Distribution of Ratio of Exponential and Gamma random variable
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A recent question asked about the distribution of the ratio of two random variables, and the answer accepted there was a reference to Wikipedia which (in simplified and restated form) claims that if $X$ is a Gamma random variable with parameters $(n, 1)$ and $Y$ is an exponential random variable with parameter $1$, then $Y/X$ is a Pareto random variable with parameters $(1, n)$. Presumably $X$ and $Y$ need to be independent for this result to hold. But, Wikipedia's page on Pareto random variables doesn't seem to include a statement as to what $(1, n)$ means, though based on what it does says, a reasonable interpretation is that the Pareto random variable takes on values in $(1,infty)$ and its complementary CDF decays away as $z^{-n}$.
My question is: what is the intuitive explanation for the ratio $Y/X$ to have value $1$ or more? It would seem that all positive values should occur, and indeed the event
${Y < X}$ should have large probability since the Gamma random variable has larger mean than the exponential random variable.
I did work out the complementary CDF of $Y/X$ and got $(1+z)^{-n}$ for $z > 0$ which is not quite what Wikipedia claims.
Added Note: Thanks to Sasha and Didier Piau for confirming my calculation that
for $z > 0$, $P{Y/X > z} = (1+z)^{-n}$ which of course implies that
$$P{Y/X + 1 > z} = P{Y/X > z-1} = (1 + z - 1)^{-n} = z^{-n}~ text{for}~ z > 1$$
and thus it is $Y/X + 1$ which is a Pareto random variable, not $Y/X$ as claimed by
Wikipedia. This leads to a simple answer to
a question posed by S Huntsman: Are there any (pairs) of simple distributions that give rise to a power law ratio?
If $W$ and $X$ are the $(n+1)$-th and $n$-th arrival times in a (homogeneous) Poisson process, then $W/X$ is a Pareto $(1,n)$ random variable: $P{W/X > a} = a^{-n}$ for $a > 1$.
I suspect that this result is quite well known in the theory
of Poisson processes but I don't have a reference for it.
probability probability-distributions
edited Apr 13 '17 at 12:21
Community♦
1
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
$endgroup$
add a comment |
$begingroup$
A recent question asked about the distribution of the ratio of two random variables, and the answer accepted there was a reference to Wikipedia which (in simplified and restated form) claims that if $X$ is a Gamma random variable with parameters $(n, 1)$ and $Y$ is an exponential random variable with parameter $1$, then $Y/X$ is a Pareto random variable with parameters $(1, n)$. Presumably $X$ and $Y$ need to be independent for this result to hold. But, Wikipedia's page on Pareto random variables doesn't seem to include a statement as to what $(1, n)$ means, though based on what it does says, a reasonable interpretation is that the Pareto random variable takes on values in $(1,infty)$ and its complementary CDF decays away as $z^{-n}$.
My question is: what is the intuitive explanation for the ratio $Y/X$ to have value $1$ or more? It would seem that all positive values should occur, and indeed the event
${Y < X}$ should have large probability since the Gamma random variable has larger mean than the exponential random variable.
I did work out the complementary CDF of $Y/X$ and got $(1+z)^{-n}$ for $z > 0$ which is not quite what Wikipedia claims.
Added Note: Thanks to Sasha and Didier Piau for confirming my calculation that
for $z > 0$, $P{Y/X > z} = (1+z)^{-n}$ which of course implies that
$$P{Y/X + 1 > z} = P{Y/X > z-1} = (1 + z - 1)^{-n} = z^{-n}~ text{for}~ z > 1$$
and thus it is $Y/X + 1$ which is a Pareto random variable, not $Y/X$ as claimed by
Wikipedia. This leads to a simple answer to
a question posed by S Huntsman: Are there any (pairs) of simple distributions that give rise to a power law ratio?
If $W$ and $X$ are the $(n+1)$-th and $n$-th arrival times in a (homogeneous) Poisson process, then $W/X$ is a Pareto $(1,n)$ random variable: $P{W/X > a} = a^{-n}$ for $a > 1$.
