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Thread: mgf

  1. #1
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    mgf

    $\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...}
    \\ 0 & \mbox{ elsewhere } \end{array}\right.
    \\ \mbox{ where } 0<p<1 \\$

    how do I show that

    i)$\displaystyle M_x(t)=\frac{pe^t}{1-(1-p)e^t}$
    ii)use the mgf to find the mean and variance of X and

    iii) find P(X + Y = 2)?

    I'm lost with mgf's, I've searched the web but just definitions are given so I'm hoping to find examples
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  2. #2
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    Hello,

    The MGF of a random variable X is :
    $\displaystyle M_X(t)=\mathbb{E}(e^{tX})$

    For discrete distributions, we have $\displaystyle \mathbb{E}(g(X))=\sum_{k=1}^\infty g(k) \mathbb{P}(X=k)$

    So here, we have :

    $\displaystyle \begin{aligned}M_X(t) &=\sum_{k=1}^\infty e^{tk} \mathbb{P}(X=k) \\
    &=\sum_{k=1}^\infty e^{tk}p(1-p)^{k-1} \\
    &=e^t \sum_{k=1}^\infty e^{t(k-1)} p(1-p)^{k-1}
    \end{aligned}$

    Now change the indice so that it starts at k=0 :

    $\displaystyle \begin{aligned} M_X(t) &=e^t \sum_{k=0}^\infty e^{tk} p(1-p)^k \\
    &=pe^t \sum_{k=0}^\infty (e^t(1-p))^k \end{aligned}$

    And this is just a geometric series
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    Thanks but how would I attempt parts (ii) and (iii)
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  4. #4
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    Quote Originally Posted by bigdoggy View Post
    $\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...}
    \\ 0 & \mbox{ elsewhere } \end{array}\right.
    \\ \mbox{ where } 0<p<1 \\$

    how do I show that

    i)$\displaystyle M_x(t)=\frac{pe^t}{1-(1-p)e^t}$

    ii)use the mgf to find the mean and variance of X and

    iii) find P(X + Y = 2)?
    [snip]
    Quote Originally Posted by bigdoggy View Post
    Thanks but how would I attempt parts (ii) [snip]
    It's called moment generating function for a reason ..... Your notes must say how to use it to calculate E(X) and E(X^2) ....?

    Quote Originally Posted by bigdoggy View Post
    [snip]
    iii) find P(X + Y = 2)?

    I'm lost with mgf's, I've searched the web but just definitions are given so I'm hoping to find examples
    What's Y? And are X and Y independent?
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  5. #5
    Moo
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    As for the steps for the mgf :

    $\displaystyle e^{tk}=(e^t)^k$ (rule of exponents)


    And for the final step, remember that $\displaystyle \sum_{n=0}^\infty x^n=\frac{1}{1-x}$ if $\displaystyle |x|<1$
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  6. #6
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    Thanks for you're help moo and mr f...sorry for double posting Iam quite new to posting the questions and I thought a question about mgf and then the solving of the gp validated another thread...lesson learnt!


    Mr f...sorry, X & Y are random and independent...
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  7. #7
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    Quote Originally Posted by bigdoggy View Post
    Thanks for you're help moo and mr f...sorry for double posting Iam quite new to posting the questions and I thought a question about mgf and then the solving of the gp validated another thread...lesson learnt!


    Mr f...sorry, X & Y are random and independent...
    I also asked what's Y? Re-phrasing: what distribution does Y follow? Things go much easier if you post the whole question.
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  8. #8
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    X&Y are independent random variables with common distribution given by:

    $\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...}
    \\ 0 & \mbox{ elsewhere } \end{array}\right.
    \\ \mbox{ where } 0<1<p \\$

    That's the info given...
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  9. #9
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    Quote Originally Posted by bigdoggy View Post
    X&Y are independent random variables with common distribution given by:

    $\displaystyle p=(X=k)=\left\{\begin{array}{cc}p(1-p)^{k-1}&\mbox{k=1,2,3...}
    \\ 0 & \mbox{ elsewhere } \end{array}\right.
    \\ \mbox{ where } 0<1<p \\$

    That's the info given...
    The choices are
    X = 0, Y = 2.
    X = 1, Y = 1.
    X = 2, Y = 0.

    It should be simple to calculate the probability of these events. Remember to multiply for 'and' and add for 'or' ....
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    The choices are
    X = 0, Y = 2.
    X = 1, Y = 1.
    X = 2, Y = 0.

    It should be simple to calculate the probability of these events. Remember to multiply for 'and' and add for 'or' ....
    P(X=1)*P(Y=1) = 2p, since k=1,2,... can't have P(X=0) nor P(Y=0)?
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