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Homework Help: Variance of Continous Random Variable

  1. Oct 23, 2012 #1
    I have the continous random variable Y, defined such that:

    and the PDF of x is zero everywhere but:
    [tex]f(x)=\frac{1}{4}e^{\frac{-x}{4}}, x>0[/tex]

    I correctly got the mean like so:
    [tex]\mu=E(h(x))=\int^{\infty}_{0} h(x)f(x)[/tex]
    and evaluated it to be 10.

    I am unsure as to how I go about getting the variance now. In previous problems I use the usual formula of:

    [tex]\sigma=\int^{b}_{a} (x-\mu)^{2}f(x)[/tex]

    or the shortcut of:

    However I'm unsure what to do in the case where the random variable is dependent on another with a given PDF function for continuous random variables.
  2. jcsd
  3. Oct 23, 2012 #2
    Start with substiting in the expression for Y into [itex]Var(Y)[/itex]

    Seperate out [itex]Var(X)[/itex]
  4. Oct 23, 2012 #3
    I'm not quite sure how to put that together. To start, I have:
    [tex]Var(Y)=\int^{\infty}_{0} (y-10)^2*f(y)[/tex]
  5. Oct 23, 2012 #4
    I would start with [itex]Var(Y)=Var(3X+2)=Var(3X)+Var(2)[/itex]
  6. Oct 23, 2012 #5
    If you want to use this form then [itex]\mu_Y=E(Y)=E(3X+2)=3E(X)+2[/itex]
  7. Oct 23, 2012 #6
    I see. So all I need to do is then find the expected value of the function defining the PDF of x?
  8. Oct 23, 2012 #7

    Ray Vickson

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    Neither X nor Y has mean 10, so you did something wrong. What is the mean of X? How is the mean of Y related to the mean of X? What is Var(X)? How is Var(Y) related to it?

  9. Oct 23, 2012 #8
    10 is what I got, and it matches the answer in the textbook...
  10. Oct 23, 2012 #9


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    I calculated E[X] = 4. Please show how you got 10.
  11. Oct 23, 2012 #10

    Ray Vickson

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    Well, EX (the mean of X) is not 10, and EY (the mean of Y) is not 10. However, E(3*X - 2) = 10, so maybe you just typed the question incorrectly, and maybe you really meant EY when you seemed to be talking about EX.

  12. Oct 23, 2012 #11
    Here is the integral I solved in order to get the mean:

    [tex]\mu=\int^{\infty}_{0} \frac{1}{4}e^{\frac{-x}{4}}(3x-2)dx=10[/tex]

    from the identity where:

    [tex]E[h(x)]=\int^{\infty}_{-\infty} h(x)f(x) dx[/tex]

    Anyhow, it turns out that in solving the variance I made an arithmetical mistake, and I did manage to get the correct answer. Thanks for the help!
  13. Oct 23, 2012 #12

    Ray Vickson

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    OK, so as I suspected, you computed EY = E(3X-2), but your question originally said Y = 3X+2!

    BTW: if 'a' and 'b' are constants, Var(AX+b) = a^2 Var(X) [that is, the constant term 'b' drops out and the factor 'a' gets squared]. Since Var(X) = 4^2 = 16, we have Var(Y) = 3^2 * 16 = 144. This is a lot easier than computing
    [tex] \int_0^{\infty} (y-10)^2 f_Y(y) \, dy.[/tex]

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