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Joint PDFs and Expected Values of 3 Cont Rand Vars

  1. Nov 13, 2012 #1
    You know how sometimes, you take a long time to write a question, and then you submit, but it takes you to a login screen and then you lose everything you wrote?

    Anyhow...I would just like to verify some work here.

    The problems states that there are 3 mutually independent random variables, X1, X2, X3, each with the pdf f(x) = 2x, 0 < x < 1.

    a) What is the joint pdf?
    For this, I'm assuming I just need to take the product of the three pdfs, giving me 8x1x2x3?

    b) Find the expected value of 5X1X2X3.
    Am I to assume that this is equal to 5 * E(X1) * E(X2) * E(X3)? If that is true, I would calculate 5 * (∫2x2dx)3 from 0 to 1?
    This gives the result 40/27 ≈ 1.48

    c) Find P(X1 < 1/2, X2 < 1/2, X3 < 1/2)
    I calculated (∫2xdx)3 from 0 to 1/2, giving me 1/64 or .015625


    I'm struggling trying to connect examples from class/book/internet with actual problems given, so any help is much appreciated.

    Thanks,
    Steve
     
    Last edited: Nov 13, 2012
  2. jcsd
  3. Nov 13, 2012 #2

    Ray Vickson

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    For (b): do not *assume*. This is something you should know. The short answer is 'yes' in this case---that is, that E(X1 X2 X3) = EX1 * EX2 * EX3----- but not in general. Do you know why?

    If you are unsure, the best way is to write out everything in detail. For example, you can certainly say that
    [tex] E(X_1 X_2 X_3) = \int_0^1 \int_0^1 \int_0^1 x_1 x_2 x_3 f(x_1,x_2,x_3) \, dx_1 \, dx_2 \, dx_3,[/tex]
    and you can go on from there.

    RGV
     
  4. Nov 13, 2012 #3
    So, in your equation, then just integrate [tex] \int_0^1 2x_3^2 dx_3 [/tex] substituting that into the integral for x2, and then that result into the integral for x1?
    Actually, the explanation helped me out with another problem:

    [tex] f(x_1) = 2x_1, 0 < x_1 < 1 and f(x_2) = 4x_2^3, 0 < x_2 < 1[/tex]

    so, [tex]E(X_1^2X_2^3) = \int_0^1 \int_0^1 x_1^2 x_2^3 f(x_1,x_2) \, dx_1 \, dx_2 [/tex]
    [tex] = \int_0^1 \frac {4}{7} 2x_1^3 \,dx_1 [/tex]
    [tex] = \frac {2}{7} [/tex]
     
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