Distribution of stochastic integral - Quantitative Finance Stack Exchange most recent 30 from quant.stackexchange.com 2019-08-23T10:55:12Z https://quant.stackexchange.com/feeds/question/18646 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://quant.stackexchange.com/q/18646 7 Distribution of stochastic integral Kian https://quant.stackexchange.com/users/6661 2015-07-02T18:25:29Z 2015-12-13T15:43:55Z <p>Suppose that $f(t)$ is a deterministic square integrable function. I want to show $$\int_{0}^{t}f(\tau)dW_{\tau}\sim N(0,\int_{0}^{t}|f(\tau)|^{2}d\tau)$$.</p> <p>I want to know if the following approach is correct and/or if there's a better approach.</p> <p>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}})) &amp; = &amp; \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}})\\ &amp; = &amp; \lim_{n\to\infty}\sum_{[t_{i-1},t_{i}]\in\pi_{n}}f(t_{i-1})\times0\\ &amp; = &amp; 0 \end{eqnarray*} due to independence of Wiener increments. Secondly: \begin{eqnarray*} var(\int_{0}^{t}f(\tau)dW_{\tau}) &amp; = &amp; E((\int_{0}^{t}f(\tau)dW_{\tau})^{2})\\ &amp; = &amp;E( \int_{0}^{t}f(\tau)^{2}d\tau)=\int_{0}^{t}f(\tau)^{2}d\tau \end{eqnarray*} by Ito isometry.</p> https://quant.stackexchange.com/questions/18646/-/18648#18648 4 Answer by Gordon for Distribution of stochastic integral Gordon https://quant.stackexchange.com/users/11678 2015-07-02T18:56:50Z 2015-07-02T19:02:37Z <p>Similar question has been discussed previously; see <a href="https://quant.stackexchange.com/questions/17841/why-does-the-short-rate-in-the-hull-white-model-follow-a-normal-distribution/17863#comment25957_17863">Why does the short rate in the Hull White model follow a normal distribution?</a>.</p> <p>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.</p> https://quant.stackexchange.com/questions/18646/-/22277#22277 1 Answer by wsw for Distribution of stochastic integral wsw https://quant.stackexchange.com/users/3188 2015-12-11T21:22:44Z 2015-12-13T15:43:55Z <p>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.</p> <p>$\text{var}\left( \int_0^t f(\tau) \; dW_\tau \right) = \mathbb{E}\left[\left( \int_0^t f(\tau) \; dW_\tau \right)^2 \right]-\mathbb{E}\left[ \int_0^t f(\tau) \; dW_\tau \right] = \int_0^t f(\tau)^2 d\tau$ using Ito's isometry as stated by others.</p>