Why is the Sum of Exponentially Distributed Variables Exponentially Distributed?

[A-fil]

Homework Statement

Let X₁, X₂, … be independent exponentially distributed stochastic variables with parameter λ. For the sum Y = X₁ + X₂ + … + X_N, where N is a geometrically distributed stochastic variable with parameter p, show that Y is exponentially distributed with parameter pλ.

Homework Equations

The expected value of exponentially distributed stochastic variables is 1/λ.
The expected value of geometrically distributed stochastic variables is 1/p.

The Attempt at a Solution

What I don’t understand is mainly why Y would be exponentially distributed. According to my notes, for N constant, Y would be Erlang(N,λ).

However, if we assume that the sum of exponentially distributed variables is exponentially distributed (how to prove this?), one should be able to calculate the expected value for this distribution as:
E(Y)=E(\sum_{i=1}^{N}X_{i})=E(\sum_{i=1}^{N} \underbrace{E(X_{i})}_{1/λ} ) =\frac{E(N)}{λ}=\frac{1}{pλ}
Implying that Y~Exp(pλ). Is this correct?

So, I would appreciate any comments or hints hints that I could get from anyone here. I got the problem in Japanese and my Japanese isn't perfect so if the problem seems incorrect, feel free to point that out.

My first time posting here so I hope that I'm not breaking any rules.

 

[micromass]

Hi A-fil!

Did you see convolutions? That's how to calculate the distribution of a sum...

 

[Ray Vickson]

Homework Statement

Let X₁, X₂, … be independent exponentially distributed stochastic variables with parameter λ. For the sum Y = X₁ + X₂ + … + X_N, where N is a geometrically distributed stochastic variable with parameter p, show that Y is exponentially distributed with parameter pλ.

Homework Equations

The expected value of exponentially distributed stochastic variables is 1/λ.
The expected value of geometrically distributed stochastic variables is 1/p.

The Attempt at a Solution

What I don’t understand is mainly why Y would be exponentially distributed. According to my notes, for N constant, Y would be Erlang(N,λ).

However, if we assume that the sum of exponentially distributed variables is exponentially distributed (how to prove this?), one should be able to calculate the expected value for this distribution as:
E(Y)=E(\sum_{i=1}^{N}X_{i})=E(\sum_{i=1}^{N} \underbrace{E(X_{i})}_{1/λ} ) =\frac{E(N)}{λ}=\frac{1}{pλ}
Implying that Y~Exp(pλ). Is this correct?

So, I would appreciate any comments or hints hints that I could get from anyone here. I got the problem in Japanese and my Japanese isn't perfect so if the problem seems incorrect, feel free to point that out.

My first time posting here so I hope that I'm not breaking any rules.

You are correct that for fixed N the sum is Erlang, so that P{Y in (y,y + dy)|N = n} = f_n(y)*dy, where f_n is the n-Erlang density. However, you want to get P{Y in (y,y+dy)}. Do you see it now?

RGV

 

[A-fil]

Thanks both of you for your help! I think I got it, but just to make sure (and for others who might have gotten stuck on something similar) I'll write it down.

f(y)=P(Y \in (y,y+dy)) = \sum_{i=1}^{\infty} P(Y \in (y,y+dy)|N=i)P(N=i)

We know that
P(N=k)=(1-p)^{k-1}p

To calculate the conditional probability one can use convultion, thus yielding:
P(Y \in (y,y+dy)|N=k) = \frac{(\lambda y)^{k-1}}{(k-1)!}\lambda e^{-\lambda y}

This yields

f(y)=\sum_{i=1}^{\infty} \frac{(\lambda y)^{k-1}}{(k-1)!}\lambda e^{-\lambda y}*(1-p)^{k-1}p=p \lambda e^{-\lambda y} \sum_{i=1}^{\infty} \frac{(\lambda y - \lambda y p)^{i-1}}{(i-1)!}=p \lambda e^{-\lambda y} \sum_{i=0}^{\infty} \frac{(\lambda y - \lambda y p)^{i}}{i!}=p \lambda e^{-\lambda y} e^{\lambda y - \lambda y p}=p \lambda e^{-p \lambda y}

Thus, Y~exp(pλ)
Q.E.D.