In Wiersema: Brownian Motion Calculus on p. 205 (in an Annex on Moment Generating Functions (mgf)) the following equation is being presented $${d^k \over d\theta^k} \left ({1\over k!}\theta^k\mathbb{E}[X^k]\right)={1\over k!}\theta^k{d^k \over d\theta^k} \mathbb{E}[X^k]$$ with $X$ being a random variable, $\theta$ the mgf-dummy variable and $\mathbb{E}$ the expectation operator (and $k$ for the $k$'th moment of $X$).
My question: Why is it possible to pull the term ${1\over k!}\theta^k$ out and in front of the $k$-times differentiation? Could anyone give me a hint or some intermediate steps? Or is this a typo? (Sorry if this is too elementary but I don't get it anyway - and I want to understand this!)
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2$\begingroup$ are you sure about this? For a start, $X$ presumably doesn't depend on $\theta$, so if anything it should be the $k$th moment of $X$ that you can pull outside the differentiation $\endgroup$– Yemon ChoiCommented Feb 20, 2010 at 19:59
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1$\begingroup$ Since this seems to be a typo, and moreover quite a localized question, I'm voting to close. $\endgroup$– Yemon ChoiCommented Feb 20, 2010 at 20:48
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That's clearly a typo. Unless I misunderstand the point of this, the right formula should be $${d^k \over d\theta^k} \left ({1\over k!}\theta^k\mathbb{E}[X^k]\right)={1\over k!}\mathbb{E}\left[{d^k \over d\theta^k} \theta^k X^k\right]$$
In fact, this is what is said in words before the formula in the book you mentioned:
The $k$th moment can be singled out by differentiating $k$ times with respect to $\theta$, and interchanging the differentiation and $\mathbb{E}$ (which is an integration)
answered Feb 20, 2010 at 20:41
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$\begingroup$ Thank you - why are you allowed to do that (interchanging differentiation and $\mathbb{E}$)? And why are you allowed to pull out the ${1\over k!}$? $\endgroup$– vonjdCommented Feb 20, 2010 at 21:06
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$\begingroup$ $\mathbb{E}$ is a linear operation. It is after all integration with respect to a probability measure. The parameter $\theta$ is auxiliary -- it is a constant from the point of view of integration. $\endgroup$ Commented Feb 20, 2010 at 21:26