91Ó°ÊÓ

According to the properties of the Brownian motion, X(s) and X(t) are normally distributed with their respective mean and variance based on the time and the coefficients. Therefore, \(X(s) \sim N(\mu s, \sigma^2 s) \) \(X(t) \sim N(\mu t, \sigma^2 t) \)

We will define the increments of the Brownian motion as follows: \(W(t) = X(t) - X(s), s<t \) Because the increments of the Brownian motion are stationary and independent, W(t) will also follow a normal distribution with a mean and variance proportional to the time (t-s), and coefficients µ and σ². Hence, \(W(t) \sim N(\mu(t - s), \sigma^2(t - s)) \)

The joint distribution of two normally distributed random variables \(X(s)\) and \(X(t)\) can be given by their bivariate normal distribution with a mean vector (\( \mu s, \mu t\)) and covariance matrix, shown below: \[Cov = \begin{pmatrix} \sigma^2 s & cov(X(s), X(t)) \\ cov(X(s), X(t)) & \sigma^2 t \end{pmatrix} \] To further analyze the joint distribution, we need to find the covariance between \(X(s)\) and \(X(t)\). However, from Step 2 above, we know that \(W(t) = X(t) - X(s)\) is independent from \(X(s)\) and follows its own normal distribution. Hence, the covariance between \(X(s)\) and \(X(t)\) is: \(cov(X(s), X(t)) = cov(X(s), X(s) + W(t)) = cov(X(s), X(s)) = \sigma^2 s\) With this covariance, our covariance matrix becomes: \[Cov = \begin{pmatrix} \sigma^2 s & \sigma^2 s \\ \sigma^2 s & \sigma^2 t \end{pmatrix} \]

Now that we have determined the mean vector and covariance matrix for the bivariate normal distribution, we can write down the joint density function: \(f_{X(s), X(t)}(x_s, x_t) = \frac{1}{2\pi\sigma^2 s\sqrt{1-t/s}} \exp \Big[-\frac{1}{2\sigma^2 s(1-t/s)} \big([x_s - \mu s]^2 - 2[x_s - \mu s][x_t - \mu t](t/s) + [x_t - \mu t]^2\big) \Big]\) This is the joint density function of X(s) and X(t) for a Brownian motion with drift coefficient µ and variance parameter σ², with s<t.