Suppose that $f(t)$ is a deterministic square integrable function. I want to show $$\int_{0}^{t}f(\tau)dW_{\tau}\sim N(0,\int_{0}^{t}|f(\tau)|^{2}d\tau)$$.
I want to know if the following approach is correct and/or if there's a better approach.
First note that $$\int_{0}^{t}f(\tau)dW_{\tau}=\lim_{n\to\infty}\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})(W_{t_{i}}-W_{t_{i-1}})$$ where $\pi_{n}$ is a sequence of partitions of $[0,t]$ with mesh going to zero. Then $\int_{0}^{t}f(\tau)dW_{\tau}$ is a sum of normal random variables and hence is normal. So all we need to do is calculate the mean and variance. Firstly: \begin{eqnarray*} E(\lim_{n\to\infty}\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})(W_{t_{i}}-W_{t_{i-1}})) & = & \lim_{n\to\infty}\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})E(W_{t_{i}}-W_{t_{i-1}})\\ & = & \lim_{n\to\infty}\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})\times0\\ & = & 0 \end{eqnarray*} due to independence of Wiener increments. Secondly: \begin{eqnarray*} var(\int_{0}^{t}f(\tau)dW_{\tau}) & = & E((\int_{0}^{t}f(\tau)dW_{\tau})^{2})\\ & = &E( \int_{0}^{t}f(\tau)^{2}d\tau)=\int_{0}^{t}f(\tau)^{2}d\tau \end{eqnarray*} by Ito isometry.
Answer
Similar question has been discussed previously; see Why does the short rate in the Hull White model follow a normal distribution?.
Basically, the probabilistic limit of normal random variables is still normal. Then, as $$\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})(W_{t_{i}}-W_{t_{i-1}})$$ is normal, the limit $$\int_{0}^{t}f(\tau)dW_{\tau},$$ in probability, is also normal, with the mean and variance as you provided.
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