Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

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

share|cite|improve this question
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
up vote 1 down vote accepted

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 $\frac{d}{dt}(f(t))^\beta=\beta f(t)^{\beta-1}\cdot f'(t)$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.