I suspect that this result is quite well known in the theory
of Poisson processes but I don't have a reference for it.
probability probability-distributions
edited Apr 13 '17 at 12:21
Community♦
1
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
$endgroup$
$begingroup$
Actually it is $Y/X + 1$ that follows $mathrm{Pareto}(1,n)$.
$endgroup$
– Sasha
Oct 26 '11 at 20:49
1
$begingroup$
This seems to be an example of a blunder on a WP page. Obviously the range of $Y/X$ is $(0,+infty)$ and not $(1,+infty)$ (as you explain) hence $Y/X$ cannot be Pareto. Note that the CDF of $Y/X$ you indicate is correct and shows that $1+(Y/X)$ is Pareto$(1,n)$.
$endgroup$
– Did
Oct 26 '11 at 20:54
$begingroup$
@DidierPiau Thanks for the comment, which I have upvoted. Would you have a reference for the result I state in the revised version of the question?
$endgroup$
– Dilip Sarwate
Oct 27 '11 at 3:36
$begingroup$
Yes, as you guessed this is standard Poisson process stuff: once the $(n+1)$th instant $W$ is known, the $n$ first instants are distributed like a uniform i.i.d. sample in $(0,W)$, hence the $k$th instant is $W$ times a Beta $(n-k,k)$ random variable. A simple argument for the $n$th instant is that $[Xleqslant xW]$ means that one chooses $n$ times a point in $(0,x)$ rather than in $(0,1)$, which obviously has probability $x^n$ (and your parameter $a$ is $1/x$).
$endgroup$
– Did
Oct 27 '11 at 7:56
add a comment |
$begingroup$
A recent question asked about the distribution of the ratio of two random variables, and the answer accepted there was a reference to Wikipedia which (in simplified and restated form) claims that if $X$ is a Gamma random variable with parameters $(n, 1)$ and $Y$ is an exponential random variable with parameter $1$, then $Y/X$ is a Pareto random variable with parameters $(1, n)$. Presumably $X$ and $Y$ need to be independent for this result to hold. But, Wikipedia's page on Pareto random variables doesn't seem to include a statement as to what $(1, n)$ means, though based on what it does says, a reasonable interpretation is that the Pareto random variable takes on values in $(1,infty)$ and its complementary CDF decays away as $z^{-n}$.
My question is: what is the intuitive explanation for the ratio $Y/X$ to have value $1$ or more? It would seem that all positive values should occur, and indeed the event
${Y < X}$ should have large probability since the Gamma random variable has larger mean than the exponential random variable.
I did work out the complementary CDF of $Y/X$ and got $(1+z)^{-n}$ for $z > 0$ which is not quite what Wikipedia claims.
Added Note: Thanks to Sasha and Didier Piau for confirming my calculation that
for $z > 0$, $P{Y/X > z} = (1+z)^{-n}$ which of course implies that
$$P{Y/X + 1 > z} = P{Y/X > z-1} = (1 + z - 1)^{-n} = z^{-n}~ text{for}~ z > 1$$
and thus it is $Y/X + 1$ which is a Pareto random variable, not $Y/X$ as claimed by
Wikipedia. This leads to a simple answer to
a question posed by S Huntsman: Are there any (pairs) of simple distributions that give rise to a power law ratio?
If $W$ and $X$ are the $(n+1)$-th and $n$-th arrival times in a (homogeneous) Poisson process, then $W/X$ is a Pareto $(1,n)$ random variable: $P{W/X > a} = a^{-n}$ for $a > 1$.
I suspect that this result is quite well known in the theory
of Poisson processes but I don't have a reference for it.
probability probability-distributions
edited Apr 13 '17 at 12:21
Community♦
1
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
$endgroup$
A recent question asked about the distribution of the ratio of two random variables, and the answer accepted there was a reference to Wikipedia which (in simplified and restated form) claims that if $X$ is a Gamma random variable with parameters $(n, 1)$ and $Y$ is an exponential random variable with parameter $1$, then $Y/X$ is a Pareto random variable with parameters $(1, n)$. Presumably $X$ and $Y$ need to be independent for this result to hold. But, Wikipedia's page on Pareto random variables doesn't seem to include a statement as to what $(1, n)$ means, though based on what it does says, a reasonable interpretation is that the Pareto random variable takes on values in $(1,infty)$ and its complementary CDF decays away as $z^{-n}$.
