How to get distribution of X1/(X1+X2+...Xn)

**hslinhc** · 12-13-2010 10:39 PM

Hi Everyone,
I've got a problem on how to find the pdf and cdf for X1/(X1+X2+...Xn) , where xi is iid exponential(1, 1/s)?
I read something about that using the fact that x2+...Xn is gamma(n-1,s), but I don't know what to do next. I think it may be associated with transformation, but I don't know how to do it?

Anyone can provide some suggestion? Really appreciated.

**Dason** · 12-14-2010 06:08 AM

It's been a while since I've done that problem but let's see... We did it for the general case of gamma distributions (which the exponential is a special case). I think we only had two gammas though (say X and Y). We can reduce your case down to this by realizing that the sum of exponentials is gamma and since $X_i$ are independent then $X_1$ and $\sum_{i=2}^n X_i$ are independent. Then I think it's just a simple bivariate transformation to get the result you want. (Hint: you might be shooting for a beta...)

**BGM** · 12-14-2010 06:23 AM

So let $X = X_1, Y = X_2 + X_3 + ... + X_n$
$Z = \frac {X} {X + Y}$

Since the support of $X, Y$ are $(0, +\infty)$ ,
the support of $Z$ is $(0, 1)$

One common way:
Since you know the pdf of the exponential distribution and the gamma
distribution, the joint pdf of $X$ and $Y$ are known. Then you can consider
an auxillary random variable $W = X$ , and then you can determine the
joint pdf of $Z$ and $W$ by taking the two-variables jacobian transformation.
After this, you integrate over all the support of $W$ to obtain
back the marginal pdf of $Z$

Another way:
Note $\forall z \in (0, 1), F_Z(z) = \Pr\left\{\frac {X} {X+Y} \leq z \right\} = \Pr\{X \leq zX + zY \} = \Pr\{(1 - z)X \leq zY \}$
$= \Pr\left\{X \leq \frac {zY} {1 - z} \right\}$

Then $F_Z(z) = \int_0^{+\infty} F_X \left(\frac {zy} {1 - z} \right) f_Y(y) dy$
by the independence of $X$ and $Y$

Differentiating both sides with respect to $z$ , we have
$f_Z(z) = \frac {\partial F_Z(z)} {\partial z} = \int_0^{+\infty} \frac {y} {(1 - z)^2} f_X \left(\frac {zy} {1 - z} \right) f_Y(y) dy$

Subsitute all the known pdf inside with your parameterization, then
do the gamma integral.

**Link** · 12-14-2010 10:29 AM

Thank you for posting BMG. That second solution was what I was thinking of, but was too lazy to post it up.

**squareandrare** · 12-15-2010 11:53 AM

Originally Posted by hslinhc

...where xi is iid exponential(1, 1/s)?

The exponential distribution only has one parameter. Did you mean Gamma(1,1/s)?

**hslinhc** · 01-05-2011 07:32 AM

Thank you Dason.
And thank you BGM. Your second solution is cool.

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