0
$\begingroup$

Suppose that $f(t)$ is a (non-random) continuous function on $[0, \infty)$. Let$$Z_t = \int_0^t f(s)\,dB_s.$$

  • How do I see that $Z_t$ is normally distributed?
  • What is the mean and variance?

I need to know these results for something I am doing with analysis, but unfortunately I do not know any statistics.

Thanks.

$\endgroup$

1 Answer 1

4
$\begingroup$

$Z_t$ is the $L^2$ limit of Riemann sums like $$Z_t^{(n)} = \sum_{i=1}^n f(it/n) (B_{it/n} - B_{(i-1)t/n}).$$ This is a sum of independent normal random variables with mean 0 and variance $\frac{t}{n}|f(it/n)|^2$, so $Z_t^{(n)}$ is normally distributed with mean 0 and variance $\frac{t}{n} \sum_{i=1}^n |f(it/n)|^2$. An $L^2$ limit (or even a limit in distribution) of normal random variables has a normal distribution; this is perhaps easiest to prove by looking at Fourier transforms. So $Z_t$ is normally distributed with mean 0 and variance $\lim_{n \to \infty} \frac{t}{n} \sum_{i=1}^n |f(it/n)|^2 = \int_0^t |f(s)|^2\,ds$. (This is a very special case of the Itô isometry.)

The same result holds if $f$ is not continuous, provided that $f \in L^2([0,t])$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.