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Math Help - Generating Functions

  1. #1
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    Generating Functions

    A random variable X assumes only the values 0,1,2,3....., and the probabilities that X assumes these values are (2/3), (2/3)(1/3) , (2/3)(1/3)^2 , (2/3)(1/3)^3 , ................ respectively. Find the E(x) and Var(x).

    Book Answer:
    E(X) = 1/2
    Var(x) = 3/4

    I dont know where to start. Can someone show me how this problem should be done?

    F(z) = P(X=0) +P(X=1)z+(p(x=2)z^2 + .......
    F(z) = (2/3) + (2/3)(1/3)(z) +(2/3)(1/3)^2 z^2 + .........
    F'(z) = (2/3)(1/3) + 2(2/3)(1/3)^2 z + ...........
    F''(z) = .................

    I know that
    E(X) = P'(1)
    and
    Var(X) = f''(1) + f'(1) - [f'(1)]^2
    [/I]
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  2. #2
    Moo
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    Hello,

    Get inspired by this : Geometric distribution - Wikipedia, the free encyclopedia

    Or consider the equality \frac{1}{1-x}=\sum_{k=0}^\infty x^k and differentiate it as you wish to compute the series you're dealing with.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Get inspired by this : Geometric distribution - Wikipedia, the free encyclopedia

    Or consider the equality \frac{1}{1-x}=\sum_{k=0}^\infty x^k and differentiate it as you wish to compute the series you're dealing with.

    Thanks alot. I was able to work out the problem.
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  4. #4
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    Hello, jjbrian!

    A random variable x assumes only the values 0,1,2,3\,\hdots
    and the probabilities that x assumes these values are
    . . (\tfrac{2}{3}),\; (\tfrac{2}{3})(\tfrac{1}{3}),\;(\tfrac{2}{3}(\tfra  c{1}{3})^2,\; (\tfrac{2}{3})(\tfrac{1}{3})^3\; \hdots resp.

    Find the E(x) and Var(x).

    Book Answer: . E(X) = \frac{1}{2},\;\;Var(x) = \frac{3}{4}

    We have: . \begin{array}{c|c} n & P(n) \\ \hline \\[-3mm] <br />
0 & \frac{2}{3} \\ \\[-3mm] 1 & (\frac{2}{3})(\frac{1}{3}) \\ \\[-3mm] 2 & (\frac{2}{3})(\frac{1}{3})^2 \\ \\[-3mm] 3 & (\frac{2}{3})(\frac{1}{3})^3 \\ \vdots & \vdots\end{array}


    \begin{array}{cccccc}\text{We have:} & E(x) &=& 0(\frac{2}{3}) + 1(\frac{2}{3})(\frac{1}{3}) + 2(\frac{2}{3})(\frac{1}{3})^2 + 3(\frac{2}{3})(\frac{1}{3})^3 + \hdots & [1] \\ \\[-4mm]<br /> <br />
\text{Multiply by }\frac{1}{3}\!:& \frac{1}{3}E(x) &=& \qquad\;\;\; 0(\frac{2}{3})(\frac{1}{3}) + 1(\frac{2}{3})(\frac{1}{3})^2 + 2(\frac{2}{3})(\frac{1}{3})^2 + \hdots & [2]\end{array}


    \text{Subtract [1] - [2]: }\;\tfrac{2}{3}E(x) \;=\;0(\tfrac{2}{3}) + (\tfrac{2}{3})(\tfrac{1}{3}) + (\tfrac{2}{3})(\tfrac{1}{3})^2 + (\tfrac{2}{3})(\tfrac{1}{3})^3 + \hdots

    . . \text{And we have: }\;\tfrac{2}{3}E(x) \;=\;(\tfrac{2}{3})(\tfrac{1}{3})\underbrace{\bigg[1 + (\tfrac{1}{3}) + (\tfrac{1}{3})^2 +(\tfrac{1}{3})^3 +  \hdots \bigg]}_{\text{Geometric series}}

    The geometric series has the sum: . \frac{1}{1-\frac{1}{3}} \:=\:\frac{3}{2}


    Hence, we have: . \frac{2}{3}E(x) \;=\;\left(\frac{2}{9}\right)\left(\frac{3}{2}\rig  ht) \quad\Rightarrow\quad \frac{2}{3}E(x)\;=\;\frac{1}{3}

    . . Therefore: . E(x) \;=\;\frac{1}{2}

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