Distribution of a stable and exponential r.v
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I am given an absolutely continuous positive r.v $X$, such that $mathbb{E}[e^{-sX}]= e^{{-s}^alpha}, s>0.$
The goal is to prove that $(Y/X)^alpha$ is an Exponential r.v with intensity $1$, provided that $Y$ is independent of $X$ and also Exp$(1)$.
My initial attempt was to prove that
$$mathbb{E}[e^{t(Y/X)^alpha}]= int_0^{infty}f_X(x)dxint_0^{infty}e^{t(y/x)^alpha}e^{-y}dy=frac{1}{1-t}=psi_Y(t),$$
which is the mgf of $Y$, but I can't evaluate that integral, so there must be some other approach. Or am I not doing it right? Any insight would be helpful..
calculus probability-theory probability-distributions laplace-transform moment-generating-functions
asked Nov 5 at 19:29
blabla zazoo
1067
add a comment |
up vote
0
down vote
favorite
I am given an absolutely continuous positive r.v $X$, such that $mathbb{E}[e^{-sX}]= e^{{-s}^alpha}, s>0.$
The goal is to prove that $(Y/X)^alpha$ is an Exponential r.v with intensity $1$, provided that $Y$ is independent of $X$ and also Exp$(1)$.
My initial attempt was to prove that
$$mathbb{E}[e^{t(Y/X)^alpha}]= int_0^{infty}f_X(x)dxint_0^{infty}e^{t(y/x)^alpha}e^{-y}dy=frac{1}{1-t}=psi_Y(t),$$
which is the mgf of $Y$, but I can't evaluate that integral, so there must be some other approach. Or am I not doing it right? Any insight would be helpful..
calculus probability-theory probability-distributions laplace-transform moment-generating-functions
asked Nov 5 at 19:29
blabla zazoo
1067
1
There is a solution using the Laplace transform here.
– Maxim
Nov 6 at 19:39
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given an absolutely continuous positive r.v $X$, such that $mathbb{E}[e^{-sX}]= e^{{-s}^alpha}, s>0.$
The goal is to prove that $(Y/X)^alpha$ is an Exponential r.v with intensity $1$, provided that $Y$ is independent of $X$ and also Exp$(1)$.
My initial attempt was to prove that
$$mathbb{E}[e^{t(Y/X)^alpha}]= int_0^{infty}f_X(x)dxint_0^{infty}e^{t(y/x)^alpha}e^{-y}dy=frac{1}{1-t}=psi_Y(t),$$
which is the mgf of $Y$, but I can't evaluate that integral, so there must be some other approach. Or am I not doing it right? Any insight would be helpful..
calculus probability-theory probability-distributions laplace-transform moment-generating-functions
asked Nov 5 at 19:29
blabla zazoo
1067
I am given an absolutely continuous positive r.v $X$, such that $mathbb{E}[e^{-sX}]= e^{{-s}^alpha}, s>0.$
The goal is to prove that $(Y/X)^alpha$ is an Exponential r.v with intensity $1$, provided that $Y$ is independent of $X$ and also Exp$(1)$.
My initial attempt was to prove that
$$mathbb{E}[e^{t(Y/X)^alpha}]= int_0^{infty}f_X(x)dxint_0^{infty}e^{t(y/x)^alpha}e^{-y}dy=frac{1}{1-t}=psi_Y(t),$$
which is the mgf of $Y$, but I can't evaluate that integral, so there must be some other approach. Or am I not doing it right? Any insight would be helpful..
calculus probability-theory probability-distributions laplace-transform moment-generating-functions
calculus probability-theory probability-distributions laplace-transform moment-generating-functions
asked Nov 5 at 19:29
blabla zazoo
1067
asked Nov 5 at 19:29
blabla zazoo
1067
edited Nov 5 at 20:01
asked Nov 5 at 19:29
blabla zazoo
1067
asked Nov 5 at 19:29
blabla zazoo
1067
asked Nov 5 at 19:29
blabla zazoo
1067
1067
1
There is a solution using the Laplace transform here.
– Maxim
Nov 6 at 19:39
add a comment |
1
There is a solution using the Laplace transform here.
