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Math Help - expected value

  1. #1
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    expected value

    If X-N(u,v) show that E((X-u)^3)=0. Hint: use symmetry.

    E(X^3-uX^2-2uX^2+2Xu^2+xu^2-u^3)

    = 3u^3-3uv+2u^3+E(X^3)
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  2. #2
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    Re: expected value

    Quote Originally Posted by Duke View Post
    If X-N(u,v) show that E((X-u)^3)=0. Hint: use symmetry.

    E(X^3-uX^2-2uX^2+2Xu^2+xu^2-u^3)

    = 3u^3-3uv+2u^3+E(X^3)
    Some words of explanation of what you are trying to do would be a great help.

    CB
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  3. #3
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    Re: expected value

    Quote Originally Posted by Duke View Post
    If X-N(u,v) show that E((X-u)^3)=0. Hint: use symmetry.

    E(X^3-uX^2-2uX^2+2Xu^2+xu^2-u^3)

    = 3u^3-3uv+2u^3+E(X^3)
    Consider the random variable Y=X-u which is \sim N(0,v)

    So:

    E((X-u)^3)=E(Y^3)

    Now the rest of the argument depends only on the parity of the normal density and of y^3.

    CB
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  4. #4
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    Re: expected value

    1) Where did you use symmetry?

    2) Are you SURE E[X^{2}] = \nu?
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    Re: expected value

    Quote Originally Posted by TKHunny View Post
    1) Where did you use symmetry?

    2) Are you SURE E[X^{2}] = \nu?
    That was a hint given in the question

    No v=E[X^2]-(E[X])^2 I have just written down final simplification. How do you find E(X^3) in terms of u,v?
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  6. #6
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    Re: expected value

    Oh, so E[X^{2}] = \nu IF E[X] = 0. That might be related to Symmetry.

    As for E[X^{3}], think really hard about the "Parity" that CaptainBlack is talking about. We know already that E[X] = 0. Hmmm... More Symmetry?
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  7. #7
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    Re: expected value

    E(X) is not 0. E(X)=u. I thought parity referered to odd and even
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    Re: expected value

    ...and since we don't care about X, but X-\mu, we did get somewhere.

    "Parity" has other meanings.

    In this case, if you cube a negative number, what do you get? Positive or Negative?

    If you cube two numbers, does their relationship change? In other words, if a > b, for a, b > 0, what can we say about a^3 vs. b^3? Same or different?
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  9. #9
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    Re: expected value

    Quote Originally Posted by Duke View Post
    E(X) is not 0. E(X)=u. I thought parity referered to odd and even
    Even and odd functions - Wikipedia, the free encyclopedia

    CB
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  10. #10
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    Re: expected value

    Quote Originally Posted by Duke View Post
    E(X) is not 0. E(X)=u. I thought parity referered to odd and even
    This thread is going nowhere because you are not taking much notice of what people are telling you, now:

    Go back to post #3 Where you were told that

    E[(X-\mu)^3)]=E(Y^3)

    and that Y \sim N(0,v)

    Now write out E(Y^3) as an integral and post what you get here.

    As a further hint you will find the definition of the expectation of a function of a continuous RV in Definition 1 >>here<<

    CB
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    Re: expected value

    The link suggest I integrate Y^3 * pdf which from wikipedia is some horrible function.
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  12. #12
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    Re: expected value

    Quote Originally Posted by Duke View Post
    The link suggest I integrate Y^3 * pdf which from wikipedia is some horrible function.
    Nobody asked you to do the integral just to write it out.

    ================================================== =

    The normal density function:

    f(x)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{- \frac{(x-\mu)^2}{2\sigma^2}}

    Making this specific to the RV Y:

    f(y)=\frac{1}{\sqrt{2\pi v}}e^{-\frac{y^2}{2v}}


    CB
    Last edited by CaptainBlack; October 15th 2011 at 10:57 AM.
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    Re: expected value

    how does this help?
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  14. #14
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    Re: expected value

    Quote Originally Posted by Duke View Post
    how does this help?
    Have you done what was asked? Do you have it in front of you scribbled down on a piece of paper?

    CB
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    Re: Show that question

    edit: removed, for some reason i didn't notice the earlier 10 replies on the thread, if anyone used my answer i think there was a sloppy change in my definition of f(x), be warned..
    Last edited by SpringFan25; October 15th 2011 at 12:34 PM.
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