Mixing
Given $f_{X,Y}$, find out the marginal distributions $f_{X}$ and $f_{Y}$ as well as the expectations...
$begingroup$
The joint probability density function of $(X,Y)$ is given by
begin{align*}
f_{X,Y}(x,y) =
begin{cases}
c(y^{2} - x^{2})e^{-y} & text{if},,,-yleq x leq y,,text{and},,0 < y < +infty\\
0 & text{otherwise}
end{cases}
end{align*}
(a) Determine the value of $c$
(b) Determine the marginal probability density functions $f_{X}$ and $f_{Y}$
(c) Calculate $textbf{E}(X)$ and $textbf{E}(Y)$
MY SOLUTION
(a) According to the definition of joint probability density function, we have
begin{align*}
cint_{0}^{infty}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}xmathrm{d}y & = cint_{0}^{infty}left(2y^{3}e^{-y} - frac{2y^{3}}{3}e^{-y}right)mathrm{d}y\
& = frac{c}{3}int_{0}^{infty}4y^{3}e^{-y}mathrm{d}y = 8c = 1
end{align*}
(b) Once again, it results immediately from the definition that
begin{align*}
f_{X}(x) & = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = frac{1}{8}int_{0}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{2-x^{2}}{8}\\
f_{Y}(y) & = int_{-y}^{y}f_{X,Y}(x,y)mathrm{d}x = frac{1}{8}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}x = frac{y^{3}e^{-y}}{6}
end{align*}
Where the support of $Y$ is given by $S_{Y} = textbf{R}_{geq0}$. Nonetheless, I am still having troubles in calculating the support of $X$. Can someome help me?
EDIT
According to Did's observation, the actual probability density function is given by
begin{align*}
f_{X}(x) = int_{|x|}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{e^{-|x|}(|x| + 1)}{4}
end{align*}
where the support of $X$ is given by $S_{X} = textbf{R}$.
(c) Finally, we have
begin{align*}
textbf{E}(X) & = int_{-infty}^{+infty}xf_{X}(x)mathrm{d}x = frac{1}{4}int_{-infty}^{+infty}xe^{-|x|}(|x|+1)mathrm{d}x = 0\\
textbf{E}(Y) & = int_{0}^{infty}yf_{Y}(y)mathrm{d}Y = frac{1}{6}int_{0}^{infty}y^{4}e^{-y}mathrm{d}y = 4
end{align*}
If there is any additional mistake, do not hesitate to comment.
probability-theory probability-distributions
asked Feb 2 at 19:51
APC89APC89
2,371720
$endgroup$
add a comment |
$begingroup$
The joint probability density function of $(X,Y)$ is given by
begin{align*}
f_{X,Y}(x,y) =
begin{cases}
c(y^{2} - x^{2})e^{-y} & text{if},,,-yleq x leq y,,text{and},,0 < y < +infty\\
0 & text{otherwise}
end{cases}
end{align*}
(a) Determine the value of $c$
(b) Determine the marginal probability density functions $f_{X}$ and $f_{Y}$
(c) Calculate $textbf{E}(X)$ and $textbf{E}(Y)$
MY SOLUTION
(a) According to the definition of joint probability density function, we have
begin{align*}
cint_{0}^{infty}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}xmathrm{d}y & = cint_{0}^{infty}left(2y^{3}e^{-y} - frac{2y^{3}}{3}e^{-y}right)mathrm{d}y\
& = frac{c}{3}int_{0}^{infty}4y^{3}e^{-y}mathrm{d}y = 8c = 1
end{align*}
(b) Once again, it results immediately from the definition that
begin{align*}
f_{X}(x) & = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = frac{1}{8}int_{0}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{2-x^{2}}{8}\\
f_{Y}(y) & = int_{-y}^{y}f_{X,Y}(x,y)mathrm{d}x = frac{1}{8}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}x = frac{y^{3}e^{-y}}{6}
end{align*}
Where the support of $Y$ is given by $S_{Y} = textbf{R}_{geq0}$. Nonetheless, I am still having troubles in calculating the support of $X$. Can someome help me?
EDIT
According to Did's observation, the actual probability density function is given by
begin{align*}
f_{X}(x) = int_{|x|}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{e^{-|x|}(|x| + 1)}{4}
end{align*}
where the support of $X$ is given by $S_{X} = textbf{R}$.
