# The variance of $S_T=\int_0^T t^2 \,X_t \,dt$

Let $$\{X_t\}$$ be a stochastic process (say $$t\in[\,0,T\,]$$), where the random variables $$X_t$$ are supposed to be iid normal distributed with mean $$= 0$$, and variance $$= 1$$.

Consider the random variable

$$S_T=\int_0^T t^2 \,X_t \,dt$$

Obviously the expectation value $$S_T$$ is equal to $$0$$. However, I would like to find the expectation value of $$S_T^2$$ or even more the probability density of $$S_T$$. An advice how to find the associated Kolmogorov-Backward-Equation would also be fine.

• There is no such process. If $X_t$'s are i.i.d. and $t \mapsto X_t$ is continuous then each $X_t$ would be a constant. Oct 24, 2022 at 8:16
• Ok, I see. I have to remove this assumption. Thank you.
– Sand
Oct 24, 2022 at 8:17
• Existence of the integral is now in question. Oct 24, 2022 at 8:26

• $$S_T$$ is a sum of (infinitely many) normal-distributed random variables. Thus it is itself normally distributed, obviously with mean equal zero.
• The variance of $$S_T$$ can be computed as \begin{align} \langle S_T^2 \rangle &= \int_0^T \int_0^T t^2 s^2\langle X_t X_s\rangle\ dt\ ds \end{align}
• Now if all the $$X_t$$ are (as you write) i.i.d. normally distributed with variance 1, then $$\langle X_t X_s \rangle$$ is 1 for $$t=s$$, and zero otherwise. This results in $$\langle S_T^2\rangle=0$$, which means $$S_T=0$$ (with probability one). I.e, your integral is almost always exactly zero.
• Alternatively, you could increase the the variance of the individual $$X_t$$, for example to $$\langle X_t X_s\rangle=\delta(t-s)$$, then the answer would be more interesing, namely $$\langle S_T^2 \rangle=T^5/5$$.
• Still another possibility is to take $$X_t$$ as as Wiener process, which means that each $$X_t$$ is still normal-distributed, but not independent. In that case, $$S_T$$ is still normal-distributed. Computing the variance is left to the reader :).