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Math Help - [SOLVED] maths expectation for random variables (how to integrate?)

  1. #1
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    Question [SOLVED] maths expectation for random variables (how to integrate?)

    for walpole qn 4.43,

    μY = E(3X − 2) = 1/4 int 0-∞ (3x − 2)e−x/4 dx = 10

    but what are the steps to get 10 ?

    how shld i integrate (3x − 2)e−x/4 ?
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  2. #2
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    Quote Originally Posted by hazel View Post
    for walpole qn 4.43,

    μY = E(3X − 2) = 1/4 int 0-∞ (3x − 2)e−x/4 dx = 10

    but what are the steps to get 10 ?

    how shld i integrate (3x − 2)e−x/4 ?
    Use integration by parts. At this level, it's a technique you should already have met. Note that you will have to deal with an improper integral using limits (again, this is something you should have already met).

    If you need more help, please show what you've done and state exactly where you're stuck in the calculation.

    By the way, E(3X - 2) = 3 E(X) - 2 .... Perhaps calculating E(X) is slightly simpler for you to do.
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  3. #3
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    μY = E(3X − 2)
    = 1/4 int 0-∞ (3x − 2)e−x/4 dx
    = 1/4 int 0-∞ (e^u) (-4 du) (3x-2)
    = int 0-∞ (3x-2) (-e^u) du
    = int 0-∞ -3x e^u du + 2 int 0-∞ e^u du
    =int 0-∞ 12 u e^u du + 2 int 0-∞ e^u du
    = 12[e^u + u.e^u] int 0-∞ + 2 [e^u] int 0-∞
    = ??

    note:
    Let U = -1/4 x
    -1/4 dx= du
    dx=-4 du
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  4. #4
    MHF Contributor matheagle's Avatar
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    I use this example all the time.
    I want students to recognize that this is an exponential rv with mean 4
    so (3)(4)-2=10.
    I even put it on the finals two weeks ago.
    This problem appears before the discussion of all the distributions.
    BUT I still explain this is the best way to do it.
    And I pass out tables of all the major densities and the means, variances, mgfs...
    that I photocopied from wackerly's book.
    Last edited by matheagle; December 30th 2009 at 07:19 PM.
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  5. #5
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    Quote Originally Posted by hazel View Post
    μY = E(3X − 2)
    = 1/4 int 0-∞ (3x − 2)e−x/4 dx
    = 1/4 int 0-∞ (e^u) (-4 du) (3x-2)
    = int 0-∞ (3x-2) (-e^u) du
    = int 0-∞ -3x e^u du + 2 int 0-∞ e^u du
    =int 0-∞ 12 u e^u du + 2 int 0-∞ e^u du
    = 12[e^u + u.e^u] int 0-∞ + 2 [e^u] int 0-∞
    = ??

    note:
    Let U = -1/4 x
    -1/4 dx= du
    dx=-4 du
    Have you been taught how to handle improper integrals using limits? I mentioned this in my first reply but you did not say.

    The very first step for your chosen method of solution should have been:

    = \frac{1}{4} {\color{red}\lim_{\alpha \to +\infty}} \int_{0}^{\color{red}\alpha} (3x - 2) e^{-x/4} \, dx.

    Calculate the integral. Then take the limit.

    Without a thorough grasp of the pre-requisite material (integration techniques, improper integrals, limits etc.) you are going to struggle. You are advised to thoroughly review this material.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Have you been taught how to handle improper integrals using limits? I mentioned this in my first reply but you did not say.

    The very first step for your chosen method of solution should have been:

    = \frac{1}{4} {\color{red}\lim_{\alpha \to +\infty}} \int_{0}^{\color{red}\alpha} (3x - 2) e^{-x/4} \, dx.

    Calculate the integral. Then take the limit.

    Without a thorough grasp of the pre-requisite material (integration techniques, improper integrals, limits etc.) you are going to struggle. You are advised to thoroughly review this material.

    Your first step is the same as my first step - if you look at it more closely :
    = 1/4 int 0-∞ (3x − 2)e−x/4 dx

    it's just that i did not code it in the fancy maths format and type out in raw format accordingly.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    I use this example all the time.
    I want students to recognize that this is an exponential rv with mean 4
    so (3)(4)-2=10.
    I even put it on the finals two weeks ago.
    This problem appears before the discussion of all the distributions.
    BUT I still explain this is the best way to do it.
    And I pass out tables of all the major densities and the means, variances, mgfs...
    that I photocopied from wackerly's book.

    I have 12 -2 = 10.
    Do you have step by step workings on how should i get to (3)(4)-2?
    i don't have wackerly's book. thks.
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  8. #8
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    Quote Originally Posted by hazel View Post
    Your first step is the same as my first step - if you look at it more closely :
    = 1/4 int 0-∞ (3x − 2)e−x/4 dx

    it's just that i did not code it in the fancy maths format and type out in raw format accordingly.
    Excuse me?

    I did look at your post closely. I don't see any limit sign or evidence that you understand that limits are to be used. Did you read the red type in my previous reply? Do you understand the importance of using limits? Go back and review the material on improper integrals.
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  9. #9
    MHF Contributor matheagle's Avatar
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    you don't need wackerly's book at all.
    you should know the mean of an exponential
    the idea is....

     \int_{0}^{\infty} (3x - 2) {e^{-x/4}\over 4} \, dx

    = 3 \int_{0}^{\infty} x {e^{-x/4}\over 4}\, dx-2  \int_{0}^{\infty}  {e^{-x/4}\over 4}\, dx

    = 3E(X)-2(1)=3(4)-2=10

    This is just E(3X-2)=3E(X)-2

    You can perform parts on the first integral if you wish, but you really should know that value.
    The second integral consists of just a density and that integrates to one.
    Last edited by matheagle; December 30th 2009 at 09:28 PM.
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  10. #10
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    Quote Originally Posted by matheagle View Post
    [snip]
    This is just E(3X-2)=3E(X)-2

    [snip]
    The funny thing is that I said this in post #2.

    Quote Originally Posted by matheagle View Post
    [snip]
    You can perform parts on the first integral if you wish, but you really should know that value.
    The second integral consists of just a density and that integrates to one.
    And the funny thing is that one of these two things is what I then thought the OP would do.

    But the OP was determined to do it his/her own way. And - apparently, because this has been completely ignored - the importance of using limits when dealing with an improper integral is not understood.
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  11. #11
    MHF Contributor matheagle's Avatar
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    I was well aware that you wanted hazel
    http://cdn1.ioffer.com/img/item/305/...o_hazel_1c.jpg
    to not do this from scratch.
    I wanted her to listen to that advice, by seconding your comments.
    AS punishment I think we need to insert that picture as her avatar.
    I suggested that to her before, but she didn't care for that picture..
    Last edited by matheagle; December 31st 2009 at 08:50 PM.
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