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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.

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  • $\begingroup$ 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

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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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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.

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    $\begingroup$ 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

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