My question is: what is the intuitive explanation for the ratio $Y/X$ to have value $1$ or more? It would seem that all positive values should occur, and indeed the event
${Y < X}$ should have large probability since the Gamma random variable has larger mean than the exponential random variable.
I did work out the complementary CDF of $Y/X$ and got $(1+z)^{-n}$ for $z > 0$ which is not quite what Wikipedia claims.
Added Note: Thanks to Sasha and Didier Piau for confirming my calculation that
for $z > 0$, $P{Y/X > z} = (1+z)^{-n}$ which of course implies that
$$P{Y/X + 1 > z} = P{Y/X > z-1} = (1 + z - 1)^{-n} = z^{-n}~ text{for}~ z > 1$$
and thus it is $Y/X + 1$ which is a Pareto random variable, not $Y/X$ as claimed by
Wikipedia. This leads to a simple answer to
a question posed by S Huntsman: Are there any (pairs) of simple distributions that give rise to a power law ratio?
If $W$ and $X$ are the $(n+1)$-th and $n$-th arrival times in a (homogeneous) Poisson process, then $W/X$ is a Pareto $(1,n)$ random variable: $P{W/X > a} = a^{-n}$ for $a > 1$.
I suspect that this result is quite well known in the theory
of Poisson processes but I don't have a reference for it.
probability probability-distributions
probability probability-distributions
edited Apr 13 '17 at 12:21
Community♦
1
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
edited Apr 13 '17 at 12:21
Community♦
1
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
asked Oct 26 '11 at 20:41
Dilip SarwateDilip Sarwate
19.1k13076
19.1k13076
$begingroup$
Actually it is $Y/X + 1$ that follows $mathrm{Pareto}(1,n)$.
$endgroup$
– Sasha
Oct 26 '11 at 20:49
1
$begingroup$
This seems to be an example of a blunder on a WP page. Obviously the range of $Y/X$ is $(0,+infty)$ and not $(1,+infty)$ (as you explain) hence $Y/X$ cannot be Pareto. Note that the CDF of $Y/X$ you indicate is correct and shows that $1+(Y/X)$ is Pareto$(1,n)$.
$endgroup$
– Did
Oct 26 '11 at 20:54
$begingroup$
@DidierPiau Thanks for the comment, which I have upvoted. Would you have a reference for the result I state in the revised version of the question?
$endgroup$
– Dilip Sarwate
Oct 27 '11 at 3:36
$begingroup$
Yes, as you guessed this is standard Poisson process stuff: once the $(n+1)$th instant $W$ is known, the $n$ first instants are distributed like a uniform i.i.d. sample in $(0,W)$, hence the $k$th instant is $W$ times a Beta $(n-k,k)$ random variable. A simple argument for the $n$th instant is that $[Xleqslant xW]$ means that one chooses $n$ times a point in $(0,x)$ rather than in $(0,1)$, which obviously has probability $x^n$ (and your parameter $a$ is $1/x$).
$endgroup$
– Did
Oct 27 '11 at 7:56
add a comment |
$begingroup$
Actually it is $Y/X + 1$ that follows $mathrm{Pareto}(1,n)$.
$endgroup$
– Sasha
Oct 26 '11 at 20:49
1
$begingroup$
This seems to be an example of a blunder on a WP page. Obviously the range of $Y/X$ is $(0,+infty)$ and not $(1,+infty)$ (as you explain) hence $Y/X$ cannot be Pareto. Note that the CDF of $Y/X$ you indicate is correct and shows that $1+(Y/X)$ is Pareto$(1,n)$.
$endgroup$
– Did
Oct 26 '11 at 20:54
$begingroup$
@DidierPiau Thanks for the comment, which I have upvoted. Would you have a reference for the result I state in the revised version of the question?
$endgroup$
– Dilip Sarwate
Oct 27 '11 at 3:36
$begingroup$
Yes, as you guessed this is standard Poisson process stuff: once the $(n+1)$th instant $W$ is known, the $n$ first instants are distributed like a uniform i.i.d. sample in $(0,W)$, hence the $k$th instant is $W$ times a Beta $(n-k,k)$ random variable. A simple argument for the $n$th instant is that $[Xleqslant xW]$ means that one chooses $n$ times a point in $(0,x)$ rather than in $(0,1)$, which obviously has probability $x^n$ (and your parameter $a$ is $1/x$).
$endgroup$
– Did
Oct 27 '11 at 7:56
$begingroup$
Actually it is $Y/X + 1$ that follows $mathrm{Pareto}(1,n)$.
$endgroup$
– Sasha
Oct 26 '11 at 20:49
$begingroup$
Actually it is $Y/X + 1$ that follows $mathrm{Pareto}(1,n)$.
$endgroup$
– Sasha
Oct 26 '11 at 20:49
1
1
$begingroup$
This seems to be an example of a blunder on a WP page. Obviously the range of $Y/X$ is $(0,+infty)$ and not $(1,+infty)$ (as you explain) hence $Y/X$ cannot be Pareto. Note that the CDF of $Y/X$ you indicate is correct and shows that $1+(Y/X)$ is Pareto$(1,n)$.
$endgroup$
– Did
Oct 26 '11 at 20:54
$begingroup$
This seems to be an example of a blunder on a WP page. Obviously the range of $Y/X$ is $(0,+infty)$ and not $(1,+infty)$ (as you explain) hence $Y/X$ cannot be Pareto. Note that the CDF of $Y/X$ you indicate is correct and shows that $1+(Y/X)$ is Pareto$(1,n)$.
$endgroup$
– Did
Oct 26 '11 at 20:54
$begingroup$
@DidierPiau Thanks for the comment, which I have upvoted. Would you have a reference for the result I state in the revised version of the question?
$endgroup$
– Dilip Sarwate
Oct 27 '11 at 3:36
$begingroup$
@DidierPiau Thanks for the comment, which I have upvoted. Would you have a reference for the result I state in the revised version of the question?
$endgroup$
– Dilip Sarwate
Oct 27 '11 at 3:36
$begingroup$
Yes, as you guessed this is standard Poisson process stuff: once the $(n+1)$th instant $W$ is known, the $n$ first instants are distributed like a uniform i.i.d. sample in $(0,W)$, hence the $k$th instant is $W$ times a Beta $(n-k,k)$ random variable. A simple argument for the $n$th instant is that $[Xleqslant xW]$ means that one chooses $n$ times a point in $(0,x)$ rather than in $(0,1)$, which obviously has probability $x^n$ (and your parameter $a$ is $1/x$).
$endgroup$
– Did
Oct 27 '11 at 7:56
$begingroup$
Yes, as you guessed this is standard Poisson process stuff: once the $(n+1)$th instant $W$ is known, the $n$ first instants are distributed like a uniform i.i.d. sample in $(0,W)$, hence the $k$th instant is $W$ times a Beta $(n-k,k)$ random variable. A simple argument for the $n$th instant is that $[Xleqslant xW]$ means that one chooses $n$ times a point in $(0,x)$ rather than in $(0,1)$, which obviously has probability $x^n$ (and your parameter $a$ is $1/x$).
$endgroup$
– Did
Oct 27 '11 at 7:56
add a comment |
1 Answer
1
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oldest
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$begingroup$
As I claimed in the comments, $Y/X + 1$ follows the $mathrm{Pareto}(1,n)$, not $Y/X$.
