Results 1 to 10 of 10

Thread: mgf

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    121

    mgf

    $\displaystyle P(X = k) = \left(\begin{array}{cc}n\\k\end{array}\right)p^kq^ {n-k}$ k=0,...,n.
    Where n is a positive integer, 0 < p < 1 , q = 1 - p.

    How would I show the mgf of X is

    $\displaystyle M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$


    Also, if $\displaystyle Y = \Sigma_{1\le i \le m} X_{i}$

    how would I derive the mgf of Y?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    The mgf of a discrete distribution is :

    $\displaystyle M_X(t)=\sum_{k=0}^\infty \mathbb{P}(X=k)e^{kt}$

    So here, we have :

    $\displaystyle \begin{aligned} M(t)
    &=\sum_{k=0}^n {n\choose k} e^{kt}p^k q^{n-k} \\
    &=\sum_{k=0}^n {n\choose k} (pe^t)^k q^{n-k}\end{aligned}$

    Now just recall Newton's binomial formula :

    $\displaystyle (a+b)^n=\sum_{k=0}^n a^k b^{n-k}$



    As for the second one, if the Xi's are independent, then the mgf of Y is the product of the mgf of the Xi.
    If they're identically distributed, this will just be $\displaystyle [M(t)]^m$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by bigdoggy View Post
    $\displaystyle P(X = k) = \left(\begin{array}{cc}n\\k\end{array}\right)p^kq^ {n-k}$ k=0,...,n.
    Where n is a positive integer, 0 < p < 1 , q = 1 - p.

    How would I show the mgf of X is

    $\displaystyle M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$


    Also, if $\displaystyle Y = \Sigma_{1\le i \le m} X_{i}$

    how would I derive the mgf of Y?
    See here: http://www.mathhelpforum.com/math-he...tribution.html
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2008
    Posts
    121
    So if the mgf is

    $\displaystyle M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$
    then the mgf of Y would be $\displaystyle (pe^t +q)^{mn}\ \mbox{where}\ Y = \Sigma_{1\le i \le m}X_{i} \mbox{ with }\ X_{i} $ independent??

    How would I find the distribution of Y,and does E[Y]= M'(0) and V[Y]= M''(0)??
    Last edited by bigdoggy; Aug 19th 2009 at 02:02 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by bigdoggy View Post
    So if the mgf is

    $\displaystyle M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty$
    then the mgf of Y would be $\displaystyle (pe^t +q)^{mn}\ \mbox{where}\ Y = \Sigma_{1\le i \le m}X_{i} \mbox{ with }\ X_{i} $ independent??

    How would I find the distribution of Y,and does E[Y]= M'(0) and V[Y]= M''(0)??

    The point of using MGF's is that you hope to recognize the resulting one.
    Here you added m INDEPENDENT binomials.
    This techniques proves that the result is a binomial with same probability of success (p) and now the number of trials is nm.


    The MGF of X+Y where they are indep bin(n,p) and bin(m,p), SAME p, different sample sizes is

    $\displaystyle (pe^t+q)^n(pe^t+q)^m=(pe^t+q)^{n+m}$

    proving that X+Y is bin(n+m,p)


    So, for you the mean is (nm)p and the variance is (nm)pq.

    $\displaystyle X_1+\cdots +X_m\sim Bin(nm,p)$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2008
    Posts
    121
    Why have you mentioned the mgf of X + Y?

    I came with the mgf of Y to be $\displaystyle (pe^t +q)^{mn}$ by using the post from Moo, which used $\displaystyle M_Y(t) = (pe^t +q)^{mn}$

    Is this correct??

    Also is E[Y]= M'(0) and V[Y]= M''(0) correct??
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by bigdoggy View Post
    Why have you mentioned the mgf of X + Y?

    I came with the mgf of Y to be $\displaystyle (pe^t +q)^{mn}$ by using the post from Moo, which used $\displaystyle M_Y(t) = (pe^t +q)^{mn}$

    Is this correct??

    Also is E[Y]= M'(0) and V[Y]= M''(0) correct??
    NO, E[Y]= M'(0)=nmp BUT E[Y^2]= M''(0) SO V(Y)=nmpq
    AS I wrote 30 minutes ago.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2008
    Posts
    121
    Quote Originally Posted by matheagle View Post
    NO, E[Y]= M'(0)=nmp BUT E[Y^2]= M''(0) SO V(Y)=nmpq
    AS I wrote 30 minutes ago.
    Do you mean no, the mgf of Y isn't $\displaystyle (pe^t +q)^{mn}$??
    What is it here??
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    It's $\displaystyle (pe^t+q)^{mn}$.. matheagle didn't deny it.

    The problem is your formula for the variance...

    We know that $\displaystyle \mathbb{E}(X^k)=M^{(k)}(0)$ (k-th derivative)

    But $\displaystyle V(X)=\mathbb{E}(X^2){\color{red}-[\mathbb{E}(X)]^2}=M''(0)-[M'(0)]^2$, and not what you wrote.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Nov 2008
    Posts
    121
    Thanks again Moo!!
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum