 20 '18 at 9:05
add a comment |
I have a homework problem, which asks:
Let X = $(X_1, X_2, X_3, X_4, X_5) sim text{Mult}_5(n, p)$ with p = $(p_1, p_2, p_3, p_4, p_5)$.
(a) Find $E(X_1 | X_2) text{ and } Var(X_1 | X_2)$.
(b) Find $E(X_1 | X_2 + X_3)$.
So here is my approach to part (a):
To find $E(X_1|X_2 = x_2)$, I think of $X_2$ being set at a fixed value and remove it from the distribution. So first I have to normalize p, so say that:
$$ p_1^{'} = frac{p_1}{1-p_2}, p_3^{'} = frac{p_3}{1-p_2}, p_4^{'} = frac{p_4}{1-p_2}, p_5^{'} = frac{p_5}{1-p_2} $$
then let $p^{'} = (p_1^{'}, p_3^{'}, p_4^{'}, p_5^{'})$
So with $X_2$ fixed, I have a new distribution: $text{Mult}_4(n-X_2, p^{'})$. And to find the probability of $(X_1|X_2)$, I can treat it like any other marginal of a multinomial distribution, and say that: $(X_1|X_2 = x_2) sim text{Bin}(n-X_2, p_1^{'})$
I know that the expected value for a binomial distribution is $n cdot p$ and the variance is $n cdot p(1-p)$, so I can say that:
$$ E(X_1|X_2) = p_1^{'}(n-X_2)$$
$$ Var(X_1|X_2) = p_1^{'}(n-X_2)(1-p_1^{'}) $$
So I have two questions.
Is this approach OK?
If so, can I simply take the same steps for (b) but adding $X_2$ and $X_3$ together?
probability statistics conditional-expectation expected-value
asked Nov 20 '18 at 4:24
JStorm
255
I have a homework problem, which asks:
Let X = $(X_1, X_2, X_3, X_4, X_5) sim text{Mult}_5(n, p)$ with p = $(p_1, p_2, p_3, p_4, p_5)$.
(a) Find $E(X_1 | X_2) text{ and } Var(X_1 | X_2)$.
(b) Find $E(X_1 | X_2 + X_3)$.
So here is my approach to part (a):
To find $E(X_1|X_2 = x_2)$, I think of $X_2$ being set at a fixed value and remove it from the distribution. So first I have to normalize p, so say that:
$$ p_1^{'} = frac{p_1}{1-p_2}, p_3^{'} = frac{p_3}{1-p_2}, p_4^{'} = frac{p_4}{1-p_2}, p_5^{'} = frac{p_5}{1-p_2} $$
then let $p^{'} = (p_1^{'}, p_3^{'}, p_4^{'}, p_5^{'})$
So with $X_2$ fixed, I have a new distribution: $text{Mult}_4(n-X_2, p^{'})$. And to find the probability of $(X_1|X_2)$, I can treat it like any other marginal of a multinomial distribution, and say that: $(X_1|X_2 = x_2) sim text{Bin}(n-X_2, p_1^{'})$
I know that the expected value for a binomial distribution is $n cdot p$ and the variance is $n cdot p(1-p)$, so I can say that:
$$ E(X_1|X_2) = p_1^{'}(n-X_2)$$
$$ Var(X_1|X_2) = p_1^{'}(n-X_2)(1-p_1^{'}) $$
So I have two questions.
Is this approach OK?
If so, can I simply take the same steps for (b) but adding $X_2$ and $X_3$ together?
probability statistics conditional-expectation expected-value
probability statistics conditional-expectation expected-value
asked Nov 20 '18 at 4:24
JStorm
255
asked Nov 20 '18 at 4:24
JStorm
255
asked Nov 20 '18 at 4:24
JStorm
255
asked Nov 20 '18 at 4:24
JStorm
255
asked Nov 20 '18 at 4:24
JStorm
255
255
It looks good to me. You may adopt a similar approach in part b).
– BGM
Nov 20 '18 at 9:05
add a comment |
It looks good to me. You may adopt a similar approach in part b).
– BGM
Nov 20 '18 at 9:05
It looks good to me. You may adopt a similar approach in part b).
– BGM
Nov 20 '18 at 9:05
It looks good to me. You may adopt a similar approach in part b).
– BGM
Nov 20 '18 at 9:05
add a comment |
active
oldest
votes
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%2f3005935%2fconditional-expectation-of-one-member-of-a-multinomial-distribution-conditioned%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3005935%2fconditional-expectation-of-one-member-of-a-multinomial-distribution-conditioned%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
It looks good to me. You may adopt a similar approach in part b).
– BGM
Nov 20 '18 at 9:05