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The second moment of a poisson distributed random variable is 2 the mean of the random variable is?
in Probability recategorized by
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Let's derive the formula then we will compute the answer,

assuming mean = $m$,

$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x^2 P (x,m)$

$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty}( x(x-1) + x)  \  P (x,m)$

$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x(x-1) P (x,m) + \sum_{0}^{\infty} x P (x,m)$

$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x(x-1) P (x,m) + m$                                                      $∵ \ \sum_{0}^{\infty} x P (x,m) = m$

 

$\Rightarrow \Large E\left ( X^2 \right ) = \sum_{0}^{\infty} x(x-1) \frac{e^{-m} m ^x}{x!} + m$

$\Rightarrow \Large E\left ( X^2 \right ) = e^{-m}\sum_{0}^{\infty} x(x-1) \frac{ m ^x}{x!} + m$

$\Rightarrow \Large E\left ( X^2 \right ) = m^2 e^{-m} \left [\sum_{2}^{\infty} \frac{ m ^{x-2}}{\left (x-2 \right )!} \right ] + m$

$\Rightarrow \Large E\left ( X^2 \right ) = m^2 e^{-m} e^{m} + m$                                                                                              $∵\left [\sum_{2}^{\infty} \frac{ m ^{x-2}}{\left (x-2 \right )!} \right ] = e ^m$

$\Rightarrow \Large E\left ( X^2 \right ) = m^2 + m$  .

Now given,

$\large E\left ( X^2 \right ) = 2$

$\large 2 = m^2 + m$

$\large m = 1$
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4 Comments

In the second line shouldn't it be (x(x-1)+x)P(x,m) ?

Also i think the line where you took m2 outside the summation..it should be mx-2 inside the summation..am i right?

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MiNiPanda thanks for the correction :)

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@Mk Utkarsh

Do we have moment and stuff. 

 

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I don't think so
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