Checking whether $X$ and $Y$ are independent given the joint pdf and calculating conditional expectation











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Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases} y^{-1}text{exp}left(-yright) & 0 < x < y \[1em]
0, & text{ otherwise.} end{cases} $$



a) Compute $mathbb{E}[X^{3} | Y = y]$



b) Are $X$ and $Y$ independent?




So, my attempt for $(a)$ was to first compute $f_{x mid y}(x mid y)$ as follows:



$$f_{x|y}(x|y) = frac{f(x,y)}{int_{-infty}^{infty}f(x,y) mathop{dx}} = frac{1}{y}.$$



So,



$$mathbb{E}[X^{3} mid Y = y] = int_{0}^{y} x^{3}frac{1}{y} mathop{dx} = boxed{y^3/4}$$




For $(b)$, I try to decompose $f(x, y)$ into the product of two marginals. We have



$$f_{Y}(y) = int_{0}^{y} f(x, y) mathop{dx} = e^{-y}.$$



But, I can't compute



$$f_{X}(x) = int_{x}^{infty} f(x,y) mathop{dy}. $$



Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?






probability probability-theory probability-distributions







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  • Since you already found in part a) that $Xmid Ysim text{Uniform}(0,Y)$ (you did not mention that $0<x<y$ in the support), this should be enough to conclude on part b).
    – StubbornAtom
    Nov 21 at 19:28

















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1
down vote

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Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases} y^{-1}text{exp}left(-yright) & 0 < x < y \[1em]
0, & text{ otherwise.} end{cases} $$



a) Compute $mathbb{E}[X^{3} | Y = y]$



b) Are $X$ and $Y$ independent?




So, my attempt for $(a)$ was to first compute $f_{x mid y}(x mid y)$ as follows:



$$f_{x|y}(x|y) = frac{f(x,y)}{int_{-infty}^{infty}f(x,y) mathop{dx}} = frac{1}{y}.$$



So,



$$mathbb{E}[X^{3} mid Y = y] = int_{0}^{y} x^{3}frac{1}{y} mathop{dx} = boxed{y^3/4}$$




For $(b)$, I try to decompose $f(x, y)$ into the product of two marginals. We have



$$f_{Y}(y) = int_{0}^{y} f(x, y) mathop{dx} = e^{-y}.$$



But, I can't compute



$$f_{X}(x) = int_{x}^{infty} f(x,y) mathop{dy}. $$



Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?






probability probability-theory probability-distributions







share|cite|improve this question






















  • Since you already found in part a) that $Xmid Ysim text{Uniform}(0,Y)$ (you did not mention that $0<x<y$ in the support), this should be enough to conclude on part b).
    – StubbornAtom
    Nov 21 at 19:28















up vote
1
down vote

favorite
1









up vote
1
down vote

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1





Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases} y^{-1}text{exp}left(-yright) & 0 < x < y \[1em]
0, & text{ otherwise.} end{cases} $$



a) Compute $mathbb{E}[X^{3} | Y = y]$



b) Are $X$ and $Y$ independent?




So, my attempt for $(a)$ was to first compute $f_{x mid y}(x mid y)$ as follows:



$$f_{x|y}(x|y) = frac{f(x,y)}{int_{-infty}^{infty}f(x,y) mathop{dx}} = frac{1}{y}.$$



So,



$$mathbb{E}[X^{3} mid Y = y] = int_{0}^{y} x^{3}frac{1}{y} mathop{dx} = boxed{y^3/4}$$




For $(b)$, I try to decompose $f(x, y)$ into the product of two marginals. We have



$$f_{Y}(y) = int_{0}^{y} f(x, y) mathop{dx} = e^{-y}.$$



But, I can't compute



$$f_{X}(x) = int_{x}^{infty} f(x,y) mathop{dy}. $$



Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?






probability probability-theory probability-distributions







share|cite|improve this question













Consider the joint pdf of $(X, Y)$ given by



$$f(x, y) = begin{cases} y^{-1}text{exp}left(-yright) & 0 < x < y \[1em]
0, & text{ otherwise.} end{cases} $$



a) Compute $mathbb{E}[X^{3} | Y = y]$



b) Are $X$ and $Y$ independent?




So, my attempt for $(a)$ was to first compute $f_{x mid y}(x mid y)$ as follows:



$$f_{x|y}(x|y) = frac{f(x,y)}{int_{-infty}^{infty}f(x,y) mathop{dx}} = frac{1}{y}.$$



So,



$$mathbb{E}[X^{3} mid Y = y] = int_{0}^{y} x^{3}frac{1}{y} mathop{dx} = boxed{y^3/4}$$




For $(b)$, I try to decompose $f(x, y)$ into the product of two marginals. We have



$$f_{Y}(y) = int_{0}^{y} f(x, y) mathop{dx} = e^{-y}.$$



But, I can't compute



$$f_{X}(x) = int_{x}^{infty} f(x,y) mathop{dy}. $$



Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?






probability probability-theory probability-distributions




probability probability-theory probability-distributions






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asked Nov 20 at 0:58









joseph

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  • Since you already found in part a) that $Xmid Ysim text{Uniform}(0,Y)$ (you did not mention that $0<x<y$ in the support), this should be enough to conclude on part b).
    – StubbornAtom
    Nov 21 at 19:28




















  • Since you already found in part a) that $Xmid Ysim text{Uniform}(0,Y)$ (you did not mention that $0<x<y$ in the support), this should be enough to conclude on part b).
    – StubbornAtom
    Nov 21 at 19:28


















Since you already found in part a) that $Xmid Ysim text{Uniform}(0,Y)$ (you did not mention that $0<x<y$ in the support), this should be enough to conclude on part b).
– StubbornAtom
Nov 21 at 19:28






Since you already found in part a) that $Xmid Ysim text{Uniform}(0,Y)$ (you did not mention that $0<x<y$ in the support), this should be enough to conclude on part b).
– StubbornAtom
Nov 21 at 19:28












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(a) is quite correct.




For (b) you have that $f_{X,Y}(x,y)=y^{-1}e^{-y}mathbf 1_{0leq xleq y}$ and $f_Y(y)=e^{-y}mathbf 1_{0leq y}$ .   Okay.



Don't worry about having trouble finding the marginal density function for $X$.  
Just ask, can there be any mono-variate function wrt $x$, $g(x)$, such that $y^{-1}e^{-y}mathbf 1_{0leq xleq y}=g(x)~e^{-y}mathbf 1_{0leq y}$ ? Therefore, can $X,Y$ be independent?




Also notice $f_{Xmid Y=2}(3)=0$ while $f_{Xmid Y=4}(3)>0$.







share|cite|improve this answer





















  • I don't understand the whole thing about monovariate functions, but from the spoiler part, I can see that the conclusion is that they're not independent. How'd you get $f_{Xmid Y = 2}(3) = 0$? Shouldn't it be $1/3?$
    – joseph
    Nov 20 at 2:33













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(a) is quite correct.




For (b) you have that $f_{X,Y}(x,y)=y^{-1}e^{-y}mathbf 1_{0leq xleq y}$ and $f_Y(y)=e^{-y}mathbf 1_{0leq y}$ .   Okay.



Don't worry about having trouble finding the marginal density function for $X$.  
Just ask, can there be any mono-variate function wrt $x$, $g(x)$, such that $y^{-1}e^{-y}mathbf 1_{0leq xleq y}=g(x)~e^{-y}mathbf 1_{0leq y}$ ? Therefore, can $X,Y$ be independent?




