What variable is the conditional probability function a function of?
0
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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
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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
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@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
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$begingroup$
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
29.7k646109
$endgroup$
add a comment |
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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
29.7k646109
$endgroup$
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
29.7k646109
$endgroup$
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
29.7k646109
$endgroup$
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
29.7k646109
answered Dec 10 '18 at 0:52
littleOlittleO
29.7k646109
answered Dec 10 '18 at 0:52
littleOlittleO
29.7k646109
answered Dec 10 '18 at 0:52
littleOlittleO
29.7k646109
29.7k646109
add a comment |
add a comment |
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$begingroup$
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