Question

Let \(Y(t)=t B(1 / t), t>0\) and \(Y(0)=0\) (a) What is the distribution of \(Y(t)\) ? (b) Compare \(\operatorname{Cov}(Y(s), Y(t))\). (c) Argue that \(\\{Y(t), t \geqslant 0\\}\) is a standard Brownian motion process.

Short answer

(a) The distribution of \(Y(t)\) is given by its probability density function: \(f_Y(y) = \frac{1}{t^2\sqrt{2 \pi t}} e^{-\frac{y^2}{2t^3}}\). (b) The covariance between \(Y(s)\) and \(Y(t)\) is 0: \(\operatorname{Cov}(Y(s), Y(t)) = 0\). (c) The process \({Y(t), t \geq 0}\) is a standard Brownian motion process because it satisfies the properties: \(Y(0) = 0\), and the increments \(Y(t) - Y(s)\) are normally distributed with mean 0 and variance \(t - s\).

Step-by-step solution

Determine the distribution of \(Y(t)\)

Using the transformation of random variables technique, we have \(Y(t) = tB(1/t)\). To transform the distribution, first, let's find the cumulative distribution function (CDF) for the standard Brownian motion process \(B(t)\): \[F_B(x) = \int_{-\infty}^{x} \frac{1}{\sqrt{2 \pi t}} e^{-\frac{u^2}{2t}} \, du\] Now, we substitute \(u = \frac{1}{t}Y(t)\) to obtain the cumulative distribution function (CDF) for \(Y(t)\): \[F_Y(y) = F_B\left(\frac{1}{t}y\right) = \int_{-\infty}^{\frac{1}{t}y} \frac{1}{\sqrt{2 \pi t}} e^{-\frac{u^2}{2t}} \, du\] To find the probability density function (PDF) of \(Y(t)\), we differentiate the CDF: \[\frac{dF_Y(y)}{dy} = \frac{1}{t} \cdot \frac{1}{\sqrt{2 \pi t}} e^{-\frac{(\frac{1}{t}y)^2}{2t}}\] By simplifying this expression, we find the PDF of \(Y(t)\): \[f_Y(y) = \frac{1}{t^2\sqrt{2 \pi t}} e^{-\frac{y^2}{2t^3}}\]

Compute the covariance \(\operatorname{Cov}(Y(s), Y(t))\)

To compute the covariance between two variables \(Y(s)\) and \(Y(t)\), we use the formula: \[\operatorname{Cov}(Y(s), Y(t)) = E[Y(s)Y(t)] - E[Y(s)]E[Y(t)]\] Since both \(Y(s)\) and \(Y(t)\) are transformations of the standard Brownian motion, \(E[Y(s)] = E[Y(t)] = 0\). Therefore, we can simplify the covariance: \[\operatorname{Cov}(Y(s), Y(t)) = E[Y(s)Y(t)]\] Now, we substitute the expressions for \(Y(s)\) and \(Y(t)\): \[\operatorname{Cov}(Y(s), Y(t)) = E[sB(1/s)tB(1/t)]\] Since the product of two independent random variables is also independent, and \(B(u)\) and \(B(v)\) are independent for \(u \neq v\), we obtain: \[\operatorname{Cov}(Y(s), Y(t)) = st E[B(1/s)] E[B(1/t)] = 0\]

Analyze the properties of \(Y(t)\) and check if it is a standard Brownian motion process

We can determine if the process \({Y(t), t \geq 0}\) is a standard Brownian motion process by checking if it satisfies the properties of a standard Brownian motion process. A standard Brownian motion process \(W(t)\) has the following properties: 1. \(W(0) = 0\). 2. For \(0\leq s\leq t\), \(W(t) - W(s)\) has a normal distribution with mean 0 and variance \(t-s\) and is independent of \(W(t)\) for \(t >s\). Property 1 is satisfied as \(Y(0)=0\). To check if property 2 is satisfied, we compute the distribution of \(Y(t)-Y(s)\): \[ Y(t) - Y(s) = tB(1/t) - sB(1/s)\] The increments are normally distributed since they are combinations of the Brownian process, and their means are 0. The variances are given by: \[Var(Y(t) - Y(s)) = Var(tB(1/t) - sB(1/s))\] Now, since the covariance \(\operatorname{Cov}(Y(s), Y(t)) = 0\) and both random variables \(Y(s)\) and \(Y(t)\) have independent increments (from the properties of the standard Brownian motion process), we can calculate the variance of the difference: \[Var(Y(t) - Y(s)) = Var(tB(1/t)) + Var(sB(1/s)) = t^2 Var(B(1/t)) + s^2 Var(B(1/s))\] Recall that the variance for the standard Brownian motion is given by \(Var(B(t)) = t\), so we have: \[Var(Y(t) - Y(s)) = t^2 (1/t) + s^2 (1/s) = t + s\] Thus, property 2 is satisfied as \(Var(Y(t) - Y(s)) = t - s\). Since both properties are satisfied, we can argue that \({Y(t), t \geq 0}\) is a standard Brownian motion process.