Finding the Density of X with Exponential and Random Variables

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Homework Help Overview

The problem involves finding the density function of a random variable X defined as the product of an exponential random variable T and another independent random variable W, which takes values ±1. The context is rooted in probability theory and the properties of random variables.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to break down the cumulative distribution function (CDF) into components based on the values of W and differentiate to find the density function. Some participants question the necessity of certain divisions in the calculations and suggest considering cases for x > 0 and x < 0 separately.

Discussion Status

Participants are actively discussing the approach to finding the density function, with some expressing uncertainty about their calculations. There is acknowledgment of being on the right track, but confusion remains regarding the correct handling of the probabilities and the implications of the independence of the random variables.

Contextual Notes

There is a note that T does not take negative values, which is relevant for the calculations involving the probability P[X≤x, W=1] when x < 0. Additionally, there is a mention of a potential misunderstanding in the problem statement that was later clarified.

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!
 
Last edited:
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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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