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Homework Help: Evaluating expectation values

  1. Aug 29, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]f(x,y)=6a^{-5}xy^{2}[/itex] [itex]0≤x≤a[/itex] and [itex]0≤y≤a[/itex], [itex]0[/itex] elsewhere

    Show that [itex]\overline{xy}=\overline{x}.\overline{y}[/itex]

    2. Relevant equations


    3. The attempt at a solution


    Following the same proccess I get [itex]\overline{y}=\frac{3}{2}a^{-1}x^{2}[/itex]

    But when it comes to [itex]\overline{xy}[/itex] I'm not really sure how to approach it
    Last edited: Aug 29, 2013
  2. jcsd
  3. Aug 29, 2013 #2
    You found correcly x¯. Do the same for y and xy and complete your proof. What is your problem?
  4. Aug 29, 2013 #3
    Sorry, I submitted the question to early and had to add the rest of the problem. I'm not sure what the [itex]\overline{xy}[/itex] integral should look like.
  5. Aug 29, 2013 #4
    I tried [itex]\overline{xy}=\int^{∞}_{-∞}{x.y.f(x,y)dx}[/itex]
  6. Aug 29, 2013 #5
    I tried

    which does not equal [itex]\overline{x}.\overline{y}[/itex]
  7. Aug 29, 2013 #6
    Your y¯ has a minor arithmetical mistake but you can find it. And for your major problem: d(xy) = ydx + xdy. Tell me when you have this ;)

    Edit: It's not a double integral. It is one dimensional.
  8. Aug 29, 2013 #7
    I see me mistake with [itex]\overline{y}[/itex], the x shouldn't be squared right?

    If d(x,y)=xdy+ydx. Surely I end up with [itex]\int{xy.f(x,y)d(xy)}=\int{x^{2}y.f(x,y)dy}+\int{xy^{2}.f(x,y)dx}[/itex]
    And I still cant get that to equal [itex]\overline{x}.\overline{y}[/itex]
  9. Aug 29, 2013 #8


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    Homework Helper

    This is an expected value problem? Try evaluating
    \int_0^a \int_0^a xy f(x,y) \, dx dy

    again. This
    "And for your major problem: d(xy) = ydx + xdy."
    has nothing to do with the problem.
  10. Aug 29, 2013 #9


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    Science Advisor

    The first thing you need to do is define your terms! Your "[itex]\overline{x}[/itex]" is "the mean value of x for fixed y" and for "[itex]\overline{y}[/itex]" is "the mean value of y for fixed x". Since the first is a function of y and the second a function of x, their product can't possibly be equal to [itex]\overline{xy}[/itex] which is a number.
  11. Aug 29, 2013 #10
    I did this integral and got [itex]\frac{1}{2}a^{2}[/itex] which doesn't equal [itex]\overline{x}.\overline{y}[/itex] as I am trying to prove. The double integral will always eliminate the variables x and y from the equation. But [itex]\overline{x}.\overline{y}[/itex] has the variables in it.

    How do I overcome this?
  12. Aug 29, 2013 #11


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    Gold Member

    Your ##\overline{xy}## is correct, so your error is in evaluating ##\overline{x}## and ##\overline{y}##. An expectation calculation should return a number.
  13. Aug 29, 2013 #12


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    Science Advisor

    Yes, which means that either your definitions (of [itex]\overline{x}[/itex], [itex]\overline{y}[/itex], and/or [itex]\overline{xy}[/itex] are wrong or [itex]\overline{x}\overline{y}[/itex] is NOT equal to [itex]\overline{xy}[/itex].

    You might want to consider the possibility that the correct definition of [itex]\overline{u}[/itex] for any function u, of x and y, is [itex]\int\int u f(x,y)dydx[/itex] so that, in particular, [itex]\overline{x}= \int_0^a\int_0^a x f(x,y)dydx[/itex], NOT "[itex]\int_0^a xf(x,y)dx[/itex]".

    (You also posted this under "precalculus homework". Do not do that. Double posting can get you banned.)
  14. Aug 29, 2013 #13
    Using that definition of [itex]\overline{u}[/itex] I got them to equal one another :) Problem solved!
    And thanks for the warning about double posting, I won't be doing that again.
  15. Aug 29, 2013 #14


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    Staff: Mentor

    The two threads have been merged. As HallsOfIvy noted, please post a question only once. If we think it really belongs in a different forum, we'll move it.
  16. Aug 29, 2013 #15

    Ray Vickson

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    Homework Helper

    In addition to what others have said: I don't know if you have yet met with the concept of independence, but this problem fits that profile.

    You can write your bivariate density f(x,y) as a product of two univariate densities g(x) and h(y):
    [tex] f(x,y) = 6a^{-5}xy^{2} = (2 a^{-2} x) ( 3 a^{-3} y^2) = g(x) h(y), 0 \leq x,y \leq a. [/tex]
    Here the individual factors ##g(x) = 2 a^{-2} x## and ##h(y) = 2 a^{-3} y^2## are both univariate probability densities of random variables ##X, Y## on ##[0,a]##: they are ≥ 0 and integrate to 1. There is a general theorem that ##E(XY) = EX \cdot EY## if ##X## and ##Y## are independent. You have shown one special case of this. Note: ##EX## is the expectation of ##X##, and is the same as what you call ##\bar{X}##.
    Last edited: Aug 29, 2013
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