Trying to integrate a stochastic RV, $\int_0^t sZ_s \, ds$

Question

I'm not taking an official class (actuarial exams), some fellow "students" created a question (forum discussion), considering the integral in title. This is my attempt at a solution with no real justifications

\begin{align} \int_0^tsZ_s \, ds & = \int_0^ts\lim_{n\to\infty}\sum_{i=1}^nY(ih)\sqrt h\,ds\\ & = \lim_{n\to\infty}\frac{1}{\sqrt n}\sum_{i=1}^nY(ih)\int_0^ts^{3/2} \, ds\\ & = \lim_{n\to\infty}\frac{1}{\sqrt n}\sum_{i=1}^nY(ih)\frac{2}{5}t^{5/2}\\ & = \frac{2}{5}t^2Z_t. \end{align}

I have strong doubts that this is correct. Or if it is correct, how should I confirm? Using Ito's lemma? I am using the simple representation of Wiener motion,

$$Z_t=\lim_{n\to\infty}\sum_{i=1}^nY(ih)\sqrt h,$$

with step size $h=t/n$, where $Y(ih)=\pm1$ each with probability $\frac{1}{2}$ (transformation of a Bernoulli RV).

I have taken the series of undergraduate real analysis courses, so one idea that pops in my head is that $Z$ is not well-behaved enough to interchange objects such as limits, sums and integrals. I would find it interesting to know which theorems' conditions I am not meeting, from various courses along the sequence of courses ascending to a course in stochastic integration.

Answer 1

The fastest way to verify your formula is to apply Ito's lemma: $$ \mathrm d(\frac25t^2Z_t) = \frac45tZ_t + \frac25t^2\mathrm dZ_t\neq tZ_t $$ so the answer is incorrect. Another point is that the representation via $Y$ amy not be applicable in case of computing integrals. Also you have to remember that in your case you have different representations for each $Z_s$ where $s$ runs in $[0,t]$ - but those representations must be clearly dependent. In particular, it is unclear to me how did you get $s^{\frac32}$ and took $Y$ outside of the integral.

In case $Z_t$ is a standard Brownian motion, sometimes it hepls to use the integration by parts: $$ \mathrm d(f_tZ_t) = f'_tZ_t\mathrm d t+f_t\mathrm dZ_t \implies f'_tZ_t\mathrm = \mathrm d(f_tZ_t) - f_t\mathrm dZ_t $$ which holds for any deterministic $C^1$ function $f_t$. Now, in your case $f'_t = t$ and so $$ \int_0^t sZ_s\mathrm ds = \frac12 t^2Z_t - \frac12\int_0^t s^2\mathrm dZ_s $$ which does not seem to be a much of simplification, though. In the end, it does not seem to me that the original integral can be simplified.