Mixing
Given the joint density function of $X$ and $Y$, find the probability density function of $Z = XY$
$begingroup$
The joint density function of $X$ and $Y$ is given by $f(x,y) = xe^{-x(y+1)}$, $x > 0, y > 0$.
(a) Find the conditional density of $X$, given $Y = y$, and that
of $Y$, given $X = x$.
(b) Find the density function of $Z = XY$.
MY SOLUTION
In the first place, we should determine the marginal distributions:
begin{align*}
f_{X}(x) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}y = xe^{-x}
end{align*}
where $x > 0$. Analogously, we have
begin{align*}
f_{Y}(y) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}x = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}x = frac{1}{(y+1)^{2}}
end{align*}
where $y > 0 $.
(a) Based on the previous results, it comes
begin{cases}
f_{X|Y}(x|y) = displaystylefrac{f_{X,Y}(x,y)}{f_{Y}(y)} = x(y+1)^{2}e^{-x(y+1)}\\
f_{Y|X}(y|x) = displaystylefrac{f_{X,Y}(x,y)}{f_{X}(x)} = e^{-xy}
end{cases}
(b) Finally, we have
begin{align*}
F_{Z}(z) & = textbf{P}(XY leq z) = int_{0}^{infty}int_{0}^{z/x}f_{X,Y}(x,y)mathrm{d}ymathrm{d}x = int_{0}^{infty}int_{0}^{z/x}xe^{-x(y+1)}mathrm{d}ymathrm{d}x\\
& = int_{0}^{infty}(e^{-x} - e^{-x-z})mathrm{d}x = 1 - e^{-z} Rightarrow f_{Z}(z) = frac{mathrm{d}}{mathrm{d}z}(1-e^{-z}) = e^{-z}
end{align*}
where $z > 0$.
probability probability-theory proof-verification probability-distributions marginal-distribution
asked Feb 3 at 2:33
APC89APC89
2,371720
$endgroup$
add a comment |
$begingroup$
The joint density function of $X$ and $Y$ is given by $f(x,y) = xe^{-x(y+1)}$, $x > 0, y > 0$.
(a) Find the conditional density of $X$, given $Y = y$, and that
of $Y$, given $X = x$.
(b) Find the density function of $Z = XY$.
MY SOLUTION
In the first place, we should determine the marginal distributions:
begin{align*}
f_{X}(x) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}y = xe^{-x}
end{align*}
where $x > 0$. Analogously, we have
begin{align*}
f_{Y}(y) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}x = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}x = frac{1}{(y+1)^{2}}
end{align*}
where $y > 0 $.
(a) Based on the previous results, it comes
begin{cases}
f_{X|Y}(x|y) = displaystylefrac{f_{X,Y}(x,y)}{f_{Y}(y)} = x(y+1)^{2}e^{-x(y+1)}\\
f_{Y|X}(y|x) = displaystylefrac{f_{X,Y}(x,y)}{f_{X}(x)} = e^{-xy}
end{cases}
(b) Finally, we have
begin{align*}
F_{Z}(z) & = textbf{P}(XY leq z) = int_{0}^{infty}int_{0}^{z/x}f_{X,Y}(x,y)mathrm{d}ymathrm{d}x = int_{0}^{infty}int_{0}^{z/x}xe^{-x(y+1)}mathrm{d}ymathrm{d}x\\
& = int_{0}^{infty}(e^{-x} - e^{-x-z})mathrm{d}x = 1 - e^{-z} Rightarrow f_{Z}(z) = frac{mathrm{d}}{mathrm{d}z}(1-e^{-z}) = e^{-z}
end{align*}
where $z > 0$.
probability probability-theory proof-verification probability-distributions marginal-distribution
asked Feb 3 at 2:33
APC89APC89
2,371720
$endgroup$
add a comment |
$begingroup$
The joint density function of $X$ and $Y$ is given by $f(x,y) = xe^{-x(y+1)}$, $x > 0, y > 0$.
(a) Find the conditional density of $X$, given $Y = y$, and that
of $Y$, given $X = x$.
(b) Find the density function of $Z = XY$.
