Finding the Density of X with Exponential and Random Variables

gajohnson
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Homework Statement



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|}

Homework Equations



Density function for exponential distribution:

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

and the CDF for the exponential distribution:

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

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!
 
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gajohnson said:
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 each;
Half each?
and let X = 2WT. Show that the density of X is:

f_{x}(x)=(\lambda/2)e^{-\lambda}
Where is x on the RHS?
 
Ok, problem statement and relevant equations are revised. Not sure how that happened, sorry! Hope it makes more sense now...
 
gajohnson said:
P[X≤x, W=1](1/2) + P[X≤x, W=-1](1/2)
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.
 
haruspex said:
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.

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.
 
gajohnson said:
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.
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.
 
haruspex said:
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.

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