2

Consider a standard Brownian Motion $X_t$ and continuous random variable $Y_t$, where $Y_t$ is defined as $$ Y_t = \int_0^t X_t \, dt $$

My goal is to compute the variance of $Y_t$. I'd like to say that $\operatorname{Var}(Y_t)=\int_0^t \operatorname{Var}(X_t) \, dt$, but I'm not sure how I'd establish that the variance of an integral is the integral of the variance, or if that's even true. Is it true, and if so, how can I show it?

Michael Hardy
  • 1
Everyone_Else
  • 289
  • Well, the variance of the sum of two random variables is the sum of their variances plus twice their covariance, so your conjecture would be true if the $X_t$ were uncorrelated. But they're not, so probably no. –  Dec 28 '15 at 01:57
  • 1
    See this answer of mine on stats.SE for ideas on how to compute $E[Y_t^2]$. Then, if you can find $E[Y_t]$, you can determine $\operatorname{var}(Y_t)$. As to your question "Is it true?" the answer is that the variance of the integral is not the integral of the variance. – Dilip Sarwate Dec 28 '15 at 03:22
  • See also this answer – saz Dec 28 '15 at 07:34

0 Answers0