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.



$\int_0^t f(s)\,dB_s$ normally distributed, mean and variance
$Z\_t$ 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](https://en.wikipedia.org/wiki/It%C3%B4_isometry).)


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