MY SOLUTION
In the first place, we should determine the marginal distributions:
begin{align*}
f_{X}(x) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}y = xe^{-x}
end{align*}
where $x > 0$. Analogously, we have
begin{align*}
f_{Y}(y) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}x = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}x = frac{1}{(y+1)^{2}}
end{align*}
where $y > 0 $.
(a) Based on the previous results, it comes
begin{cases}
f_{X|Y}(x|y) = displaystylefrac{f_{X,Y}(x,y)}{f_{Y}(y)} = x(y+1)^{2}e^{-x(y+1)}\\
f_{Y|X}(y|x) = displaystylefrac{f_{X,Y}(x,y)}{f_{X}(x)} = e^{-xy}
end{cases}
(b) Finally, we have
begin{align*}
F_{Z}(z) & = textbf{P}(XY leq z) = int_{0}^{infty}int_{0}^{z/x}f_{X,Y}(x,y)mathrm{d}ymathrm{d}x = int_{0}^{infty}int_{0}^{z/x}xe^{-x(y+1)}mathrm{d}ymathrm{d}x\\
& = int_{0}^{infty}(e^{-x} - e^{-x-z})mathrm{d}x = 1 - e^{-z} Rightarrow f_{Z}(z) = frac{mathrm{d}}{mathrm{d}z}(1-e^{-z}) = e^{-z}
end{align*}
where $z > 0$.
probability probability-theory proof-verification probability-distributions marginal-distribution
asked Feb 3 at 2:33
APC89APC89
2,371720
$endgroup$
The joint density function of $X$ and $Y$ is given by $f(x,y) = xe^{-x(y+1)}$, $x > 0, y > 0$.
(a) Find the conditional density of $X$, given $Y = y$, and that
of $Y$, given $X = x$.
(b) Find the density function of $Z = XY$.
MY SOLUTION
In the first place, we should determine the marginal distributions:
begin{align*}
f_{X}(x) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}y = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}y = xe^{-x}
end{align*}
where $x > 0$. Analogously, we have
begin{align*}
f_{Y}(y) = int_{0}^{infty}f_{X,Y}(x,y)mathrm{d}x = int_{0}^{infty}xe^{-x(y+1)}mathrm{d}x = frac{1}{(y+1)^{2}}
end{align*}
where $y > 0 $.
(a) Based on the previous results, it comes
begin{cases}
f_{X|Y}(x|y) = displaystylefrac{f_{X,Y}(x,y)}{f_{Y}(y)} = x(y+1)^{2}e^{-x(y+1)}\\
f_{Y|X}(y|x) = displaystylefrac{f_{X,Y}(x,y)}{f_{X}(x)} = e^{-xy}
end{cases}
(b) Finally, we have
begin{align*}
F_{Z}(z) & = textbf{P}(XY leq z) = int_{0}^{infty}int_{0}^{z/x}f_{X,Y}(x,y)mathrm{d}ymathrm{d}x = int_{0}^{infty}int_{0}^{z/x}xe^{-x(y+1)}mathrm{d}ymathrm{d}x\\
& = int_{0}^{infty}(e^{-x} - e^{-x-z})mathrm{d}x = 1 - e^{-z} Rightarrow f_{Z}(z) = frac{mathrm{d}}{mathrm{d}z}(1-e^{-z}) = e^{-z}
end{align*}
where $z > 0$.
probability probability-theory proof-verification probability-distributions marginal-distribution
probability probability-theory proof-verification probability-distributions marginal-distribution
asked Feb 3 at 2:33
APC89APC89
2,371720
asked Feb 3 at 2:33
APC89APC89
2,371720
edited Feb 3 at 16:32
APC89
asked Feb 3 at 2:33
APC89APC89
2,371720
asked Feb 3 at 2:33
APC89APC89
2,371720
asked Feb 3 at 2:33
APC89APC89
2,371720
2,371720
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
The joint density $f(x,y)$ factors in the following ways:
$$f(x,y)=xe^{-xy}mathbf1_{y>0},e^{-x}mathbf1_{x>0}tag{1}$$
$$f(x,y)=(1+y)^2xe^{-(1+y)x}mathbf1_{x>0},frac{1}{(1+y)^2}mathbf1_{y>0}tag{2}$$
From $(1)$ it follows that the conditional density of $Ymid X$ is
$$f_{Ymid X=x}(y)=xe^{-xy}mathbf1_{y>0}$$
, and the marginal density of $X$ is $$f_X(x)=e^{-x}mathbf1_{x>0}$$
From $(2)$, your answer for the conditional and marginal densities is correct.
So your answer for $f_X(x)$, though seemingly okay as it integrates to unity, is not correct. As a result, $f_{Ymid X=x}(y)$ (the variable is $y$ here because it is a function of $y$ for a given $x$, not $(x,y)$ as you have written) is also out of line. The rest looks okay.
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
$endgroup$
$begingroup$
In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$.
$endgroup$
– APC89
Feb 3 at 16:34
$begingroup$
That is the notation used in books, to emphasize the "given $x$" part.
$endgroup$
– StubbornAtom
Feb 3 at 16:36
2
$begingroup$
(+1) A minor quibble: the notation $f_{Ymid X}(ymid x)$ is common, one could even argue it is preferable to $f_{Ymid X=x}(y)$.
$endgroup$
– Did
Feb 3 at 16:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The joint density $f(x,y)$ factors in the following ways:
$$f(x,y)=xe^{-xy}mathbf1_{y>0},e^{-x}mathbf1_{x>0}tag{1}$$
$$f(x,y)=(1+y)^2xe^{-(1+y)x}mathbf1_{x>0},frac{1}{(1+y)^2}mathbf1_{y>0}tag{2}$$
From $(1)$ it follows that the conditional density of $Ymid X$ is
$$f_{Ymid X=x}(y)=xe^{-xy}mathbf1_{y>0}$$
, and the marginal density of $X$ is $$f_X(x)=e^{-x}mathbf1_{x>0}$$
From $(2)$, your answer for the conditional and marginal densities is correct.
So your answer for $f_X(x)$, though seemingly okay as it integrates to unity, is not correct. As a result, $f_{Ymid X=x}(y)$ (the variable is $y$ here because it is a function of $y$ for a given $x$, not $(x,y)$ as you have written) is also out of line. The rest looks okay.
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
$endgroup$
$begingroup$
In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$.
$endgroup$
– APC89
Feb 3 at 16:34
$begingroup$
That is the notation used in books, to emphasize the "given $x$" part.
$endgroup$
– StubbornAtom
Feb 3 at 16:36
2
$begingroup$
(+1) A minor quibble: the notation $f_{Ymid X}(ymid x)$ is common, one could even argue it is preferable to $f_{Ymid X=x}(y)$.
$endgroup$
– Did
Feb 3 at 16:36
add a comment |
$begingroup$
The joint density $f(x,y)$ factors in the following ways:
$$f(x,y)=xe^{-xy}mathbf1_{y>0},e^{-x}mathbf1_{x>0}tag{1}$$
$$f(x,y)=(1+y)^2xe^{-(1+y)x}mathbf1_{x>0},frac{1}{(1+y)^2}mathbf1_{y>0}tag{2}$$
From $(1)$ it follows that the conditional density of $Ymid X$ is
$$f_{Ymid X=x}(y)=xe^{-xy}mathbf1_{y>0}$$
, and the marginal density of $X$ is $$f_X(x)=e^{-x}mathbf1_{x>0}$$
From $(2)$, your answer for the conditional and marginal densities is correct.
So your answer for $f_X(x)$, though seemingly okay as it integrates to unity, is not correct. As a result, $f_{Ymid X=x}(y)$ (the variable is $y$ here because it is a function of $y$ for a given $x$, not $(x,y)$ as you have written) is also out of line. The rest looks okay.
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
$endgroup$
$begingroup$
In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$.
$endgroup$
– APC89
Feb 3 at 16:34
$begingroup$
That is the notation used in books, to emphasize the "given $x$" part.
$endgroup$
– StubbornAtom
Feb 3 at 16:36
2
$begingroup$
(+1) A minor quibble: the notation $f_{Ymid X}(ymid x)$ is common, one could even argue it is preferable to $f_{Ymid X=x}(y)$.
