Mixing
Finding density function of $operatorname{E}[Xmid Y]$ and $operatorname{E}[Ymid X]$
$begingroup$
$newcommand{E}{operatorname{E}}$
Finding density function of $E[Xmid Y]$ and $E[Ymid X]$, where $E[Xmid Y]=Y/2$, $E[Ymid X]=X+1,$ $f(x,y)=e^{-y}$, and $0 leq x leq y$.
I am unsure on how to find the density of an expected value, could anyone please help me with the formula or method to use to find it?
conditional-expectation density-function expected-value
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
$endgroup$
|
show 1 more comment
$begingroup$
$newcommand{E}{operatorname{E}}$
Finding density function of $E[Xmid Y]$ and $E[Ymid X]$, where $E[Xmid Y]=Y/2$, $E[Ymid X]=X+1,$ $f(x,y)=e^{-y}$, and $0 leq x leq y$.
I am unsure on how to find the density of an expected value, could anyone please help me with the formula or method to use to find it?
conditional-expectation density-function expected-value
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
$endgroup$
1
$begingroup$
Or $mathsf{E}[Ymid X=x]=x+1$, and $mathsf{E}[Xmid Y=y]=y/2$.
$endgroup$
– d.k.o.
Oct 5 '17 at 0:19
$begingroup$
@MichaelHardy Sorry I didn't notice while typing, edited
$endgroup$
– Silvia Rossi
Oct 5 '17 at 0:20
$begingroup$
Then is should be clear that you need to find the densities of $X+1$ and $Y/2$ given the joint distribution of $X$ and $Y$...
$endgroup$
– d.k.o.
Oct 5 '17 at 0:20
$begingroup$
@d.k.o. Could I use the transformation method? Let $W=X+1$. $f(W)=f(x) lvert{frac{dX}{dW}}rvert=e^{-x}(1)=e^{-x}$ And let $T=Y/2$. $f(T)=f(y) lvert{frac{dY}{dT}}rvert=ye^{-y}(2)=2ye^{-y}$
$endgroup$
– Silvia Rossi
Oct 5 '17 at 2:28
$begingroup$
$X+1ge 1$. $f(T)$ is ok.
$endgroup$
– d.k.o.
Oct 5 '17 at 3:02
|
show 1 more comment
$begingroup$
$newcommand{E}{operatorname{E}}$
Finding density function of $E[Xmid Y]$ and $E[Ymid X]$, where $E[Xmid Y]=Y/2$, $E[Ymid X]=X+1,$ $f(x,y)=e^{-y}$, and $0 leq x leq y$.
I am unsure on how to find the density of an expected value, could anyone please help me with the formula or method to use to find it?
conditional-expectation density-function expected-value
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
$endgroup$
$newcommand{E}{operatorname{E}}$
Finding density function of $E[Xmid Y]$ and $E[Ymid X]$, where $E[Xmid Y]=Y/2$, $E[Ymid X]=X+1,$ $f(x,y)=e^{-y}$, and $0 leq x leq y$.
I am unsure on how to find the density of an expected value, could anyone please help me with the formula or method to use to find it?
conditional-expectation density-function expected-value
conditional-expectation density-function expected-value
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
edited Feb 1 at 15:17
Martin Sleziak
45k10122277
45k10122277
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
asked Oct 5 '17 at 0:14
Silvia RossiSilvia Rossi
452417
452417
1
$begingroup$
Or $mathsf{E}[Ymid X=x]=x+1$, and $mathsf{E}[Xmid Y=y]=y/2$.
$endgroup$
– d.k.o.
Oct 5 '17 at 0:19
$begingroup$
@MichaelHardy Sorry I didn't notice while typing, edited
$endgroup$
– Silvia Rossi
Oct 5 '17 at 0:20
$begingroup$
Then is should be clear that you need to find the densities of $X+1$ and $Y/2$ given the joint distribution of $X$ and $Y$...
$endgroup$
– d.k.o.
Oct 5 '17 at 0:20
$begingroup$
@d.k.o. Could I use the transformation method? Let $W=X+1$. $f(W)=f(x) lvert{frac{dX}{dW}}rvert=e^{-x}(1)=e^{-x}$ And let $T=Y/2$. $f(T)=f(y) lvert{frac{dY}{dT}}rvert=ye^{-y}(2)=2ye^{-y}$
$endgroup$
– Silvia Rossi
Oct 5 '17 at 2:28
$begingroup$
$X+1ge 1$. $f(T)$ is ok.
$endgroup$
– d.k.o.
Oct 5 '17 at 3:02
|
show 1 more comment
1
$begingroup$
Or $mathsf{E}[Ymid X=x]=x+1$, and $mathsf{E}[Xmid Y=y]=y/2$.
$endgroup$
– d.k.o.
Oct 5 '17 at 0:19
$begingroup$
@MichaelHardy Sorry I didn't notice while typing, edited
$endgroup$
– Silvia Rossi
Oct 5 '17 at 0:20
$begingroup$
Then is should be clear that you need to find the densities of $X+1$ and $Y/2$ given the joint distribution of $X$ and $Y$...
$endgroup$
– d.k.o.
