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Math Help - Straightforward PDF/CDF question

  1. #1
    r45
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    Straightforward PDF/CDF question

    Hi there, I am taking a Statistics course at University, but it's been a while since I did any of this stuff so I'm a bit rusty. Here's a question I'm trying to do:

    Let X denote a random variable with probability density function

    f(x) = 2e^{-2x} when 0<x<inf, and = 0 elsewhere

    (i) Derive the CDF of X
    (ii) Calculate E[X^2 + 3X]
    (iii) Let Y = min(V,W), where V and W are independent and each have the same distribution as X. Find the CDF of Y and hence the PDF of Y.
    Well for part (i) I simply integrated 2e^{-2x} to get -e^{-2x} for when x is between 0 and infinity. But then I remembered, isn't 0 <= F(x) <= 1 a property of the CDF? And in this case, if x is for example 3, then F(3) = -e^{-6} which certainly isn't between 0 and 1. So where have I wrong here?

    I then thought I could do part (ii) without successfully having done part (i). I used the fact that E[X^2 + 3X] = E[X^2] + 3E[X] and ended up getting an answer of 2, can anyone verify this for me please? Just so I know whether I'm doing things right!

    And for part (iii) I decided I needed to know about part (i) first...once I get part (i), I think I might be able to do it, I have something in my notes and 2 random variables being "independent and identically distributed" which I think may apply here.

    Many thanks for anyone who can help me out, my exam is on Friday and I need to sort out a few kinks!

    Best,
    Ruaraidh
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  2. #2
    MHF Contributor
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    I think you made an integration error in part 1

     F(x) =  \int_0^{x} 2e^{-2x} = \left[-e^{-2x} \right]^{x}_0 = \left( -e^{-2x} - -e^{0}} \right) = \left(-e^{-2x} + 1} \right) = (1-e^{-2x})


    i dont think your part b is quite right either
    E(X) = 0.5
    E(X^2) = 0.25

    (note, it is just a coincident that E(X^2) = E(X) * E(X) in this case)

    You can get the above answers by integration

    E(X) = \int x f(x) ~dx
    E(X^2) = \int x^2 f(x) ~dx
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  3. #3
    r45
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    Hi Springfan25, many thanks for your response. I see where I went wrong with part (i) now, I messed up the limits basically.

    But with part (ii), I don't see how you get E(X^2) = 0.25? I get E(X^2) = 0.5, and E(X) = 0.5 which gives my value of E(X^2 + 3X) to be 0.5 + 3*0.5 = 2.

    To get E(X^2) = 0.5 I did:

     <br />
E(X^2) = \int^{\infty}_{0} x^2 f(x) ~dx<br />
\\ = [(-x^2 -x - \frac{1}{2})e^{-2x}]^{\infty}_{0} \\ = 0 - (-\frac{1}{2}) = \frac{1}{2}<br />

    Which one of us has made a mistake?
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  5. #5
    r45
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    But f(x) isn't exp(-2x), it's 2*exp(-2x). So from that I take it that E[X^2] = 2*0.25 = 0.5 is correct?

    Also, I have had a go at part (iii), can anyone verify this?

    CDF:
    F(Y) = 0 for v \leq 0, w \leq 0

    F(Y) = 1 - e^{-2v} for 0<v \leq w

    F(Y) = 1 - e^{-w} for 0<w \leq v

    And thus the PDF:

    f(y) = 2*e^{-2v} 0<v \leq w

    f(y) = 2*e^{-2w} 0<w \leq v

    f(y) = 0 elsewhere

    I don't think this is right though because the integral of f(y) between the limits should equal 1, but this doesn't...
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  6. #6
    r45
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    OK I'm really stuck on part (iii). I've done similar questions that are like "Let Y = 1 - X. Find the CDF and PDF of Y", which I can do by using a technique like this (arbitrary example):

    F_Y(y) = P(Y \leq y) = P(1-X \leq y) = P(1-y \leq X) = 1 - P(X \leq 1 -y) = 1 - F_X(1-y)

    Which can then be solved using the CDF for X, but with this specific question, I don't really know how to do that cos we have Y=min(V,W) which isn't an explicit function of X. I thought I had something in my notes about independent and identically distributed (iid) random variables, but I can't seem to find it. Can anyone explain to me how to solve part (iii)?
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