I am curious whether the following theorem is true.
Let $Z=X \cdot Y$ where $X$ and $Y$ are two independent positive random variables. Moreover, suppose that $E[Z^k],E[Y^k]<\infty$ for all $k \in (a,\infty)$, and $E[X^k] <\infty$ for all $k \in (b,\infty)$. Then, \begin{align*} E[X^k]= \lim_{n \to k} \frac{ E[Z^n]}{ E[Y^n]}, \forall k \in (b,\infty). \end{align*}
First, note that we are not assuming that $a$ and $b$ are positive. In fact, the interesting case happens when $a$ and $b$ are negative.
Note that the above result is true if $a\le b$ and $k \in (a,b)$, then by independence we have that \begin{align*} E[Z^k]=E[X^k] E[Y^k] \Rightarrow E[X^k]=\frac{E[Z^k]}{E[Y^k]}. \end{align*}
Therefore, an interesting case is when $a>b$. Note that in this case we can not write $E[Z^k]=E[X^k] E[Y^k] $ as $E[Z^k],E[Y^k]=\infty$.
Let me also give you an example of $Z,X,Y$. Take $X$ to be Rayleigh and $Y=|W|$ where $W$ is a standard normal. Then $Z$ has an exponential distribution.
Moreover, note that
- $E[Y^k]=E[|W|^k] <\infty$ iff $k \in (-1,\infty)$,
- $E[Z^k] <\infty$ iff $k \in (-1,\infty)$,
but $E[X^k] <\infty$ iff $k \in (-2,\infty)$.
I also wonder if continuity of $f(k)=E[X^k]$ is an important element of the proof. See this related question.