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!)



Moment Generating Function: Pulling a term out of k-times differentiation
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:



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> 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)
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