Oct 5 '17 at 0:20
$begingroup$
@d.k.o. Could I use the transformation method? Let $W=X+1$. $f(W)=f(x) lvert{frac{dX}{dW}}rvert=e^{-x}(1)=e^{-x}$ And let $T=Y/2$. $f(T)=f(y) lvert{frac{dY}{dT}}rvert=ye^{-y}(2)=2ye^{-y}$
$endgroup$
– Silvia Rossi
Oct 5 '17 at 2:28
$begingroup$
$X+1ge 1$. $f(T)$ is ok.
$endgroup$
– d.k.o.
Oct 5 '17 at 3:02
1
1
$begingroup$
Or $mathsf{E}[Ymid X=x]=x+1$, and $mathsf{E}[Xmid Y=y]=y/2$.
$endgroup$
– d.k.o.
Oct 5 '17 at 0:19
$begingroup$
Or $mathsf{E}[Ymid X=x]=x+1$, and $mathsf{E}[Xmid Y=y]=y/2$.
$endgroup$
– d.k.o.
Oct 5 '17 at 0:19
$begingroup$
@MichaelHardy Sorry I didn't notice while typing, edited
$endgroup$
– Silvia Rossi
Oct 5 '17 at 0:20
$begingroup$
@MichaelHardy Sorry I didn't notice while typing, edited
$endgroup$
– Silvia Rossi
Oct 5 '17 at 0:20
$begingroup$
Then is should be clear that you need to find the densities of $X+1$ and $Y/2$ given the joint distribution of $X$ and $Y$...
$endgroup$
– d.k.o.
Oct 5 '17 at 0:20
$begingroup$
Then is should be clear that you need to find the densities of $X+1$ and $Y/2$ given the joint distribution of $X$ and $Y$...
$endgroup$
– d.k.o.
Oct 5 '17 at 0:20
$begingroup$
@d.k.o. Could I use the transformation method? Let $W=X+1$. $f(W)=f(x) lvert{frac{dX}{dW}}rvert=e^{-x}(1)=e^{-x}$ And let $T=Y/2$. $f(T)=f(y) lvert{frac{dY}{dT}}rvert=ye^{-y}(2)=2ye^{-y}$
$endgroup$
– Silvia Rossi
Oct 5 '17 at 2:28
$begingroup$
@d.k.o. Could I use the transformation method? Let $W=X+1$. $f(W)=f(x) lvert{frac{dX}{dW}}rvert=e^{-x}(1)=e^{-x}$ And let $T=Y/2$. $f(T)=f(y) lvert{frac{dY}{dT}}rvert=ye^{-y}(2)=2ye^{-y}$
$endgroup$
– Silvia Rossi
Oct 5 '17 at 2:28
$begingroup$
$X+1ge 1$. $f(T)$ is ok.
$endgroup$
– d.k.o.
Oct 5 '17 at 3:02
$begingroup$
$X+1ge 1$. $f(T)$ is ok.
$endgroup$
– d.k.o.
Oct 5 '17 at 3:02
|
show 1 more comment
1 Answer
1
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oldest
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$begingroup$
$newcommand{E}{operatorname{E}}$
$$
iintlimits_{{,(x,y),:, 0,le,x,le,y,}} e^{-y} , d(x,y) = int_0^infty left( int_x^infty e^{-y} , dy right) , dx = int_0^infty e^{-x} , dx = 1
$$
and thus this is indeed a probability density.
The conditional density of $y$ given the event $X=x$ is $big( text{constant}cdot e^{-y} big)$ for $yge x.$ Since
$$
int_x^infty e^{-y},dy = e^{-x},
$$
the "constant" must be $1/e^{-x} = e^x.$ Thus the density is $e^x e^{-y} = e^{x-y} text{ for } yge x.$ The conditional expect value is therefore
$$
E(Ymid X=x) = int_x^infty e^{x-y} , dy = x+1,
$$
so we can say
$$
E(Ymid X) = X+1.
$$
For the conditional density of $X$ given $Y=y,$ note that the joint density does not depend on $x$ beyond the fact that $0le xle y,$ so the density is constant on the interval from $0$ to $y;$ in other words, given $Y=y,$ the conditional distribution of $X$ is the uniform distribution on that interval.
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
$endgroup$
add a comment |
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$begingroup$
$newcommand{E}{operatorname{E}}$
$$
iintlimits_{{,(x,y),:, 0,le,x,le,y,}} e^{-y} , d(x,y) = int_0^infty left( int_x^infty e^{-y} , dy right) , dx = int_0^infty e^{-x} , dx = 1
$$
and thus this is indeed a probability density.
The conditional density of $y$ given the event $X=x$ is $big( text{constant}cdot e^{-y} big)$ for $yge x.$ Since
$$
int_x^infty e^{-y},dy = e^{-x},
$$
the "constant" must be $1/e^{-x} = e^x.$ Thus the density is $e^x e^{-y} = e^{x-y} text{ for } yge x.$ The conditional expect value is therefore
$$
E(Ymid X=x) = int_x^infty e^{x-y} , dy = x+1,
$$
so we can say
$$
E(Ymid X) = X+1.
$$
For the conditional density of $X$ given $Y=y,$ note that the joint density does not depend on $x$ beyond the fact that $0le xle y,$ so the density is constant on the interval from $0$ to $y;$ in other words, given $Y=y,$ the conditional distribution of $X$ is the uniform distribution on that interval.
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
$endgroup$
add a comment |
$begingroup$
$newcommand{E}{operatorname{E}}$
$$
iintlimits_{{,(x,y),:, 0,le,x,le,y,}} e^{-y} , d(x,y) = int_0^infty left( int_x^infty e^{-y} , dy right) , dx = int_0^infty e^{-x} , dx = 1
$$
and thus this is indeed a probability density.
The conditional density of $y$ given the event $X=x$ is $big( text{constant}cdot e^{-y} big)$ for $yge x.$ Since
$$
int_x^infty e^{-y},dy = e^{-x},
$$
the "constant" must be $1/e^{-x} = e^x.$ Thus the density is $e^x e^{-y} = e^{x-y} text{ for } yge x.$ The conditional expect value is therefore
$$
E(Ymid X=x) = int_x^infty e^{x-y} , dy = x+1,
$$
so we can say
$$
E(Ymid X) = X+1.
$$
For the conditional density of $X$ given $Y=y,$ note that the joint density does not depend on $x$ beyond the fact that $0le xle y,$ so the density is constant on the interval from $0$ to $y;$ in other words, given $Y=y,$ the conditional distribution of $X$ is the uniform distribution on that interval.
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
$endgroup$
add a comment |
$begingroup$
$newcommand{E}{operatorname{E}}$
$$
iintlimits_{{,(x,y),:, 0,le,x,le,y,}} e^{-y} , d(x,y) = int_0^infty left( int_x^infty e^{-y} , dy right) , dx = int_0^infty e^{-x} , dx = 1
$$
and thus this is indeed a probability density.
The conditional density of $y$ given the event $X=x$ is $big( text{constant}cdot e^{-y} big)$ for $yge x.$ Since
$$
int_x^infty e^{-y},dy = e^{-x},
$$
the "constant" must be $1/e^{-x} = e^x.$ Thus the density is $e^x e^{-y} = e^{x-y} text{ for } yge x.$ The conditional expect value is therefore
$$
E(Ymid X=x) = int_x^infty e^{x-y} , dy = x+1,
$$
so we can say
$$
E(Ymid X) = X+1.
$$
For the conditional density of $X$ given $Y=y,$ note that the joint density does not depend on $x$ beyond the fact that $0le xle y,$ so the density is constant on the interval from $0$ to $y;$ in other words, given $Y=y,$ the conditional distribution of $X$ is the uniform distribution on that interval.
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
$endgroup$
$newcommand{E}{operatorname{E}}$
$$
iintlimits_{{,(x,y),:, 0,le,x,le,y,}} e^{-y} , d(x,y) = int_0^infty left( int_x^infty e^{-y} , dy right) , dx = int_0^infty e^{-x} , dx = 1
$$
and thus this is indeed a probability density.
The conditional density of $y$ given the event $X=x$ is $big( text{constant}cdot e^{-y} big)$ for $yge x.$ Since
$$
int_x^infty e^{-y},dy = e^{-x},
$$
the "constant" must be $1/e^{-x} = e^x.$ Thus the density is $e^x e^{-y} = e^{x-y} text{ for } yge x.$ The conditional expect value is therefore
$$
E(Ymid X=x) = int_x^infty e^{x-y} , dy = x+1,
$$
so we can say
$$
E(Ymid X) = X+1.
$$
For the conditional density of $X$ given $Y=y,$ note that the joint density does not depend on $x$ beyond the fact that $0le xle y,$ so the density is constant on the interval from $0$ to $y;$ in other words, given $Y=y,$ the conditional distribution of $X$ is the uniform distribution on that interval.
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
answered Oct 5 '17 at 0:35
Michael HardyMichael Hardy
1
1
add a comment |
add a comment |
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1
$begingroup$
Or $mathsf{E}[Ymid X=x]=x+1$, and $mathsf{E}[Xmid Y=y]=y/2$.
$endgroup$
– d.k.o.
Oct 5 '17 at 0:19
$begingroup$
@MichaelHardy Sorry I didn't notice while typing, edited
$endgroup$
– Silvia Rossi
Oct 5 '17 at 0:20
$begingroup$
Then is should be clear that you need to find the densities of $X+1$ and $Y/2$ given the joint distribution of $X$ and $Y$...
$endgroup$
– d.k.o.
Oct 5 '17 at 0:20
$begingroup$
@d.k.o. Could I use the transformation method? Let $W=X+1$. $f(W)=f(x) lvert{frac{dX}{dW}}rvert=e^{-x}(1)=e^{-x}$ And let $T=Y/2$. $f(T)=f(y) lvert{frac{dY}{dT}}rvert=ye^{-y}(2)=2ye^{-y}$
$endgroup$
– Silvia Rossi
Oct 5 '17 at 2:28
$begingroup$
$X+1ge 1$. $f(T)$ is ok.
$endgroup$
– d.k.o.
Oct 5 '17 at 3:02