– Maxim
Nov 6 at 19:39
1
1
There is a solution using the Laplace transform here.
– Maxim
Nov 6 at 19:39
There is a solution using the Laplace transform here.
– Maxim
Nov 6 at 19:39
add a comment |
1 Answer
1
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Here is a solution from first principles.
$Z = (Y/X)^a$.
$P(Z leq z) = P(Y/X leq z^{1/a})$
$ = int_0^infty P(Y leq xz^{1/a})f_X(x)dx$
$ = int_0^infty (1-e^{-xz^{1/a}})f_X(x)dx$
$ = 1 - int_0^{infty}e^{-xz^{1/a}}f_X(x)dx$
$ = 1 - mathbb{E}[e^{-z^{1/a}X}]$
$ = 1 - e^{-z}$ (using the property of $X$ given in the problem and setting $s = z^{1/a}$).
This shows that $Z sim exp(1)$.
answered Nov 22 at 8:55
Aditya Dua
5008
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here is a solution from first principles.
$Z = (Y/X)^a$.
$P(Z leq z) = P(Y/X leq z^{1/a})$
$ = int_0^infty P(Y leq xz^{1/a})f_X(x)dx$
$ = int_0^infty (1-e^{-xz^{1/a}})f_X(x)dx$
$ = 1 - int_0^{infty}e^{-xz^{1/a}}f_X(x)dx$
$ = 1 - mathbb{E}[e^{-z^{1/a}X}]$
$ = 1 - e^{-z}$ (using the property of $X$ given in the problem and setting $s = z^{1/a}$).
This shows that $Z sim exp(1)$.
answered Nov 22 at 8:55
Aditya Dua
5008
add a comment |
up vote
0
down vote
Here is a solution from first principles.
$Z = (Y/X)^a$.
$P(Z leq z) = P(Y/X leq z^{1/a})$
$ = int_0^infty P(Y leq xz^{1/a})f_X(x)dx$
$ = int_0^infty (1-e^{-xz^{1/a}})f_X(x)dx$
$ = 1 - int_0^{infty}e^{-xz^{1/a}}f_X(x)dx$
$ = 1 - mathbb{E}[e^{-z^{1/a}X}]$
$ = 1 - e^{-z}$ (using the property of $X$ given in the problem and setting $s = z^{1/a}$).
This shows that $Z sim exp(1)$.
answered Nov 22 at 8:55
Aditya Dua
5008
add a comment |
up vote
0
down vote
up vote
0
down vote
Here is a solution from first principles.
$Z = (Y/X)^a$.
$P(Z leq z) = P(Y/X leq z^{1/a})$
$ = int_0^infty P(Y leq xz^{1/a})f_X(x)dx$
$ = int_0^infty (1-e^{-xz^{1/a}})f_X(x)dx$
$ = 1 - int_0^{infty}e^{-xz^{1/a}}f_X(x)dx$
$ = 1 - mathbb{E}[e^{-z^{1/a}X}]$
$ = 1 - e^{-z}$ (using the property of $X$ given in the problem and setting $s = z^{1/a}$).
This shows that $Z sim exp(1)$.
answered Nov 22 at 8:55
Aditya Dua
5008
Here is a solution from first principles.
$Z = (Y/X)^a$.
$P(Z leq z) = P(Y/X leq z^{1/a})$
$ = int_0^infty P(Y leq xz^{1/a})f_X(x)dx$
$ = int_0^infty (1-e^{-xz^{1/a}})f_X(x)dx$
$ = 1 - int_0^{infty}e^{-xz^{1/a}}f_X(x)dx$
$ = 1 - mathbb{E}[e^{-z^{1/a}X}]$
$ = 1 - e^{-z}$ (using the property of $X$ given in the problem and setting $s = z^{1/a}$).
This shows that $Z sim exp(1)$.
answered Nov 22 at 8:55
Aditya Dua
5008
answered Nov 22 at 8:55
Aditya Dua
5008
answered Nov 22 at 8:55
Aditya Dua
5008
answered Nov 22 at 8:55
Aditya Dua
5008
5008
add a comment |
add a comment |
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There is a solution using the Laplace transform here.
– Maxim
Nov 6 at 19:39