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.

  • 1
    $\begingroup$ 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. $\endgroup$ Oct 24, 2022 at 8:16
  • $\begingroup$ Ok, I see. I have to remove this assumption. Thank you. $\endgroup$
    – Sand
    Oct 24, 2022 at 8:17
  • 2
    $\begingroup$ Existence of the integral is now in question. $\endgroup$ Oct 24, 2022 at 8:26

1 Answer 1


Ignoring all mathematical rigor, this is easy enough to answer actually.

  • $S_T$ 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 :).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .