# $\frac{\partial}{\partial t} e^{\lambda e^{it-1}}$ and $\left( 1-\frac{it}{\alpha}\right)^{-\beta}$

$\frac{\partial}{\partial t} e^{\lambda e^{it-1}}$

$\frac{\partial}{\partial t}\left( 1-\frac{it}{\alpha}\right)^{-\beta}$

I calculated the characteristic functions of the Poisson and Gamma distributions to be the above function and am trying to calculate expectation and variance from them.

It has been years since I took calculus and I am pretty unsure how to take the derivatives.

I do know that I can find $E(X)$ by taking the first derivative of the characteristic function and evaluating it at t=0, (and similar for higher moments) and I know what the answers for expectation and variance are for these distributions, but I am struggling with the calculus.

Any help is appreciated

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The notation is a little ambiguous. The function $\frac{\partial \phi_X(t)}{\partial t}e^{\lambda e^{it-1}}$ already has a derivative in it. Are you asking how to do $\frac{\partial \phi_X(t)}{\partial t}$ or $\frac{\partial}{\partial t}$ of the whole thing? –  Matt Aug 15 '12 at 0:07
I apologize...I am trying to take the derivative with respect to "t" of the whole thing in both situations. –  Justin Aug 15 '12 at 0:40
@Matt, I believe I fixed it now –  Justin Aug 15 '12 at 0:45

You can break this into a chain rule by writing it as the composition of two functions. Let $f(x)=e^x$ and $g(t)=\lambda e^{it-1}$. Then $f(g(t))=e^{\lambda e^{it-1}}$.
The chain rule says $\frac{d}{dt}(f(g(t)))=f'(g(t))\cdot g'(t)$. The derivative $f'(x)=e^x$ and you need another chain rule for $g'(t)=\lambda ie^{it-1}$. Thus $\frac{d}{dt}\left(e^{\lambda e^{it-1}}\right)=\lambda ie^{it-1}e^{\lambda e^{it-1}}$
Does this help? You should just use the chain rule for the second one as well and the fact that $\frac{d}{dt}(f(t))^\beta=\beta f(t)^{\beta-1}\cdot f'(t)$
If you evaluate the first one at t=0, you should be able to eliminate all of the exponentials. I believe the answer should be just $\lambda i$ to get the first moment of the Poisson distribution. –  Justin Aug 15 '12 at 1:15
Then one should start with the correct characteristic function, which has $e^{it}-1$ in the exponent instead of $e^{it-1}$. –  Did Aug 15 '12 at 9:17