Normal Distribution of a Time-Weighted Brownian Integral

# Probability distribution of the stochastic process $\int_{0} ^{t}\frac{u}{t}dW_{u}$


# Probability distribution of the stochastic process $\int_{0} ^{t}\frac{u}{t}dW_{u}$












I am wondering about the probability distribution of the stochastic process

$$X_t=\int_0^t \frac{u} {t} dW_{u}$$ I thought of using the Kolmogorov equation but after converting this into An SDE $$dX_t=dW_t-\frac{1}{t^2}(\int_{0}^{t}udW_{u})dt$$ $$X_0=0$$

I found that I couldn't apply the forward Kolmogrov equation to it since the $\mu$ here is itself a random variable. Is there some other equation for finding the probability distribution of such processes?

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/49927

Here, we use the time-changed Brownian motion technique to show the normality of \begin{align*} Y_t = \int_0^t u\, dW_u, \end{align*} where $\{W_t, \, t \ge 0\}$ is a standard Brownian motion with respect to the filtration $\{\mathscr{F}_t,\, t \ge 0\}$. For $t\ge 0$, let $\mathscr{G}_t = \mathscr{F}_{\sqrt[3]{3t}}$. Consider the process $M=\{M_t, \, t\ge 0\}$, where \begin{align*} M_t = \int_0^{\sqrt[3]{3t}} u\, dW_u. \end{align*} Then, it is clear that $M$ is a continuous martingale with respect to the filtration $\{\mathscr{G}_t,\, t \ge 0\}$. Moreover, we have the quadratic variation $\langle M, M\rangle_t = t$. By Levy's martingale characterization of Brownian motion, $\{M_t, t \ge 0\}$ is a Brownian motion. That is, for $t> 0$, $M_t$ is normally distributed. Consequently, \begin{align*} Y_t &= \int_0^t u\, dW_u\\ &=M_{\frac{1}{3}t^3} \end{align*} is normally distributed, and $X_t = \frac{1}{t}Y_t$ is also normally distributed.

> Comments

Note that, for $t>0$, $X_t \sim N\big(0, \frac{1}{3}t\big)$. Then, for any $\delta >0$, \begin{align*} \lim_{t \rightarrow 0} P(|X_t|>\delta) &=\lim_{t \rightarrow 0}2P(X_t > \delta)\\ &=\lim_{t \rightarrow 0}2P\left(\sqrt{\frac{3}{t}}X_t > \sqrt{\frac{3}{t}}\delta\right)\\ &=\lim_{t \rightarrow 0}\frac{2}{\sqrt{2\pi}}\int_{\sqrt{\frac{3}{t}}\delta}^{\infty}e^{-\frac{x^2}{2}}dx\\ &=0. \end{align*} That is, as $t$ approaches $0$, $X_t$ approaches $0$ in probability.

## Answer by AXH (score 2)

https://quant.stackexchange.com/a/49901

It is better to express $X$ as $X_t = \frac{1}{t} \int_{0}^{t} u \, d W_u$. The mean of $X$ is given by $$ \mathbb{E}[X_t]=\frac{1}{t} \mathbb{E} \left[ \int_{0}^{t} u \, d W_u \right] = \frac{1}{t} 0 = 0 $$ and the variance of $X$ is given by $$ \mathbb{E}[X^2_t]=\frac{1}{t^2} \mathbb{E} \left[ \left( \int_{0}^{t} u \, d W_u \right)^2 \right] = \frac{1}{t^2} \int_{0}^{t} u^2 \, d u = \frac{t}{3} $$ To show that $X$ is Normally distributed, it is sufficient to calculate the moment generating function and show that it is that of a Normal distribution with mean zero and variance as expressed above. Since I am extremely lazy, let me write $X_t = \int_{0}^{t} f_u d W_u$ where in your case $f_u=u/t$. For our proposition to be correct, it must be true that $$ \mathbb{E} \left[ e^{\lambda X_t} \right] = e^{ \frac{1}{2} \lambda^2 \int_{0}^{t} f_u^2 du } $$ Since $f$ is not random, we can express this equation as $$ \mathbb{E} \left[ e^{\lambda X_t - \frac{1}{2} \lambda^2 \int_{0}^{t} f_u^2 du } \right] = 1 $$ Equivalently, $$ \mathbb{E} \left[ e^{ \lambda \int_{0}^{t} f_u d W_u - \frac{1}{2} \lambda^2 \int_{0}^{t} f_u^2 du } \right] = 1 $$ The process $$ Z_t = e^{ \int_{0}^{t} \left[ \lambda f_u \right] d W_u - \frac{1}{2} \int_{0}^{t} \left[ \lambda f_u \right]^2 du } $$ is a martingale - it is the stochastic exponential. We know that $Z_0=1$ and that $\mathbb{E} \left[ Z_t \right] = Z_0 = 1$, which proves that $X$ has the desired moment generating function, i.e., $X_t \sim \mathcal{N}\left(0,\frac{t}{3}\right) $.