(c) Finally, we have
begin{align*}
textbf{E}(X) & = int_{-infty}^{+infty}xf_{X}(x)mathrm{d}x = frac{1}{4}int_{-infty}^{+infty}xe^{-|x|}(|x|+1)mathrm{d}x = 0\\
textbf{E}(Y) & = int_{0}^{infty}yf_{Y}(y)mathrm{d}Y = frac{1}{6}int_{0}^{infty}y^{4}e^{-y}mathrm{d}y = 4
end{align*}
If there is any additional mistake, do not hesitate to comment.
probability-theory probability-distributions
asked Feb 2 at 19:51
APC89APC89
2,371720
$endgroup$
2
$begingroup$
Actually, $$f_X(x) = int_0^infty f_{X,Y}(x,y)mathrm dy = frac18int_{|x|}^infty(y^2e^{-y} - x^2e^{-y})mathrm dy = cdots$$
$endgroup$
– Did
Feb 2 at 20:53
$begingroup$
Thanks for the comment, Did. I have already edited the answer.
$endgroup$
– APC89
Feb 2 at 21:16
$begingroup$
@Did Why is the lower limit of the integral $|x|$?
$endgroup$
– David
Apr 9 at 4:22
add a comment |
$begingroup$
The joint probability density function of $(X,Y)$ is given by
begin{align*}
f_{X,Y}(x,y) =
begin{cases}
c(y^{2} - x^{2})e^{-y} & text{if},,,-yleq x leq y,,text{and},,0 < y < +infty\\
0 & text{otherwise}
end{cases}
end{align*}
(a) Determine the value of $c$
(b) Determine the marginal probability density functions $f_{X}$ and $f_{Y}$
(c) Calculate $textbf{E}(X)$ and $textbf{E}(Y)$
MY SOLUTION
(a) According to the definition of joint probability density function, we have
begin{align*}
cint_{0}^{infty}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}xmathrm{d}y & = cint_{0}^{infty}left(2y^{3}e^{-y} - frac{2y^{3}}{3}e^{-y}right)mathrm{d}y\
& = frac{c}{3}int_{0}^{infty}4y^{3}e^{-y}mathrm{d}y = 8c = 1
end{align*}
(b) Once again, it results immediately from the definition that
begin{align*}
f_{X}(x) & = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = frac{1}{8}int_{0}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{2-x^{2}}{8}\\
f_{Y}(y) & = int_{-y}^{y}f_{X,Y}(x,y)mathrm{d}x = frac{1}{8}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}x = frac{y^{3}e^{-y}}{6}
end{align*}
Where the support of $Y$ is given by $S_{Y} = textbf{R}_{geq0}$. Nonetheless, I am still having troubles in calculating the support of $X$. Can someome help me?
EDIT
According to Did's observation, the actual probability density function is given by
begin{align*}
f_{X}(x) = int_{|x|}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{e^{-|x|}(|x| + 1)}{4}
end{align*}
where the support of $X$ is given by $S_{X} = textbf{R}$.
(c) Finally, we have
begin{align*}
textbf{E}(X) & = int_{-infty}^{+infty}xf_{X}(x)mathrm{d}x = frac{1}{4}int_{-infty}^{+infty}xe^{-|x|}(|x|+1)mathrm{d}x = 0\\
textbf{E}(Y) & = int_{0}^{infty}yf_{Y}(y)mathrm{d}Y = frac{1}{6}int_{0}^{infty}y^{4}e^{-y}mathrm{d}y = 4
end{align*}
If there is any additional mistake, do not hesitate to comment.
probability-theory probability-distributions
asked Feb 2 at 19:51
APC89APC89
2,371720
$endgroup$
The joint probability density function of $(X,Y)$ is given by
begin{align*}
f_{X,Y}(x,y) =
begin{cases}
c(y^{2} - x^{2})e^{-y} & text{if},,,-yleq x leq y,,text{and},,0 < y < +infty\\
0 & text{otherwise}
end{cases}
end{align*}
(a) Determine the value of $c$
(b) Determine the marginal probability density functions $f_{X}$ and $f_{Y}$
(c) Calculate $textbf{E}(X)$ and $textbf{E}(Y)$
MY SOLUTION
(a) According to the definition of joint probability density function, we have
begin{align*}
cint_{0}^{infty}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}xmathrm{d}y & = cint_{0}^{infty}left(2y^{3}e^{-y} - frac{2y^{3}}{3}e^{-y}right)mathrm{d}y\
& = frac{c}{3}int_{0}^{infty}4y^{3}e^{-y}mathrm{d}y = 8c = 1
end{align*}
(b) Once again, it results immediately from the definition that
begin{align*}
f_{X}(x) & = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = frac{1}{8}int_{0}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{2-x^{2}}{8}\\
f_{Y}(y) & = int_{-y}^{y}f_{X,Y}(x,y)mathrm{d}x = frac{1}{8}int_{-y}^{y}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}x = frac{y^{3}e^{-y}}{6}
end{align*}
Where the support of $Y$ is given by $S_{Y} = textbf{R}_{geq0}$. Nonetheless, I am still having troubles in calculating the support of $X$. Can someome help me?
