Finding pdf of function of independent random variables
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Suppose $X$ and $Y$ are i.i.d with a common pdf
$$f(t) = begin{cases} text{exp}(-t) & text{ if } t > 0 \ 0, &
> text{ otherwise}. end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
I think to solve this problem, the approach will be to compute the joint density of $X$ and $Y$. Then, make a transformation using a Jacobian and compute the joint density of $X + Y $ and $X/Y$, and use those to find the marginal densities of $X + Y$ and $X/Y$, and see if their products equals the joint density.
I believe the joint densities of $X$ and $Y$ are given by
$$text{exp}(-x) cdot text{exp}(-y) = text{exp}(-(x + y)) text{ if } t > 0$$
and $0$ otherwise.
Define $h(X, Y) = X + Y$ and define $g(X, Y) = X/Y$. Then, the Jacobian is given by
$$J(x, y)= frac{partial h}{partial x} frac{partial g}{partial y} - frac{partial h}{partial y}frac{partial g}{partial x} = left(1right)left(-frac{X}{Y^{2}}right) - left(1right)left(frac{1}{Y}right) = -frac{X + Y}{Y^{2}}.$$
The joint density of $H$ and $G$ is given by
$$f_{HG}(h, g) = f_{XY}(x, y) cdot |J(x, y)|^{-1}$$
$$= text{exp}left(-(x + y)right) cdot frac{Y^2}{X + Y}.$$
I don't understand what I'm doing wrong though, because I think that it should be in terms of $h$ and $g$.
probability multivariable-calculus
asked Nov 15 at 0:05
joseph
958
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Suppose $X$ and $Y$ are i.i.d with a common pdf
$$f(t) = begin{cases} text{exp}(-t) & text{ if } t > 0 \ 0, &
> text{ otherwise}. end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
I think to solve this problem, the approach will be to compute the joint density of $X$ and $Y$. Then, make a transformation using a Jacobian and compute the joint density of $X + Y $ and $X/Y$, and use those to find the marginal densities of $X + Y$ and $X/Y$, and see if their products equals the joint density.
I believe the joint densities of $X$ and $Y$ are given by
$$text{exp}(-x) cdot text{exp}(-y) = text{exp}(-(x + y)) text{ if } t > 0$$
and $0$ otherwise.
Define $h(X, Y) = X + Y$ and define $g(X, Y) = X/Y$. Then, the Jacobian is given by
$$J(x, y)= frac{partial h}{partial x} frac{partial g}{partial y} - frac{partial h}{partial y}frac{partial g}{partial x} = left(1right)left(-frac{X}{Y^{2}}right) - left(1right)left(frac{1}{Y}right) = -frac{X + Y}{Y^{2}}.$$
The joint density of $H$ and $G$ is given by
$$f_{HG}(h, g) = f_{XY}(x, y) cdot |J(x, y)|^{-1}$$
$$= text{exp}left(-(x + y)right) cdot frac{Y^2}{X + Y}.$$
I don't understand what I'm doing wrong though, because I think that it should be in terms of $h$ and $g$.
probability multivariable-calculus
asked Nov 15 at 0:05
joseph
958
2
Express $x=x(h,g)$ and $y=y(h,g)$ in terms of $h$ and $g$ and substitute.
– NCh
Nov 15 at 0:26
An alternative approach is to show that, given $X+Y=h$, you have $X$ uniformly distributed on $[0,h]$ and so $G=frac{X}{Y}=frac{X}{h-X}$ has a cumulative distribution function of $Pleft(frac{X}{Y} le gright)= frac{g}{1+g}$ and a density of $frac{1}{(1+g)^2}$ on the positive reals, neither of which are affected by the particular value of $h$
– Henry
Nov 15 at 1:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $X$ and $Y$ are i.i.d with a common pdf
$$f(t) = begin{cases} text{exp}(-t) & text{ if } t > 0 \ 0, &
> text{ otherwise}. end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
I think to solve this problem, the approach will be to compute the joint density of $X$ and $Y$. Then, make a transformation using a Jacobian and compute the joint density of $X + Y $ and $X/Y$, and use those to find the marginal densities of $X + Y$ and $X/Y$, and see if their products equals the joint density.
I believe the joint densities of $X$ and $Y$ are given by
$$text{exp}(-x) cdot text{exp}(-y) = text{exp}(-(x + y)) text{ if } t > 0$$
and $0$ otherwise.
Define $h(X, Y) = X + Y$ and define $g(X, Y) = X/Y$. Then, the Jacobian is given by
$$J(x, y)= frac{partial h}{partial x} frac{partial g}{partial y} - frac{partial h}{partial y}frac{partial g}{partial x} = left(1right)left(-frac{X}{Y^{2}}right) - left(1right)left(frac{1}{Y}right) = -frac{X + Y}{Y^{2}}.$$
The joint density of $H$ and $G$ is given by
$$f_{HG}(h, g) = f_{XY}(x, y) cdot |J(x, y)|^{-1}$$
$$= text{exp}left(-(x + y)right) cdot frac{Y^2}{X + Y}.$$
I don't understand what I'm doing wrong though, because I think that it should be in terms of $h$ and $g$.
probability multivariable-calculus
asked Nov 15 at 0:05
joseph
958
Suppose $X$ and $Y$ are i.i.d with a common pdf
$$f(t) = begin{cases} text{exp}(-t) & text{ if } t > 0 \ 0, &
> text{ otherwise}. end{cases} $$
Show that $X + Y$ and $X/Y$ are independent.
I think to solve this problem, the approach will be to compute the joint density of $X$ and $Y$. Then, make a transformation using a Jacobian and compute the joint density of $X + Y $ and $X/Y$, and use those to find the marginal densities of $X + Y$ and $X/Y$, and see if their products equals the joint density.
I believe the joint densities of $X$ and $Y$ are given by
$$text{exp}(-x) cdot text{exp}(-y) = text{exp}(-(x + y)) text{ if } t > 0$$
and $0$ otherwise.
Define $h(X, Y) = X + Y$ and define $g(X, Y) = X/Y$. Then, the Jacobian is given by
$$J(x, y)= frac{partial h}{partial x} frac{partial g}{partial y} - frac{partial h}{partial y}frac{partial g}{partial x} = left(1right)left(-frac{X}{Y^{2}}right) - left(1right)left(frac{1}{Y}right) = -frac{X + Y}{Y^{2}}.$$
The joint density of $H$ and $G$ is given by
$$f_{HG}(h, g) = f_{XY}(x, y) cdot |J(x, y)|^{-1}$$
$$= text{exp}left(-(x + y)right) cdot frac{Y^2}{X + Y}.$$
I don't understand what I'm doing wrong though, because I think that it should be in terms of $h$ and $g$.
probability multivariable-calculus
probability multivariable-calculus
asked Nov 15 at 0:05
joseph
958
asked Nov 15 at 0:05
joseph
958
asked Nov 15 at 0:05
joseph
958
asked Nov 15 at 0:05
joseph
958
asked Nov 15 at 0:05
joseph
958
958
2
Express $x=x(h,g)$ and $y=y(h,g)$ in terms of $h$ and $g$ and substitute.
– NCh
Nov 15 at 0:26
An alternative approach is to show that, given $X+Y=h$, you have $X$ uniformly distributed on $[0,h]$ and so $G=frac{X}{Y}=frac{X}{h-X}$ has a cumulative distribution function of $Pleft(frac{X}{Y} le gright)= frac{g}{1+g}$ and a density of $frac{1}{(1+g)^2}$ on the positive reals, neither of which are affected by the particular value of $h$
– Henry
Nov 15 at 1:07
add a comment |
2
Express $x=x(h,g)$ and $y=y(h,g)$ in terms of $h$ and $g$ and substitute.
– NCh
Nov 15 at 0:26
An alternative approach is to show that, given $X+Y=h$, you have $X$ uniformly distributed on $[0,h]$ and so $G=frac{X}{Y}=frac{X}{h-X}$ has a cumulative distribution function of $Pleft(frac{X}{Y} le gright)= frac{g}{1+g}$ and a density of $frac{1}{(1+g)^2}$ on the positive reals, neither of which are affected by the particular value of $h$
– Henry
Nov 15 at 1:07
2
2
Express $x=x(h,g)$ and $y=y(h,g)$ in terms of $h$ and $g$ and substitute.
– NCh
Nov 15 at 0:26
Express $x=x(h,g)$ and $y=y(h,g)$ in terms of $h$ and $g$ and substitute.
– NCh
Nov 15 at 0:26
An alternative approach is to show that, given $X+Y=h$, you have $X$ uniformly distributed on $[0,h]$ and so $G=frac{X}{Y}=frac{X}{h-X}$ has a cumulative distribution function of $Pleft(frac{X}{Y} le gright)= frac{g}{1+g}$ and a density of $frac{1}{(1+g)^2}$ on the positive reals, neither of which are affected by the particular value of $h$
– Henry
Nov 15 at 1:07
An alternative approach is to show that, given $X+Y=h$, you have $X$ uniformly distributed on $[0,h]$ and so $G=frac{X}{Y}=frac{X}{h-X}$ has a cumulative distribution function of $Pleft(frac{X}{Y} le gright)= frac{g}{1+g}$ and a density of $frac{1}{(1+g)^2}$ on the positive reals, neither of which are affected by the particular value of $h$
– Henry
Nov 15 at 1:07
add a comment |
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Express $x=x(h,g)$ and $y=y(h,g)$ in terms of $h$ and $g$ and substitute.
– NCh
Nov 15 at 0:26
An alternative approach is to show that, given $X+Y=h$, you have $X$ uniformly distributed on $[0,h]$ and so $G=frac{X}{Y}=frac{X}{h-X}$ has a cumulative distribution function of $Pleft(frac{X}{Y} le gright)= frac{g}{1+g}$ and a density of $frac{1}{(1+g)^2}$ on the positive reals, neither of which are affected by the particular value of $h$
– Henry
Nov 15 at 1:07