What variable is the conditional probability function a function of?
0
$begingroup$
I've got the following excerpt from my notes in my statistics course:
In the second line, the professor put $x=1$ in the Poisson distribution equation. My question, how do I know that the conditional distribution function is a function of $X_2$, and not $X_1$?
probability statistics conditional-probability
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
$endgroup$
|
show 2 more comments
0
$begingroup$
I've got the following excerpt from my notes in my statistics course:
In the second line, the professor put $x=1$ in the Poisson distribution equation. My question, how do I know that the conditional distribution function is a function of $X_2$, and not $X_1$?
probability statistics conditional-probability
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
$endgroup$
$begingroup$
No you can't. The conditional distribution function $X_2mid X_1$ is by defintion a function of $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:46
$begingroup$
@GNUSupporter8964民主女神地下教會: But then how can the professor have substituted 1 into the equation, if it's not a function of $X_2$?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:01
$begingroup$
It's a function of both $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:02
$begingroup$
@GNUSupporter8964民主女神地下教會: But in the second line, he substituted, $P(X_2 = x_2 | X_1 = x_1)$ with $frac{e^{-k}k^1}{1!}$. So in the Poisson distribution, in the place of the variable $x$, he put $1$. So that means he made it a function of $X_2$, right?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:07
$begingroup$
In your previous comment, I don't understand what $x$ is. $P(X_2 = x_2 | X_1 = x_1)$ is a function of $x_1$ and $x_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:15
|
show 2 more comments
0
0
0
$begingroup$
I've got the following excerpt from my notes in my statistics course:
In the second line, the professor put $x=1$ in the Poisson distribution equation. My question, how do I know that the conditional distribution function is a function of $X_2$, and not $X_1$?
probability statistics conditional-probability
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
$endgroup$
I've got the following excerpt from my notes in my statistics course:
In the second line, the professor put $x=1$ in the Poisson distribution equation. My question, how do I know that the conditional distribution function is a function of $X_2$, and not $X_1$?
probability statistics conditional-probability
probability statistics conditional-probability
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
asked Dec 9 '18 at 20:44
The_QuestionerThe_Questioner
4771514
4771514
$begingroup$
No you can't. The conditional distribution function $X_2mid X_1$ is by defintion a function of $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:46
$begingroup$
@GNUSupporter8964民主女神地下教會: But then how can the professor have substituted 1 into the equation, if it's not a function of $X_2$?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:01
$begingroup$
It's a function of both $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:02
$begingroup$
@GNUSupporter8964民主女神地下教會: But in the second line, he substituted, $P(X_2 = x_2 | X_1 = x_1)$ with $frac{e^{-k}k^1}{1!}$. So in the Poisson distribution, in the place of the variable $x$, he put $1$. So that means he made it a function of $X_2$, right?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:07
$begingroup$
In your previous comment, I don't understand what $x$ is. $P(X_2 = x_2 | X_1 = x_1)$ is a function of $x_1$ and $x_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:15
|
show 2 more comments
$begingroup$
No you can't. The conditional distribution function $X_2mid X_1$ is by defintion a function of $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:46
$begingroup$
@GNUSupporter8964民主女神地下教會: But then how can the professor have substituted 1 into the equation, if it's not a function of $X_2$?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:01
$begingroup$
It's a function of both $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:02
$begingroup$
@GNUSupporter8964民主女神地下教會: But in the second line, he substituted, $P(X_2 = x_2 | X_1 = x_1)$ with $frac{e^{-k}k^1}{1!}$. So in the Poisson distribution, in the place of the variable $x$, he put $1$. So that means he made it a function of $X_2$, right?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:07
$begingroup$
In your previous comment, I don't understand what $x$ is. $P(X_2 = x_2 | X_1 = x_1)$ is a function of $x_1$ and $x_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:15
$begingroup$
No you can't. The conditional distribution function $X_2mid X_1$ is by defintion a function of $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:46
