I am self-learning introductory stochastic calculus from the text, A first course in stochastic calculus, by L.P.Arguin. Exercise 2.6 asks to find the PDF of a brownian bridge.
We have:
Definition (Brownian Bridge). The Brownian Bridge is a Gaussian process $(Z_t)_{t\in[0,1]}$ such that, if $(B_t)$ is a brownian motion on $[0,1]$, then
$$Z_t = B_t – tB_1$$
I would like for someone to verify, if my solution to this exercise problem is correct.
Problem. Let $(M_{t}),t\in[0,1]$ be a Brownian bridge.
(a) Write down the PDF of $(M_{1/4},M_{3/4})$.
My Solution attempt.
Firstly, $(M_t)$ is a Gaussian process. By definition of a Gaussian process, for any finite set of times $t_1<\ldots<t_n$, the random vector $(M_{t_1},\ldots,M_{t_n})$ is Gaussian (jointly normal).
So, $(M_{1/4},M_{3/4})$ is jointly normal.
The expectation of the random variable $M_{t}$ is given by:
\begin{align}
\mathbb{E}M_{t} =\mathbb{E}B_{t}-t\mathbb{E}B_{1}
=0
\end{align}
The variance of the random variable $M_{t}$ is given by:
\begin{align}
\mathbb{E}M_{t}^{2}&=\mathbb{E}\left[(B_{t}-tB_{1})^{2}\right] \\
&=\mathbb{E}[B_{t}^{2}-2tB_{t}B_{1}+t^{2}B_{1}^{2}]\\
&=t – 2t \cdot t+t^{2}\\
&=t(1-t)
\end{align}
So, $\mathbb{E}M_{1/4}^{2}=\mathbb{E}M_{3/4}^{2}=\frac{3}{16}$.
The covariance of the random variables $M_{s}$ and $M_{t}$ where $0<s\leq t\leq 1$ is given by:
\begin{align}
\mathbb{E}(M_{s}M_{t})&=\mathbb{E}\left[(B_{s}-sB_{1})(B_{t}-tB_{1})\right]\\
&=\mathbb{E}\left[B_{s}B_{t}-tB_{s}B_{1}-sB_{t}B_{1}+stB_{1}^{2}\right]\\
&=s-st-st+st\\
&=s(1-t)
\end{align}
So, $\mathbb{E}M_{1/4}M_{3/4}=\frac{1}{16}$. So, the covariance matrix $C$ of the random vector $(M_{1/4},M_{3/4})$ is :
\begin{align}
C=\left[\begin{array}{cc}
3/16 & 1/16\\
1/16 & 3/16
\end{array}\right]
\end{align}
The inverse of $C$, by performing row operations on the augmented matrix $[C|I]$ is:
\begin{align}
C^{-1}=\left[\begin{array}{cc}
6 & -2\\
-2 & 6
\end{array}\right]
\end{align}
Also, $|\det C| = \frac{1}{32}$. So, the joint PDF of $(M_{1/4},M_{3/4})$ is:
$$f_{(M_{1/4},M_{3/4})}(x_1,x_2)=\frac{1}{\sqrt{(2\pi)^2 (1/32)}}\exp\left[-\frac{1}{2}(6x_1^2 – 4x_1 x_2 + 6x_2^2)\right]$$
Best Answer
Yes, that all looks reasonable.
As an alternative approach, consider $X_a=B_{1/4}-B_{0}$, $X_b=B_{3/4}-B_{1/4}$, $X_c=B_{1}-B_{3/4}$
so $X_a\sim N(0,\frac 14), X_b\sim N(0,\frac 12), X_c\sim N(0,\frac 14)$ independent normal distributions
and thus $M_{1/4}=\frac34X_a-\frac14X_b-\frac14X_c$ and $M_{3/4}=\frac14X_a+\frac14X_b-\frac34X_c$ are jointly bivariate normal with mean $(0,0)$ and the covariance matrix (I think $\Sigma$ is more usual notation than $C$) and its inverse as you found.