$$
phi_{Y/X}(t)= mathbb{E}left(expleft(i t frac{Y}{X}right)right) = mathbb{E}left( phi_Yleft( frac{t}{X} right) right) = mathbb{E}left( frac{X}{X-i t} right)
$$
Writing the last expectation explicitly:
$$ begin{eqnarray}
phi_{Y/X}(t) &=& frac{1}{(n-1)!} int_0^infty frac{x^n}{x - i t} mathrm{e}^{-x} mathrm{d} x = frac{1}{(n-1)!} int_0^infty mathrm{d} u int_0^infty x^n mathrm{e}^{-x} mathrm{e}^{-u(x - i t)} mathrm{d} x \
&=& frac{1}{(n-1)!} int_0^infty mathrm{e}^{i t u} n! (1+u)^{-1-n} , mathrm{d} u =
mathrm{e}^{-i t} int_0^infty mathrm{e}^{i t (u+1)} , frac{n}{(1+u)^{n+1}} mathrm{d} u \ &=& mathrm{e}^{-i t} phi_{mathrm{Pareto}(1,n)}(t)
end{eqnarray}
$$
This proves that $Y/X stackrel{d}{=} Z - 1$, where $Z$ follows $mathrm{Pareto}(1,n)$
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
$endgroup$
$begingroup$
@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $Xsimoperatorname{Exp}(alpha)$ and $YsimGamma(n,beta)$).
$endgroup$
– M.B.M.
Oct 3 '12 at 5:11
1
$begingroup$
@M.B.M. This is just the conditional tower law: $$ mathbb{E}left(exp(i t Y/X) right) = mathbb{E}left(mathbb{E}left(exp(i t Y/X) | X right)right) = mathbb{E}left( phi_Yleft(t/X right)right) $$ where $phi_Y(t)$ is the characteristic function of $Y$ random variable.
$endgroup$
– Sasha
Oct 3 '12 at 5:24
add a comment |
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1 Answer
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$begingroup$
As I claimed in the comments, $Y/X + 1$ follows the $mathrm{Pareto}(1,n)$, not $Y/X$.
$$
phi_{Y/X}(t)= mathbb{E}left(expleft(i t frac{Y}{X}right)right) = mathbb{E}left( phi_Yleft( frac{t}{X} right) right) = mathbb{E}left( frac{X}{X-i t} right)
$$
Writing the last expectation explicitly:
$$ begin{eqnarray}
phi_{Y/X}(t) &=& frac{1}{(n-1)!} int_0^infty frac{x^n}{x - i t} mathrm{e}^{-x} mathrm{d} x = frac{1}{(n-1)!} int_0^infty mathrm{d} u int_0^infty x^n mathrm{e}^{-x} mathrm{e}^{-u(x - i t)} mathrm{d} x \
&=& frac{1}{(n-1)!} int_0^infty mathrm{e}^{i t u} n! (1+u)^{-1-n} , mathrm{d} u =
mathrm{e}^{-i t} int_0^infty mathrm{e}^{i t (u+1)} , frac{n}{(1+u)^{n+1}} mathrm{d} u \ &=& mathrm{e}^{-i t} phi_{mathrm{Pareto}(1,n)}(t)
end{eqnarray}
$$
This proves that $Y/X stackrel{d}{=} Z - 1$, where $Z$ follows $mathrm{Pareto}(1,n)$
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
$endgroup$
$begingroup$
@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $Xsimoperatorname{Exp}(alpha)$ and $YsimGamma(n,beta)$).
$endgroup$
– M.B.M.
Oct 3 '12 at 5:11
1
$begingroup$
@M.B.M. This is just the conditional tower law: $$ mathbb{E}left(exp(i t Y/X) right) = mathbb{E}left(mathbb{E}left(exp(i t Y/X) | X right)right) = mathbb{E}left( phi_Yleft(t/X right)right) $$ where $phi_Y(t)$ is the characteristic function of $Y$ random variable.
$endgroup$
– Sasha
Oct 3 '12 at 5:24
add a comment |
$begingroup$
As I claimed in the comments, $Y/X + 1$ follows the $mathrm{Pareto}(1,n)$, not $Y/X$.
