Results 1 to 1 of 1

Math Help - [SOLVED] Contradiction between expectation and MGF

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    38

    [SOLVED] Contradiction between expectation and MGF

    Sorry, it was a stupid question. You can ignore it. But it was a good LATEX practice for me though.





    In the text, for
    <br /> <br />
P\left (X=x \right )=\binom{x-1}{r-1}p^r\left (1-p \right )^{x-r}; Supp\left (X \right )=\left \{r,r+1,r+2,... \right \}<br />
    We have
    E(X)=\frac{r}{p} and Var(X)=\frac{r(1-P)}{p^2} ; therefore, E(X^2)=\frac{r^2+r-rp}{p^2}
    I then did that transformation [tex]Y=X-r{/math] to obtain
    P\left (Y=y \right )=\binom{y+r-1}{r-1}p^r\left (1-p \right )^y; Supp\left (Y \right )=\left \{0,1,2,3,... \right \}<br />
    I then found the MGF to be:
    M_Y(t)=E(e^{tY})=\sum_{y=0}^{\infty }e^{ty}\binom{y+r-1}{r-1}p^r\left (1-p \right )^y= \frac{p^r}{(1-(1-p)e^t)^r};t\neq -\ln (1-p)<br />
    I then tried us use the MGF to find  E(X) and     E(X^2)

    E(Y)=\frac{d }{d x}M_Y(0)=\frac{r(1-p)}{p}
    and
    E(Y^2)=\frac{d^2 }{d x^2}M_Y(0)=\frac{r(1-p)[(r(1-p)+1]}{p^2}
    therefore
    Var(Y)=\frac{r(1-P)}{p^2}

    So here's my question:
    Why does
    E(X)\neq E(Y)
    and
    E(X^2) \neq E(Y^2)
    but
    Var(Y)=Var(X)?
    Last edited by synclastica_86; February 15th 2009 at 10:41 AM. Reason: I found out why by looking at it.... sorry
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Expectation - geometric series?
    Posted in the Statistics Forum
    Replies: 2
    Last Post: June 2nd 2010, 02:10 AM
  2. [SOLVED] Proving a theorem without using contradiction
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: November 16th 2009, 03:42 PM
  3. [SOLVED] Probability question - Expectation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 23rd 2009, 02:51 PM
  4. [SOLVED] Expectation and Variance
    Posted in the Statistics Forum
    Replies: 1
    Last Post: December 10th 2008, 11:45 AM
  5. [SOLVED] Please help with conditional expectation
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: January 19th 2008, 11:50 AM

Search Tags


/mathhelpforum @mathhelpforum