Also notice $f_{Xmid Y=2}(3)=0$ while $f_{Xmid Y=4}(3)>0$.







share|cite|improve this answer





















  • I don't understand the whole thing about monovariate functions, but from the spoiler part, I can see that the conclusion is that they're not independent. How'd you get $f_{Xmid Y = 2}(3) = 0$? Shouldn't it be $1/3?$
    – joseph
    Nov 20 at 2:33

















up vote
1
down vote













(a) is quite correct.




For (b) you have that $f_{X,Y}(x,y)=y^{-1}e^{-y}mathbf 1_{0leq xleq y}$ and $f_Y(y)=e^{-y}mathbf 1_{0leq y}$ .   Okay.



Don't worry about having trouble finding the marginal density function for $X$.  
Just ask, can there be any mono-variate function wrt $x$, $g(x)$, such that $y^{-1}e^{-y}mathbf 1_{0leq xleq y}=g(x)~e^{-y}mathbf 1_{0leq y}$ ? Therefore, can $X,Y$ be independent?




Also notice $f_{Xmid Y=2}(3)=0$ while $f_{Xmid Y=4}(3)>0$.







share|cite|improve this answer





















  • I don't understand the whole thing about monovariate functions, but from the spoiler part, I can see that the conclusion is that they're not independent. How'd you get $f_{Xmid Y = 2}(3) = 0$? Shouldn't it be $1/3?$
    – joseph
    Nov 20 at 2:33















up vote
1
down vote










up vote
1
down vote









(a) is quite correct.




For (b) you have that $f_{X,Y}(x,y)=y^{-1}e^{-y}mathbf 1_{0leq xleq y}$ and $f_Y(y)=e^{-y}mathbf 1_{0leq y}$ .   Okay.



Don't worry about having trouble finding the marginal density function for $X$.  
Just ask, can there be any mono-variate function wrt $x$, $g(x)$, such that $y^{-1}e^{-y}mathbf 1_{0leq xleq y}=g(x)~e^{-y}mathbf 1_{0leq y}$ ? Therefore, can $X,Y$ be independent?




Also notice $f_{Xmid Y=2}(3)=0$ while $f_{Xmid Y=4}(3)>0$.







share|cite|improve this answer












(a) is quite correct.




For (b) you have that $f_{X,Y}(x,y)=y^{-1}e^{-y}mathbf 1_{0leq xleq y}$ and $f_Y(y)=e^{-y}mathbf 1_{0leq y}$ .   Okay.



Don't worry about having trouble finding the marginal density function for $X$.  
Just ask, can there be any mono-variate function wrt $x$, $g(x)$, such that $y^{-1}e^{-y}mathbf 1_{0leq xleq y}=g(x)~e^{-y}mathbf 1_{0leq y}$ ? Therefore, can $X,Y$ be independent?




Also notice $f_{Xmid Y=2}(3)=0$ while $f_{Xmid Y=4}(3)>0$.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 1:57









Graham Kemp

84.6k43378




84.6k43378












  • I don't understand the whole thing about monovariate functions, but from the spoiler part, I can see that the conclusion is that they're not independent. How'd you get $f_{Xmid Y = 2}(3) = 0$? Shouldn't it be $1/3?$
    – joseph
    Nov 20 at 2:33




















  • I don't understand the whole thing about monovariate functions, but from the spoiler part, I can see that the conclusion is that they're not independent. How'd you get $f_{Xmid Y = 2}(3) = 0$? Shouldn't it be $1/3?$
    – joseph
    Nov 20 at 2:33


















I don't understand the whole thing about monovariate functions, but from the spoiler part, I can see that the conclusion is that they're not independent. How'd you get $f_{Xmid Y = 2}(3) = 0$? Shouldn't it be $1/3?$
– joseph
Nov 20 at 2:33






I don't understand the whole thing about monovariate functions, but from the spoiler part, I can see that the conclusion is that they're not independent. How'd you get $f_{Xmid Y = 2}(3) = 0$? Shouldn't it be $1/3?$
– joseph
Nov 20 at 2:33




















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