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Exponential Distribution

  1. Mar 26, 2012 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    If we have the exponential distribution [tex]f_X(x)=\frac{1}{2}e^{-x/2}[/tex] then show that the cumulative distribution function of [tex]Y=\sqrt{X}[/tex] is given by [tex]F_Y(y)=1-e^{-y^2/2}[/tex]


    2. Relevant equations
    [tex]F_Y(y)=f_X(x)\cdot\left| \frac{dx}{dy}\right|[/tex]

    [tex]F_Y(y)=f_X(h^{-1}(y))\cdot\left| \frac{d(h^{-1}(y))}{dy}\right|[/tex]


    3. The attempt at a solution
    [tex]Y=\sqrt{X}=h(x)[/tex]

    [tex]\therefore h^{-1}(y)=y^2[/tex]

    [tex]\frac{d(h^{-1}(y))}{dy}=2y[/tex]

    [tex]f_X(h^{-1}(y))=\frac{1}{2}e^{-y^2/2}[/tex]

    [itex]\therefore[/itex] after plugging these values into the formula in the relevant equations,

    [tex]F_Y(y)=y\cdot e^{-y^2/2}[/tex]

    Which is not what I was meant to show. I only had one example in my text book to go off of and I (from what I can tell) think I applied it correctly to my question, but clearly I haven't. Can someone please guide me in the right direction, and also if you can see anything in my steps that need to be scrutinized, don't be afraid to speak out.
     
  2. jcsd
  3. Mar 26, 2012 #2
    You used an incorrect equation.

    [itex]f_Y(y)=f_X(x)\cdot\left| \frac{dx}{dy}\right|[/itex].

    What you found was the PDF not the CDF. You have to integrate.
     
  4. Mar 26, 2012 #3

    Mentallic

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    Oh, cheers :biggrin:
     
  5. Mar 26, 2012 #4

    Ray Vickson

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    Rather than plugging in formulas, I think it is better to proceed from first principles: [tex] P\{Y \leq y\} = P\{ \sqrt{X} \leq y \} = P\{ X \leq y^2 \} = \left. (1 - e^{-x/2})\right|_{x=y^2}. [/tex]

    RGV
     
    Last edited: Mar 26, 2012
  6. Mar 26, 2012 #5

    Mentallic

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    Doesn't the jump between these two steps defeat the purpose? I believe I'm meant to use the integral to show that.

    Now, if I were to calculate P(Y<x) for example, wouldn't I need to integrate again? Which means that I'd need to find a numerical solution.

    edit: Never mind, I wouldn't integrate again.
     
    Last edited: Mar 26, 2012
  7. Mar 26, 2012 #6

    Ray Vickson

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    There is no such a thing as "meant to" (unless stated explicitly). Your are "meant to" do anything that is correct and that you feel comfortable doing; aside from that, there are no rules. If you prefer to use the formulas you wrote before, go ahead and do that, but there are other ways to proceed, too, and I showed you one of them (which happens to be the one I personally prefer).

    RGV
     
  8. Mar 26, 2012 #7

    Mentallic

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    Sorry, I guess I wasn't clear. What I was trying to say was, didn't you arrive at that answer by integrating? It seems that in between those two steps there was some integration involved. And yes, I've also done essentially what you have there.
     
  9. Mar 26, 2012 #8

    Ray Vickson

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    It all depends on where you start. I am so used to the exponential CDF, F(x) = 1 - exp(-ax), that I write it down almost without thinking. Of course it was obtained way in the past by integration. Still, doing its integral is a little bit easier than integrating your f_Y(y).

    RGV
     
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