# Double Exponential Density

1. Feb 13, 2013

### gajohnson

1. The problem statement, all variables and given/known data

Let T be an exponential random variable with parameter λ; let W be a random
variable independent of T , which is ±1 with probability 1/2 each; and let X =
WT. Show that the density of X is:

$f_{x}(x)=(\lambda/2)e^{-(\lambda)\left|x\right|}$

2. Relevant equations

Density function for exponential distribution:

$(\lambda)e^{-(\lambda)x}$

and the CDF for the exponential distribution:

$1-e^{-(\lambda)x}$

3. The attempt at a solution

Well I wanted to break the equation into:

$P[X≤x, W=1](1/2) + P[X≤x, W=-1](1/2)$

and then differentiate the result to find the density function for X.

Will this work? If not, is there another approach someone can suggest. If so, a little help with the computation would be helpful because I've been having a tough time getting anywhere with it (I know it shouldn't be hard, that's why I'm stumped). Thanks!

Last edited: Feb 13, 2013
2. Feb 13, 2013

### haruspex

Half each?
Where is x on the RHS?

3. Feb 13, 2013

### gajohnson

Ok, problem statement and relevant equations are revised. Not sure how that happened, sorry! Hope it makes more sense now...

4. Feb 13, 2013

### haruspex

Nearly right. You don't need the divisions by 2.
You then need to consider what the above reduces to separately for x > 0, < 0.

5. Feb 13, 2013

### gajohnson

Good to know I'm on the right track. The issue I'm having now is proceeding with the calculation.

I get led to this:
$(1-e^{-(\lambda)x}) + (-1+e^{-(\lambda)x})$ = 0, which is clearly incorrect.

6. Feb 14, 2013

### haruspex

As I said, you need to consider x > 0 and x < 0 separately. For x < 0, what is P[X≤x,W=1]? Remember that T does not take negative values.

7. Feb 15, 2013

### gajohnson

I believe I got it, thanks for bearing with me!