$$
phi_{Y/X}(t)= mathbb{E}left(expleft(i t frac{Y}{X}right)right) = mathbb{E}left( phi_Yleft( frac{t}{X} right) right) = mathbb{E}left( frac{X}{X-i t} right)
$$
Writing the last expectation explicitly:
$$ begin{eqnarray}
phi_{Y/X}(t) &=& frac{1}{(n-1)!} int_0^infty frac{x^n}{x - i t} mathrm{e}^{-x} mathrm{d} x = frac{1}{(n-1)!} int_0^infty mathrm{d} u int_0^infty x^n mathrm{e}^{-x} mathrm{e}^{-u(x - i t)} mathrm{d} x \
&=& frac{1}{(n-1)!} int_0^infty mathrm{e}^{i t u} n! (1+u)^{-1-n} , mathrm{d} u =
mathrm{e}^{-i t} int_0^infty mathrm{e}^{i t (u+1)} , frac{n}{(1+u)^{n+1}} mathrm{d} u \ &=& mathrm{e}^{-i t} phi_{mathrm{Pareto}(1,n)}(t)
end{eqnarray}
$$
This proves that $Y/X stackrel{d}{=} Z - 1$, where $Z$ follows $mathrm{Pareto}(1,n)$
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
$endgroup$
$begingroup$
@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $Xsimoperatorname{Exp}(alpha)$ and $YsimGamma(n,beta)$).
$endgroup$
– M.B.M.
Oct 3 '12 at 5:11
1
$begingroup$
@M.B.M. This is just the conditional tower law: $$ mathbb{E}left(exp(i t Y/X) right) = mathbb{E}left(mathbb{E}left(exp(i t Y/X) | X right)right) = mathbb{E}left( phi_Yleft(t/X right)right) $$ where $phi_Y(t)$ is the characteristic function of $Y$ random variable.
$endgroup$
– Sasha
Oct 3 '12 at 5:24
add a comment |
$begingroup$
As I claimed in the comments, $Y/X + 1$ follows the $mathrm{Pareto}(1,n)$, not $Y/X$.
$$
phi_{Y/X}(t)= mathbb{E}left(expleft(i t frac{Y}{X}right)right) = mathbb{E}left( phi_Yleft( frac{t}{X} right) right) = mathbb{E}left( frac{X}{X-i t} right)
$$
Writing the last expectation explicitly:
$$ begin{eqnarray}
phi_{Y/X}(t) &=& frac{1}{(n-1)!} int_0^infty frac{x^n}{x - i t} mathrm{e}^{-x} mathrm{d} x = frac{1}{(n-1)!} int_0^infty mathrm{d} u int_0^infty x^n mathrm{e}^{-x} mathrm{e}^{-u(x - i t)} mathrm{d} x \
&=& frac{1}{(n-1)!} int_0^infty mathrm{e}^{i t u} n! (1+u)^{-1-n} , mathrm{d} u =
mathrm{e}^{-i t} int_0^infty mathrm{e}^{i t (u+1)} , frac{n}{(1+u)^{n+1}} mathrm{d} u \ &=& mathrm{e}^{-i t} phi_{mathrm{Pareto}(1,n)}(t)
end{eqnarray}
$$
This proves that $Y/X stackrel{d}{=} Z - 1$, where $Z$ follows $mathrm{Pareto}(1,n)$
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
$endgroup$
As I claimed in the comments, $Y/X + 1$ follows the $mathrm{Pareto}(1,n)$, not $Y/X$.
$$
phi_{Y/X}(t)= mathbb{E}left(expleft(i t frac{Y}{X}right)right) = mathbb{E}left( phi_Yleft( frac{t}{X} right) right) = mathbb{E}left( frac{X}{X-i t} right)
$$
Writing the last expectation explicitly:
$$ begin{eqnarray}
phi_{Y/X}(t) &=& frac{1}{(n-1)!} int_0^infty frac{x^n}{x - i t} mathrm{e}^{-x} mathrm{d} x = frac{1}{(n-1)!} int_0^infty mathrm{d} u int_0^infty x^n mathrm{e}^{-x} mathrm{e}^{-u(x - i t)} mathrm{d} x \
&=& frac{1}{(n-1)!} int_0^infty mathrm{e}^{i t u} n! (1+u)^{-1-n} , mathrm{d} u =
mathrm{e}^{-i t} int_0^infty mathrm{e}^{i t (u+1)} , frac{n}{(1+u)^{n+1}} mathrm{d} u \ &=& mathrm{e}^{-i t} phi_{mathrm{Pareto}(1,n)}(t)
end{eqnarray}
$$
This proves that $Y/X stackrel{d}{=} Z - 1$, where $Z$ follows $mathrm{Pareto}(1,n)$
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
answered Oct 26 '11 at 21:01
SashaSasha
60.7k5108180
60.7k5108180
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@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $Xsimoperatorname{Exp}(alpha)$ and $YsimGamma(n,beta)$).
