Suppose that $f(t)$ is a (non-random) continuous function on $[0, \infty)$. Let$$Z_t = \int_0^t f(s)\,dB_s.$$
I need to know these results for something I am doing with analysis, but unfortunately I do not know any statistics.
Thanks.
$Z_t$ 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 $f \in L^2([0,t])$.
