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I'm trying to derive the mean and variance for the Poisson distribution but I'm encountering a problem and I believe its due to my derivatives.

So the mgf for poisson is: $M_x(t)=e^{\lambda e^t-\lambda}$ where $x=0,1,2..$ and $\lambda \geq 0 $

So to find the mean, i need to plug in $0$ in the derivative of the mgf.

Here is what I did:

$M'x(t)=e^{\lambda e^t-\lambda} \lambda e^t=(\lambda e^t)(e^{\lambda e^t-\lambda})=\lambda e^{\lambda e^t-\lambda+t}$

So when i plug in $t=0$ i actually get the mean to be $\lambda$, however I realize my error when I started calculating the variance.

Variance=$E(x^2)-\mu^2$

$\mu^2=\lambda^2$

$E(x^2)=M''x(0)$

When I calculate the second derivative, I get:

$M''x(t)=\lambda e^{\lambda e^t-\lambda+t} \lambda e^t +1$

That 1 is what is making my answer wrong since the variance is also $\lambda$ but i cant seem to find my error.

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  • $\begingroup$ You need to be a bit more careful while differentiating the MGF. I got $M^{\prime\prime}_X(0) = \lambda^2 + \lambda$. $\endgroup$
    – Calculon
    Sep 18, 2015 at 21:15

1 Answer 1

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The second derivative is $$M''_x(t)=\lambda e^{e^t \lambda +t-\lambda } \left(e^t \lambda +1\right)$$ hence $$M''_x(0)=\lambda e^{e^0 \lambda +0-\lambda } \left(e^0 \lambda +1\right)=\lambda^2+\lambda$$

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  • $\begingroup$ I SEE IT! THANKS! $\endgroup$ Sep 18, 2015 at 21:23
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    $\begingroup$ you are welcome :-) it happens to almost everybody to miss some parts here and there ... $\endgroup$
    – Math-fun
    Sep 18, 2015 at 21:25

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