Why a Deterministic Brownian Integral Has a Normal Distribution
Summary
The document asks how to establish the distribution of an Itô integral with a deterministic, square-integrable integrand. It proposes approximating the integral by sums of weighted Brownian increments. Each sum is normally distributed because Brownian increments are normal, and its mean and variance can be calculated from the increments. The replies explain that convergence in probability preserves normality in this setting, so the integral is normal with mean zero and variance equal to the integral of the squared integrand.
The variance follows from Itô isometry, while the mean follows from the zero means of Brownian increments. The document gives a concise outline rather than a fully detailed proof: passing expectations and variances through limits requires appropriate convergence arguments, and the normality claim relies on the approximating normal distributions having convergent parameters. Its result applies to deterministic square-integrable integrands and does not by itself cover random integrands.
Key ideas
- A deterministic square-integrable integrand produces a normally distributed Itô integral against Brownian motion.
- The integral can be approximated by sums of deterministic weights times Brownian increments.
- The integral has mean zero because each Brownian increment has mean zero.
- Itô isometry gives the variance as the time integral of the squared integrand.
- Convergence in probability supports the normal limit when the approximating sums' distributions converge.
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# Distribution of stochastic integral
# Distribution of stochastic integral
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 by Gordon (score 8, accepted)
https://quant.stackexchange.com/a/18648
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.
## Answer by wsw (score 1)
https://quant.stackexchange.com/a/22277
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.