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– M.B.M.
Oct 3 '12 at 5:11
1
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@M.B.M. This is just the conditional tower law: $$ mathbb{E}left(exp(i t Y/X) right) = mathbb{E}left(mathbb{E}left(exp(i t Y/X) | X right)right) = mathbb{E}left( phi_Yleft(t/X right)right) $$ where $phi_Y(t)$ is the characteristic function of $Y$ random variable.
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– Sasha
Oct 3 '12 at 5:24
add a comment |
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@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $Xsimoperatorname{Exp}(alpha)$ and $YsimGamma(n,beta)$).
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– M.B.M.
Oct 3 '12 at 5:11
1
$begingroup$
@M.B.M. This is just the conditional tower law: $$ mathbb{E}left(exp(i t Y/X) right) = mathbb{E}left(mathbb{E}left(exp(i t Y/X) | X right)right) = mathbb{E}left( phi_Yleft(t/X right)right) $$ where $phi_Y(t)$ is the characteristic function of $Y$ random variable.
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– Sasha
Oct 3 '12 at 5:24
$begingroup$
@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $Xsimoperatorname{Exp}(alpha)$ and $YsimGamma(n,beta)$).
$endgroup$
– M.B.M.
Oct 3 '12 at 5:11
$begingroup$
@Sasha, I am wondering about how you got from first expectation to second (in the first line of math). I am trying to figure out how to write down the cdf for the general case with two different scaling parameters (i.e. $Xsimoperatorname{Exp}(alpha)$ and $YsimGamma(n,beta)$).
$endgroup$
– M.B.M.
Oct 3 '12 at 5:11
1
1
$begingroup$
@M.B.M. This is just the conditional tower law: $$ mathbb{E}left(exp(i t Y/X) right) = mathbb{E}left(mathbb{E}left(exp(i t Y/X) | X right)right) = mathbb{E}left( phi_Yleft(t/X right)right) $$ where $phi_Y(t)$ is the characteristic function of $Y$ random variable.
$endgroup$
– Sasha
Oct 3 '12 at 5:24
$begingroup$
@M.B.M. This is just the conditional tower law: $$ mathbb{E}left(exp(i t Y/X) right) = mathbb{E}left(mathbb{E}left(exp(i t Y/X) | X right)right) = mathbb{E}left( phi_Yleft(t/X right)right) $$ where $phi_Y(t)$ is the characteristic function of $Y$ random variable.
$endgroup$
– Sasha
Oct 3 '12 at 5:24
add a comment |
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Actually it is $Y/X + 1$ that follows $mathrm{Pareto}(1,n)$.
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– Sasha
Oct 26 '11 at 20:49
1
$begingroup$
This seems to be an example of a blunder on a WP page. Obviously the range of $Y/X$ is $(0,+infty)$ and not $(1,+infty)$ (as you explain) hence $Y/X$ cannot be Pareto. Note that the CDF of $Y/X$ you indicate is correct and shows that $1+(Y/X)$ is Pareto$(1,n)$.
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– Did
Oct 26 '11 at 20:54
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@DidierPiau Thanks for the comment, which I have upvoted. Would you have a reference for the result I state in the revised version of the question?
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– Dilip Sarwate
Oct 27 '11 at 3:36
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Yes, as you guessed this is standard Poisson process stuff: once the $(n+1)$th instant $W$ is known, the $n$ first instants are distributed like a uniform i.i.d. sample in $(0,W)$, hence the $k$th instant is $W$ times a Beta $(n-k,k)$ random variable. A simple argument for the $n$th instant is that $[Xleqslant xW]$ means that one chooses $n$ times a point in $(0,x)$ rather than in $(0,1)$, which obviously has probability $x^n$ (and your parameter $a$ is $1/x$).
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– Did
Oct 27 '11 at 7:56