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Double Exponential Density

  1. Feb 13, 2013 #1
    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:

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

    2. Relevant equations

    Density function for exponential distribution:

    [itex](\lambda)e^{-(\lambda)x}[/itex]

    and the CDF for the exponential distribution:

    [itex]1-e^{-(\lambda)x}[/itex]

    3. The attempt at a solution

    Well I wanted to break the equation into:

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

    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. jcsd
  3. Feb 13, 2013 #2

    haruspex

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    Half each?
    Where is x on the RHS?
     
  4. Feb 13, 2013 #3
    Ok, problem statement and relevant equations are revised. Not sure how that happened, sorry! Hope it makes more sense now...
     
  5. Feb 13, 2013 #4

    haruspex

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    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.
     
  6. Feb 13, 2013 #5
    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:
    [itex](1-e^{-(\lambda)x}) + (-1+e^{-(\lambda)x})[/itex] = 0, which is clearly incorrect.
     
  7. Feb 14, 2013 #6

    haruspex

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    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.
     
  8. Feb 15, 2013 #7
    I believe I got it, thanks for bearing with me!
     
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