How to Prove the Sum of Exponential Distributions is Erlang?

[ghostyc]

Hi all,

I am now doing revision for one of the statistics module.

I am having some difficulty to proove the following:

Given n iid Exponential distribution with rate parameter \mu,

using convolution to show that the sum of them is Erlang distribution with density

f(x) = \mu \frac{(\mu x)^{k-1}} {(k-1)!} \exp(-\mu x)

I have read many book, which all have seen to ommitted the proof or
let as an exercise.

Can someone help?

Thanks!

 

[chiro]

Hi all,

I am now doing revision for one of the statistics module.

I am having some difficulty to proove the following:

Given n iid Exponential distribution with rate parameter \mu,

using convolution to show that the sum of them is Erlang distribution with density

f(x) = \mu \frac{(\mu x)^{k-1}} {(k-1)!} \exp(-\mu x)

I have read many book, which all have seen to ommitted the proof or
let as an exercise.

Can someone help?

Thanks!

Do you know the definition of convolution? The application of how it applies to finding the distribution of P(X1 + X2 + X3 + ... XN < s) (ie the CDF) is given by finding the convolution of pdf's.

If you want a detailed example step-by-step (for exponential random variables) visit Page 298 of "Introduction to Probability Models 9th Edition" by Sheldon M. Ross published by Academic Press.

Other versions of the book may have the same step-by-step proof, but if you can't find it just use the convolution theorem to obtain the results and take into account the relevant domains of the variables.

 

[ghostyc]

Hi there,

Thank you for pointing the right direction.

In fact. I have tried that already,

<br /> <br /> f(x)=\int_0^x \mu \exp(-\mu s) \mu \exp(-\mu (x-s)) \, \mathrm{d} s =x\mu^2\exp(-\mu x)<br /> <br />

which is in the form of \mu (\mu x) \exp(-\mu x), is something we would expect to get.

Then I have some problems to proceed to generalize it.

If I integrate it again with
<br /> \int_0^x s\mu^2\exp(-\mu s) \mu \exp(-\mu (x-s)) \, \mathrm{d} s<br /> =<br /> -\mu^2 x \exp(-2\mu x) + x \mu ^2 \exp(-\mu x)<br />
which is hard to spot the parttern and justify the general result to the erlang density..

 

[sfs01]

Hi there,

Thank you for pointing the right direction.

In fact. I have tried that already,

<br /> <br /> f(x)=\int_0^x \mu \exp(-\mu s) \mu \exp(-\mu (x-s)) \, \mathrm{d} s =x\mu^2\exp(-\mu x)<br /> <br />

which is in the form of \mu (\mu x) \exp(-\mu x), is something we would expect to get.

Then I have some problems to proceed to generalize it.

If I integrate it again with
<br /> \int_0^x s\mu^2\exp(-\mu s) \mu \exp(-\mu (x-s)) \, \mathrm{d} s<br /> =<br /> -\mu^2 x \exp(-2\mu x) + x \mu ^2 \exp(-\mu x)<br />
which is hard to spot the parttern and justify the general result to the erlang density..

Your second integral actually evaluates to

<br /> 0.5 \mu^3 x^2 \exp(-\mu x)<br />

All you're really integrating is s. The exponential is untouched because it doesn't contain s anymore after you multiply the 2 exponentials.

 

[ghostyc]

Your second integral actually evaluates to

<br /> 0.5 \mu^3 x^2 \exp(-\mu x)<br />

All you're really integrating is s. The exponential is untouched because it doesn't contain s anymore after you multiply the 2 exponentials.

You are absolutely right. I was doing that for a long time and I got messed up with my integration. Now I have double checked with Maple.

Thanks!