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?

  • $\begingroup$ I think it'll be easier to just use typical tools of stochastic calculus to get the distribution. For example, calculating the mean and variance and seeing that the integral is normally distributed. $\endgroup$ – Slade Nov 21 '19 at 2:30
  • $\begingroup$ @Slade I can calculate the mean and variance of this integral but how do I check that the integral is normally distributed? $\endgroup$ – ben tenyson Nov 21 '19 at 2:32
  • $\begingroup$ Something along the lines of: quant.stackexchange.com/questions/18646/… $\endgroup$ – Slade Nov 21 '19 at 2:34
  • $\begingroup$ @Slade Thanks.This way is much easier but I wonder if there is some equation similar to Kolmogorov for such processes. $\endgroup$ – ben tenyson Nov 21 '19 at 3:53

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.


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.

  • $\begingroup$ X_t in the question has a 1/t term in the denominator $\endgroup$ – ben tenyson Nov 26 '19 at 14:23
  • $\begingroup$ Changed the notation. As $\frac{1}{t}$ is a scalar, it has no impact on the distributional property. $\endgroup$ – Gordon Nov 26 '19 at 14:30
  • $\begingroup$ what about the value of X_t at t=0 how can I find the limit at t=0?it seems to be divergent $\endgroup$ – ben tenyson Nov 26 '19 at 14:34
  • $\begingroup$ See the added comments. $\endgroup$ – Gordon Nov 26 '19 at 14:53
  • $\begingroup$ @Gordon quick question, but how did you calculate the quadratic variation of $M_t$. I actually haven't seen an ito integral with the upper limit other than $t$, so when trying to calculate $dM_t$ I am unsure how to proceed. Is there some form of Leibniz rule for this? $\endgroup$ – Slade Nov 26 '19 at 15:20

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) $.

  • $\begingroup$ how do we know $Z_0=1 $ there is a 1/t factor in the exponential are you sure it doesn't make it diverge? $\endgroup$ – ben tenyson Nov 25 '19 at 11:11
  • $\begingroup$ Good question. Try it yourself! To get some intuition you can discretise the integral $\int_{0}^{t} f_u^2 du$ and examine what happens when $t$ is small. $\endgroup$ – AXH Nov 25 '19 at 11:21
  • $\begingroup$ This limit is not so hard to cacluate it can be done with L'Hôpital's rule without having to evaluate the integral itself but I can't figure out the limit of the first integral $\int_0^t f_u dW_u$ as t approaches 0 $\endgroup$ – ben tenyson Nov 26 '19 at 10:25

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