$endgroup$
– Did
Feb 3 at 16:36
add a comment |
$begingroup$
The joint density $f(x,y)$ factors in the following ways:
$$f(x,y)=xe^{-xy}mathbf1_{y>0},e^{-x}mathbf1_{x>0}tag{1}$$
$$f(x,y)=(1+y)^2xe^{-(1+y)x}mathbf1_{x>0},frac{1}{(1+y)^2}mathbf1_{y>0}tag{2}$$
From $(1)$ it follows that the conditional density of $Ymid X$ is
$$f_{Ymid X=x}(y)=xe^{-xy}mathbf1_{y>0}$$
, and the marginal density of $X$ is $$f_X(x)=e^{-x}mathbf1_{x>0}$$
From $(2)$, your answer for the conditional and marginal densities is correct.
So your answer for $f_X(x)$, though seemingly okay as it integrates to unity, is not correct. As a result, $f_{Ymid X=x}(y)$ (the variable is $y$ here because it is a function of $y$ for a given $x$, not $(x,y)$ as you have written) is also out of line. The rest looks okay.
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
$endgroup$
The joint density $f(x,y)$ factors in the following ways:
$$f(x,y)=xe^{-xy}mathbf1_{y>0},e^{-x}mathbf1_{x>0}tag{1}$$
$$f(x,y)=(1+y)^2xe^{-(1+y)x}mathbf1_{x>0},frac{1}{(1+y)^2}mathbf1_{y>0}tag{2}$$
From $(1)$ it follows that the conditional density of $Ymid X$ is
$$f_{Ymid X=x}(y)=xe^{-xy}mathbf1_{y>0}$$
, and the marginal density of $X$ is $$f_X(x)=e^{-x}mathbf1_{x>0}$$
From $(2)$, your answer for the conditional and marginal densities is correct.
So your answer for $f_X(x)$, though seemingly okay as it integrates to unity, is not correct. As a result, $f_{Ymid X=x}(y)$ (the variable is $y$ here because it is a function of $y$ for a given $x$, not $(x,y)$ as you have written) is also out of line. The rest looks okay.
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
answered Feb 3 at 6:52
StubbornAtomStubbornAtom
6,42731440
6,42731440
$begingroup$
In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$.
$endgroup$
– APC89
Feb 3 at 16:34
$begingroup$
That is the notation used in books, to emphasize the "given $x$" part.
$endgroup$
– StubbornAtom
Feb 3 at 16:36
2
$begingroup$
(+1) A minor quibble: the notation $f_{Ymid X}(ymid x)$ is common, one could even argue it is preferable to $f_{Ymid X=x}(y)$.
$endgroup$
– Did
Feb 3 at 16:36
add a comment |
$begingroup$
In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$.
$endgroup$
– APC89
Feb 3 at 16:34
$begingroup$
That is the notation used in books, to emphasize the "given $x$" part.
$endgroup$
– StubbornAtom
Feb 3 at 16:36
2
$begingroup$
(+1) A minor quibble: the notation $f_{Ymid X}(ymid x)$ is common, one could even argue it is preferable to $f_{Ymid X=x}(y)$.
$endgroup$
– Did
Feb 3 at 16:36
$begingroup$
In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$.
$endgroup$
– APC89
Feb 3 at 16:34
$begingroup$
In the first place, thank you very much for the contribution. As to the notation of conditional probability, I think the most common way to denote it is $f_{Y|X}(y|x)$.
$endgroup$
– APC89
Feb 3 at 16:34
$begingroup$
That is the notation used in books, to emphasize the "given $x$" part.
$endgroup$
– StubbornAtom
Feb 3 at 16:36
$begingroup$
That is the notation used in books, to emphasize the "given $x$" part.
$endgroup$
– StubbornAtom
Feb 3 at 16:36
2
2
$begingroup$
(+1) A minor quibble: the notation $f_{Ymid X}(ymid x)$ is common, one could even argue it is preferable to $f_{Ymid X=x}(y)$.
$endgroup$
– Did
Feb 3 at 16:36
$begingroup$
(+1) A minor quibble: the notation $f_{Ymid X}(ymid x)$ is common, one could even argue it is preferable to $f_{Ymid X=x}(y)$.
$endgroup$
– Did
Feb 3 at 16:36
add a comment |
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