EDIT
According to Did's observation, the actual probability density function is given by
begin{align*}
f_{X}(x) = int_{|x|}^{infty}(y^{2}e^{-y} - x^{2}e^{-y})mathrm{d}y = frac{e^{-|x|}(|x| + 1)}{4}
end{align*}
where the support of $X$ is given by $S_{X} = textbf{R}$.
(c) Finally, we have
begin{align*}
textbf{E}(X) & = int_{-infty}^{+infty}xf_{X}(x)mathrm{d}x = frac{1}{4}int_{-infty}^{+infty}xe^{-|x|}(|x|+1)mathrm{d}x = 0\\
textbf{E}(Y) & = int_{0}^{infty}yf_{Y}(y)mathrm{d}Y = frac{1}{6}int_{0}^{infty}y^{4}e^{-y}mathrm{d}y = 4
end{align*}
If there is any additional mistake, do not hesitate to comment.
probability-theory probability-distributions
probability-theory probability-distributions
asked Feb 2 at 19:51
APC89APC89
2,371720
asked Feb 2 at 19:51
APC89APC89
2,371720
edited Feb 2 at 21:14
APC89
asked Feb 2 at 19:51
APC89APC89
2,371720
asked Feb 2 at 19:51
APC89APC89
2,371720
asked Feb 2 at 19:51
APC89APC89
2,371720
2,371720
2
$begingroup$
Actually, $$f_X(x) = int_0^infty f_{X,Y}(x,y)mathrm dy = frac18int_{|x|}^infty(y^2e^{-y} - x^2e^{-y})mathrm dy = cdots$$
$endgroup$
– Did
Feb 2 at 20:53
$begingroup$
Thanks for the comment, Did. I have already edited the answer.
$endgroup$
– APC89
Feb 2 at 21:16
$begingroup$
@Did Why is the lower limit of the integral $|x|$?
$endgroup$
– David
Apr 9 at 4:22
add a comment |
2
$begingroup$
Actually, $$f_X(x) = int_0^infty f_{X,Y}(x,y)mathrm dy = frac18int_{|x|}^infty(y^2e^{-y} - x^2e^{-y})mathrm dy = cdots$$
$endgroup$
– Did
Feb 2 at 20:53
$begingroup$
Thanks for the comment, Did. I have already edited the answer.
$endgroup$
– APC89
Feb 2 at 21:16
$begingroup$
@Did Why is the lower limit of the integral $|x|$?
$endgroup$
– David
Apr 9 at 4:22
2
2
$begingroup$
Actually, $$f_X(x) = int_0^infty f_{X,Y}(x,y)mathrm dy = frac18int_{|x|}^infty(y^2e^{-y} - x^2e^{-y})mathrm dy = cdots$$
$endgroup$
– Did
Feb 2 at 20:53
$begingroup$
Actually, $$f_X(x) = int_0^infty f_{X,Y}(x,y)mathrm dy = frac18int_{|x|}^infty(y^2e^{-y} - x^2e^{-y})mathrm dy = cdots$$
$endgroup$
– Did
Feb 2 at 20:53
$begingroup$
Thanks for the comment, Did. I have already edited the answer.
$endgroup$
– APC89
Feb 2 at 21:16
$begingroup$
Thanks for the comment, Did. I have already edited the answer.
$endgroup$
– APC89
Feb 2 at 21:16
$begingroup$
@Did Why is the lower limit of the integral $|x|$?
$endgroup$
– David
Apr 9 at 4:22
$begingroup$
@Did Why is the lower limit of the integral $|x|$?
$endgroup$
– David
Apr 9 at 4:22
add a comment |
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$begingroup$
Actually, $$f_X(x) = int_0^infty f_{X,Y}(x,y)mathrm dy = frac18int_{|x|}^infty(y^2e^{-y} - x^2e^{-y})mathrm dy = cdots$$
$endgroup$
– Did
Feb 2 at 20:53
$begingroup$
Thanks for the comment, Did. I have already edited the answer.
$endgroup$
– APC89
Feb 2 at 21:16
$begingroup$
@Did Why is the lower limit of the integral $|x|$?
$endgroup$
– David
Apr 9 at 4:22