If $f$ is the pdf of a random variable, show that $g(x, y) = f(x+y)/(x + y)$ is a density function in the...
$begingroup$
Let $f$ be the pdf of a positive random variable and write
$$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$
Show that $g$ is the density function in the plane.
Clearly, $g(x, y) geq 0$, but I need help evaluating the integral
$$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$
One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.
Does anyone have any suggestions?
probability probability-theory probability-distributions
asked Dec 9 '18 at 7:27
josephjoseph
500111
$endgroup$
add a comment |
$begingroup$
Let $f$ be the pdf of a positive random variable and write
$$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$
Show that $g$ is the density function in the plane.
Clearly, $g(x, y) geq 0$, but I need help evaluating the integral
$$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$
One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.
Does anyone have any suggestions?
probability probability-theory probability-distributions
asked Dec 9 '18 at 7:27
josephjoseph
500111
$endgroup$
add a comment |
$begingroup$
Let $f$ be the pdf of a positive random variable and write
$$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$
Show that $g$ is the density function in the plane.
Clearly, $g(x, y) geq 0$, but I need help evaluating the integral
$$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$
One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.
Does anyone have any suggestions?
probability probability-theory probability-distributions
asked Dec 9 '18 at 7:27
josephjoseph
500111
$endgroup$
Let $f$ be the pdf of a positive random variable and write
$$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$
Show that $g$ is the density function in the plane.
Clearly, $g(x, y) geq 0$, but I need help evaluating the integral
$$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$
One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.
Does anyone have any suggestions?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Dec 9 '18 at 7:27
josephjoseph
500111
asked Dec 9 '18 at 7:27
josephjoseph
500111
asked Dec 9 '18 at 7:27
josephjoseph
500111
asked Dec 9 '18 at 7:27
josephjoseph
500111
asked Dec 9 '18 at 7:27
josephjoseph
500111
500111
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
$$dv = dy,$$ and hence
$$dudv = dxdy.
$$ Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
$$
int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
$$
$textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
$$left( {begin{array}{c}
du \
dv\
end{array} } right) =
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right)left( {begin{array}{c}
dx \
dy \
end{array} } right),
$$ and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.
answered Dec 9 '18 at 8:13
SongSong
12.6k631
$endgroup$
$begingroup$
don't we need to do a jacobian ?
$endgroup$
– joseph
Dec 9 '18 at 19:07
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
$$dv = dy,$$ and hence
$$dudv = dxdy.
$$ Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
$$
int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
$$
$textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
$$left( {begin{array}{c}
du \
dv\
end{array} } right) =
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right)left( {begin{array}{c}
dx \
dy \
end{array} } right),
$$ and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.
answered Dec 9 '18 at 8:13
SongSong
12.6k631
$endgroup$
$begingroup$
don't we need to do a jacobian ?
$endgroup$
– joseph
Dec 9 '18 at 19:07
add a comment |
$begingroup$
You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
$$dv = dy,$$ and hence
$$dudv = dxdy.
$$ Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
$$
int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
$$
$textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
$$left( {begin{array}{c}
du \
dv\
end{array} } right) =
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right)left( {begin{array}{c}
dx \
dy \
end{array} } right),
$$ and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.
answered Dec 9 '18 at 8:13
SongSong
12.6k631
$endgroup$
$begingroup$
don't we need to do a jacobian ?
$endgroup$
– joseph
Dec 9 '18 at 19:07
add a comment |
$begingroup$
You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
$$dv = dy,$$ and hence
$$dudv = dxdy.
$$ Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
$$
int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
$$
$textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
$$left( {begin{array}{c}
du \
dv\
end{array} } right) =
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right)left( {begin{array}{c}
dx \
dy \
end{array} } right),
$$ and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.
answered Dec 9 '18 at 8:13
SongSong
12.6k631
$endgroup$
You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
$$dv = dy,$$ and hence
$$dudv = dxdy.
$$ Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
$$
int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
$$
$textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
$$left( {begin{array}{c}
du \
dv\
end{array} } right) =
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right)left( {begin{array}{c}
dx \
dy \
end{array} } right),
$$ and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.
answered Dec 9 '18 at 8:13
SongSong
12.6k631
edited Dec 9 '18 at 23:02
answered Dec 9 '18 at 8:13
SongSong
12.6k631
answered Dec 9 '18 at 8:13
SongSong
12.6k631
answered Dec 9 '18 at 8:13
SongSong
12.6k631
12.6k631
$begingroup$
don't we need to do a jacobian ?
$endgroup$
– joseph
Dec 9 '18 at 19:07
add a comment |
$begingroup$
don't we need to do a jacobian ?
$endgroup$
– joseph
Dec 9 '18 at 19:07
$begingroup$
don't we need to do a jacobian ?
$endgroup$
– joseph
Dec 9 '18 at 19:07
$begingroup$
don't we need to do a jacobian ?
$endgroup$
– joseph
Dec 9 '18 at 19:07
add a comment |
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