Conditional mean and variance of $X$ given that $Y=6$ for $X$ normal $N(0,1)$ and $Y$ conditionally on $X=x$...
$begingroup$
Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
Find the conditional mean and variance of $Xmid Y=6$.
I proceeded as follows:
$$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
But
$${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$
And that is where I am stuck.
probability-theory probability-distributions conditional-expectation variance expected-value
edited Dec 13 '18 at 14:29
Did
248k23224463
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
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add a comment |
$begingroup$
Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
Find the conditional mean and variance of $Xmid Y=6$.
I proceeded as follows:
$$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
But
$${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$
And that is where I am stuck.
probability-theory probability-distributions conditional-expectation variance expected-value
edited Dec 13 '18 at 14:29
Did
248k23224463
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
$endgroup$
add a comment |
$begingroup$
Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
Find the conditional mean and variance of $Xmid Y=6$.
I proceeded as follows:
$$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
But
$${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$
And that is where I am stuck.
probability-theory probability-distributions conditional-expectation variance expected-value
edited Dec 13 '18 at 14:29
Did
248k23224463
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
$endgroup$
Assume that $Xsim N(0,1)$ and that, given $X=x$, the conditional distribution of $Ymid X=x$ is $N(x,1)$.
Find the conditional mean and variance of $Xmid Y=6$.
I proceeded as follows:
$$f_{X|Y}(X|Y=6)= frac{f_X(x)f_{Y|X}(Y|X=x)}{int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}$$
I know that $$f_X(x)f_{Y|X}(Y|X=x)=frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}$$
But
$${int^{infty}_{-infty}f_X(x)f_{Y|X}(Y|X=x)dx}=int^{infty}_{-infty}frac{1}{2pi}e^{frac{-x^2}{2}}e^{frac{-(y-x)^2}{2}}dx$$
And that is where I am stuck.
probability-theory probability-distributions conditional-expectation variance expected-value
probability-theory probability-distributions conditional-expectation variance expected-value
edited Dec 13 '18 at 14:29
Did
248k23224463
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
edited Dec 13 '18 at 14:29
Did
248k23224463
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
edited Dec 13 '18 at 14:29
Did
248k23224463
edited Dec 13 '18 at 14:29
Did
248k23224463
edited Dec 13 '18 at 14:29
Did
248k23224463
248k23224463
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
asked Dec 5 '18 at 5:08
Mahamad A. KanoutéMahamad A. Kanouté
371113
371113
add a comment |
add a comment |
1 Answer
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$begingroup$
One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.
Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.
Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.
This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.
answered Dec 13 '18 at 14:24
DidDid
248k23224463
$endgroup$
$begingroup$
Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
$endgroup$
– Mahamad A. Kanouté
Dec 14 '18 at 2:25
1
$begingroup$
This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
$endgroup$
– Did
Dec 14 '18 at 6:10
add a comment |
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$begingroup$
One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.
Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.
Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.
This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.
answered Dec 13 '18 at 14:24
DidDid
248k23224463
$endgroup$
$begingroup$
Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
$endgroup$
– Mahamad A. Kanouté
Dec 14 '18 at 2:25
1
$begingroup$
This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
$endgroup$
– Did
Dec 14 '18 at 6:10
add a comment |
$begingroup$
One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.
Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.
Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.
This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.
answered Dec 13 '18 at 14:24
DidDid
248k23224463
$endgroup$
$begingroup$
Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
$endgroup$
– Mahamad A. Kanouté
Dec 14 '18 at 2:25
1
$begingroup$
This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
$endgroup$
– Did
Dec 14 '18 at 6:10
add a comment |
$begingroup$
One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.
Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.
Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.
This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.
answered Dec 13 '18 at 14:24
DidDid
248k23224463
$endgroup$
One is given that, for every $x$, the conditional distribution of $Y$ conditionally on $X=x$ is normal with mean $x$ and variance $1$, hence $Y=X+Z$ where $(X,Z)$ is i.i.d. standard normal. Thus, $X=frac12Y+T$ where $T=frac12(X-Z)$.
Since $(Y,T)$ is jointly normal, the fact that $mathrm{Cov}(Y,T)=0$ alone, implies that $T$ is independent of $Y$.
Now, $T$ is centered normal with variance $frac12$ hence, for every $y$, the conditional distribution of $X$ conditionally on $Y=y$ is normal with mean $frac12y$ and variance $frac12$.
This result can be recovered, using conditional PDFs and neglecting the inherent structure of families of normal random variables, only, more clumsily and more lengthily.
answered Dec 13 '18 at 14:24
DidDid
248k23224463
answered Dec 13 '18 at 14:24
DidDid
248k23224463
answered Dec 13 '18 at 14:24
DidDid
248k23224463
answered Dec 13 '18 at 14:24
DidDid
248k23224463
248k23224463
$begingroup$
Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
$endgroup$
– Mahamad A. Kanouté
Dec 14 '18 at 2:25
1
$begingroup$
This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
$endgroup$
– Did
Dec 14 '18 at 6:10
add a comment |
$begingroup$
Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
$endgroup$
– Mahamad A. Kanouté
Dec 14 '18 at 2:25
1
$begingroup$
This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
$endgroup$
– Did
Dec 14 '18 at 6:10
$begingroup$
Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
$endgroup$
– Mahamad A. Kanouté
Dec 14 '18 at 2:25
$begingroup$
Why does Y=X+Z ? do you have some reference on what you wrote. I'm not familiar at all with this way.
$endgroup$
– Mahamad A. Kanouté
Dec 14 '18 at 2:25
1
1
$begingroup$
This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
$endgroup$
– Did
Dec 14 '18 at 6:10
$begingroup$
This is simply because, if $Y$ conditionally on $X=x$ is $N(x,1)$, for every $x$, then $Y-X$ conditionally on $X=x$ is $N(0,1)$, for every $x$, that is, $Y-X$ is $N(0,1)$ and independent of $X$, qed.
$endgroup$
– Did
Dec 14 '18 at 6:10
add a comment |
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