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.

  • $\begingroup$ Your solution is correct. $\endgroup$
    – user16651
    Commented Jul 2, 2015 at 18:56
  • $\begingroup$ The mean is implied by the martingale property of a stochastic integral. $\endgroup$
    – Gordon
    Commented Jul 2, 2015 at 20:33
  • $\begingroup$ I think in your very last equation, you can remove sign of expectation($E$), because variance is no longer stochastic. $\endgroup$
    – Neeraj
    Commented Dec 12, 2015 at 6:23

2 Answers 2


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.


Since $\mathbb{E}\left[ \int_0^t f(\tau) \; dW_\tau \right] = \int_0^t f(\tau) \; \mathbb{E}\left[dW_\tau \right] = 0$, $\int_0^t f(\tau) \; dW_\tau$ has zero mean.

$\text{var}\left( \int_0^t f(\tau) \; dW_\tau \right) = \mathbb{E}\left[\left( \int_0^t f(\tau) \; dW_\tau \right)^2 \right]-\mathbb{E}\left[ \int_0^t f(\tau) \; dW_\tau \right] = \int_0^t f(\tau)^2 d\tau$ using Ito's isometry as stated by others.

  • 1
    $\begingroup$ It is not clear what do you want to say. How is your answer related to the question? $\endgroup$
    – Gordon
    Commented Dec 12, 2015 at 20:44
  • $\begingroup$ I think that this reasoning is true but too short and it is only about the expected value ... we would need more as a full answer $\endgroup$
    – Richi Wa
    Commented Dec 13, 2015 at 15:38

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.