Question

3. Let U-Bt- tB be Brownian bridge on [0, 1], where {BiJosesi is a Brownian process (i) Show E(Ut0 (ii) Show Cov(U,, Ut) s(1- t) for 0 s ts1. (ii) Let Xg(t)B Find functions g and h such that X, has the same covariance as a Brownian bridge.

Answer 1

Since, B_t is a Brownian Process

=> B_t ~ N(0,t)

(i)

$\begin{align*} E(U_t) &= E(B_t - tB_1) \\ &= E(B_t) - tE(B_1) \\ &= 0 - t*0\ \ \ \ \ \ \text{[Since, B}_t \sim N(0,t)] \\ &= 0 \end{align*}$

(ii)

Here we use the following property of a Brownian Motion:

For all $\begin{align*} s \le t \end{align*}$ :

$\begin{align*} Cov(B_s, B_t) &= Cov(B_s, B_s+ B_t - B_s) \\ &= Cov(B_s,B_s) + Cov(B_s,B_t-B_s) \\ \text{Now, the se} &\text{cond term is zero since a Brownian process has independent increments.} \\ \text{Thus, we get:} \\ Cov(B_s,B_t) &= Var(B_s) \\ &= s \end{align*}$Now, consider for $\begin{align*} 0 \le s \le t \le 1 \end{align*}$ :

$\begin{align*} Cov(U_s,U_t) &= Cov(B_s-sB_1, B_t-tB_1) \\ &= Cov(B_s,B_t) - tCov(B_s,B_1) - sCov(B_1,B_t) + stCov(B_1,B_1) \\ &= s - ts-st+st \\ &= s(1-t) \end{align*}$

(iii)

We need to find functions g and h which satisfy:

$\begin{align*} Cov(g(s)B_{h(s)},g(t)B_{h(t)}) &= s(1-t) \ \ \ \ \forall \ 0 \le s \le t \le 1 \end{align*}$

Taking s = t and solving, we get:

$\begin{align*} Cov(g(t)B_{h(t)},g(t)B_{h(t)}) &= t(1-t) \\ \Rightarrow [g(t)]^2 Var(h(t)) &= t(1-t) \\ \Rightarrow [g(t)]^2 h(t) &= t(1-t) \\ \Rightarrow h(t) &= \frac{t(1-t)}{[g(t)]^2}..........................(1) \end{align*}$

Now, consider for arbitrary $0 \le s \le t \le 1$ :

$\begin{align*} Cov(g(s)B_{h(s)},g(t)B_{h(t)}) &= s(1-t) \\ \Rightarrow g(s)g(t) Cov(B_{h(s)},B_{h(t)}) &= s(1-t) \\ \text{Let us consider the case w} &\text{here h() is a monotonic increasing function.} \\ \text{Then, since } s\le t \ \Rightarrow & \ h(s) \le h(t) \ \ \Rightarrow Cov(B_{h(s)},B_{h(t)}) = h(s) \\ \text{Thus, we get:} \\ g(s)g(t)h(s) &= s(1-t) \\ \Rightarrow g(s)g(t) \frac{s(1-s)}{[g(s)]^2} &= s(1-t) \ \ \ \ \ \ \ \text{[Using (1)]} \\ \Rightarrow \frac{g(t)}{g(s)} &= \frac{1-t}{1-s} \end{align*}$

Thus, we can take g(t) = 1 - t

From (1), we get:

$\begin{align*} \textbf{h(t)} &= \frac{t(1-t)}{(1-t)^2} \\ &= \mathbf{\frac{t}{1-t}} \end{align*}$

If we had assumed h() to be a monotonic decreasing function above then we would get another set of functions g and h which would satisfy the covariance equation, but since the question asks for only one set of functions, we can stop here.

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