$begingroup$
No you can't. The conditional distribution function $X_2mid X_1$ is by defintion a function of $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:46
$begingroup$
@GNUSupporter8964民主女神地下教會: But then how can the professor have substituted 1 into the equation, if it's not a function of $X_2$?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:01
$begingroup$
@GNUSupporter8964民主女神地下教會: But then how can the professor have substituted 1 into the equation, if it's not a function of $X_2$?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:01
$begingroup$
It's a function of both $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:02
$begingroup$
It's a function of both $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:02
$begingroup$
@GNUSupporter8964民主女神地下教會: But in the second line, he substituted, $P(X_2 = x_2 | X_1 = x_1)$ with $frac{e^{-k}k^1}{1!}$. So in the Poisson distribution, in the place of the variable $x$, he put $1$. So that means he made it a function of $X_2$, right?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:07
$begingroup$
@GNUSupporter8964民主女神地下教會: But in the second line, he substituted, $P(X_2 = x_2 | X_1 = x_1)$ with $frac{e^{-k}k^1}{1!}$. So in the Poisson distribution, in the place of the variable $x$, he put $1$. So that means he made it a function of $X_2$, right?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:07
$begingroup$
In your previous comment, I don't understand what $x$ is. $P(X_2 = x_2 | X_1 = x_1)$ is a function of $x_1$ and $x_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:15
$begingroup$
In your previous comment, I don't understand what $x$ is. $P(X_2 = x_2 | X_1 = x_1)$ is a function of $x_1$ and $x_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:15
|
show 2 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
You are given that
$$
P(X_2 = j | X_1 = k) =
e^{-k} frac{k^j}{j!}
$$ for any nonnegative integers $j, k$. Now just plug in $j=1$ to obtain
$$
P(X_2 = 1 | X_1 = k) =
e^{-k} frac{k^1}{1!}.
$$
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
$endgroup$
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
});
}
});
draft saved
draft discarded
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%2f3032980%2fwhat-variable-is-the-conditional-probability-function-a-function-of%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
0
$begingroup$
You are given that
$$
P(X_2 = j | X_1 = k) =
e^{-k} frac{k^j}{j!}
$$ for any nonnegative integers $j, k$. Now just plug in $j=1$ to obtain
$$
P(X_2 = 1 | X_1 = k) =
e^{-k} frac{k^1}{1!}.
$$
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
$endgroup$
add a comment |
0
$begingroup$
You are given that
$$
P(X_2 = j | X_1 = k) =
e^{-k} frac{k^j}{j!}
$$ for any nonnegative integers $j, k$. Now just plug in $j=1$ to obtain
$$
P(X_2 = 1 | X_1 = k) =
e^{-k} frac{k^1}{1!}.
$$
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
$endgroup$
add a comment |
0
0
0
$begingroup$
You are given that
$$
P(X_2 = j | X_1 = k) =
e^{-k} frac{k^j}{j!}
$$ for any nonnegative integers $j, k$. Now just plug in $j=1$ to obtain
$$
P(X_2 = 1 | X_1 = k) =
e^{-k} frac{k^1}{1!}.
$$
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
$endgroup$
You are given that
$$
P(X_2 = j | X_1 = k) =
e^{-k} frac{k^j}{j!}
$$ for any nonnegative integers $j, k$. Now just plug in $j=1$ to obtain
$$
P(X_2 = 1 | X_1 = k) =
e^{-k} frac{k^1}{1!}.
$$
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
answered Dec 10 '18 at 0:52
littleOlittleO
29.6k646109
29.6k646109
add a comment |
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3032980%2fwhat-variable-is-the-conditional-probability-function-a-function-of%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
$begingroup$
No you can't. The conditional distribution function $X_2mid X_1$ is by defintion a function of $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 20:46
$begingroup$
@GNUSupporter8964民主女神地下教會: But then how can the professor have substituted 1 into the equation, if it's not a function of $X_2$?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:01
$begingroup$
It's a function of both $X_1$ and $X_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:02
$begingroup$
@GNUSupporter8964民主女神地下教會: But in the second line, he substituted, $P(X_2 = x_2 | X_1 = x_1)$ with $frac{e^{-k}k^1}{1!}$. So in the Poisson distribution, in the place of the variable $x$, he put $1$. So that means he made it a function of $X_2$, right?
$endgroup$
– The_Questioner
Dec 9 '18 at 21:07
$begingroup$
In your previous comment, I don't understand what $x$ is. $P(X_2 = x_2 | X_1 = x_1)$ is a function of $x_1$ and $x_2$.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 9 '18 at 21:15