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Math Help - please>>>>

  1. #1
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    please>>>>

    depending on the moment generating function of the random variable X : M(t)=e^(t^2 ) :find

    1. p(x=2)
    2.find the mean of X
    3. FIND THE VARIANCE of -2x
    4.find the the second moment about the origin

    i don't know how i starting ??
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  2. #2
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    Quote Originally Posted by flower3 View Post
    depending on the moment generating function of the random variable X : M(t)=e^(t^2 ) :find

    1. p(x=2)
    2.find the mean of X
    3. FIND THE VARIANCE of -2x
    4.find the the second moment about the origin

    i don't know how i starting ??
    You're expected to know the following formula: E\left( X^n \right) = \left. \frac{d^n M}{d t^n}\right| _{t=0}

    2. \mu = E(X) = \left. \frac{d M}{d t}\right| _{t=0} = \, ....

    4. E \left( X^2\right) = \left. \frac{d^2 M}{d t^2}\right| _{t=0} = \, ....

    3. Var(-2X) = (-2)^2 Var(X) = 4 Var(X) and Var(X) = E \left( X^2\right) - \mu^2.

    Now substitute the results from 2. and 4.

    1. If X is a continuous random variable then Pr(X = a) = 0.

    I have more to say and when I have the time I'll say it.
    Last edited by mr fantastic; August 7th 2008 at 03:29 PM. Reason: Fixed a simple careless error.
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  3. #3
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    Quote Originally Posted by flower3 View Post
    depending on the moment generating function of the random variable X : M(t)=e^(t^2 )

    [snip]
    Note that the normal distribution with mean \mu and variance \sigma^2 has the moment generating function e^{\mu t + \frac{t^2 \sigma^2}{2}}.

    Compare this with the given moment generating function and you see that X must follow a normal distribution with \mu = 0 and variance \sigma^2 = 2.

    This gives you a way of checking the answers to 2. and 3.

    It also lets you calculate Pr(a < X < b) were it to be asked.
    Last edited by mr fantastic; August 6th 2008 at 08:25 PM.
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  4. #4
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    thank you i think you're the best "mr fantastic"
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  5. #5
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    wait >>>>>>

    you say the distribution is continuous :

    when i find the variance by the 1st and the 2nd derivative for moment generating function i get var(x)=0 but you say that is = 2 how???
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  6. #6
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    Quote Originally Posted by flower3 View Post
    wait >>>>>>

    you say the distribution is continuous :

    when i find the variance by the 1st and the 2nd derivative for moment generating function i get var(x)=0 but you say that is = 2 how???
    If you showed your working I could identify the mistake(s) you made in getting Var(X) = 0.

    \frac{d^2 M}{dt^2} = 2 e^{t^2} + 4t^2 e^{t^2}. Substitute t = 0 and you get E\left(X^2\right) = \left.\frac{d^2 M}{dt^2} \right|_{t=0} = 2.

    Therefore \sigma^2 = Var(X) = E(X^2) - \mu^2 = 2 - 0 = 2, as expected.

    Therefore Var(-2X) = 4 Var(X) = 8.

    Feel free to repeat post #4 now lol!
    Last edited by mr fantastic; August 7th 2008 at 03:30 PM. Reason: Fixed a simple careless errror.
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  7. #7
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    i'm so sorry i solve it before i open my mail and see the solution >>>

    i forgot that the derivative for 2te^(t^2 ) is equal to "derivative for two multiply function " so the answer is false >>>>>

    but i solve it >>>>>then i verify from your saying

    thank you>>>>you're a nice man
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  8. #8
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    Quote Originally Posted by mr fantastic View Post

    Therefore Var(-2X) = 2 Var(X) = 4.
    this answer is false the truth is :


    Therefore Var(-2X) = 4 Var(X) = 8.
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  9. #9
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    Quote Originally Posted by flower3 View Post
    this answer is false the truth is :


    Therefore Var(-2X) = 4 Var(X) = 8.
    Whoops. My mistake. I was thinking sd while typing it. Congratulations for realising the mistake instead of getting all worried and antsy.
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  10. #10
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    thank you !

    all people maybe miscalculate since we're human



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