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}. $$
Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?
probability probability-theory probability-distributions
asked Nov 20 at 0:58
joseph
3809
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1
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Consider the joint pdf of $(X, Y)$ given by
$$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}. $$
Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?
probability probability-theory probability-distributions
asked Nov 20 at 0:58
joseph
3809
Since you already found in part a) that $Xmid Ysim text{Uniform}(0,Y)$ (you did not mention that $0<x<y$ in the support), this should be enough to conclude on part b).
– StubbornAtom
Nov 21 at 19:28
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the joint pdf of $(X, Y)$ given by
$$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}. $$
Wolfram Alpha can't compute it in a nice way either. Is my answer for $(a)$ correct? How can I do $(b)$?
probability probability-theory probability-distributions
asked Nov 20 at 0:58
joseph
3809
Consider the joint pdf of $(X, Y)$ given by
$$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
asked Nov 20 at 0:58
joseph
3809
asked Nov 20 at 0:58
joseph
3809
asked Nov 20 at 0:58
joseph
3809
asked Nov 20 at 0:58
joseph
3809
asked Nov 20 at 0:58
joseph
3809
3809
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
add a comment |
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
add a comment |
1 Answer
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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 $f_{Xmid Y=4}(3)>0$.
answered Nov 20 at 1:57
Graham Kemp
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
add a comment |
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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 $f_{Xmid Y=4}(3)>0$.
answered Nov 20 at 1:57
Graham Kemp
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
add a comment |
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 $f_{Xmid Y=4}(3)>0$.
answered Nov 20 at 1:57
Graham Kemp
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
add a comment |
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 $f_{Xmid Y=4}(3)>0$.
answered Nov 20 at 1:57
Graham Kemp
84.6k43378
(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 $f_{Xmid Y=4}(3)>0$.
answered Nov 20 at 1:57
Graham Kemp
84.6k43378
answered Nov 20 at 1:57
Graham Kemp
84.6k43378
answered Nov 20 at 1:57
Graham Kemp
84.6k43378
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
add a comment |
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
add a comment |
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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