Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 38 / 1

Write $B=\beta$ for the driving Brownian motion throughout; the two symbols in the displayed formulas refer to the same process. For $t<1$ the deterministic integrand is square-integrable on $[0,t]$, sois well defined. The Itô product rule, with zero quadratic covariation against the deterministic factor, givesFor the other form, apply the same Itô product rule to $B_t/(1-t)$:Multiplying by $1-t$ gives the stochastic integral representation of a Brownian bridgeThe drift coefficient is singular only at the terminal endpoint, so this calculation is performed on compact subintervals of $[0,1)$.

For times $t_1,\dots ,t_n<1$ and real coefficients $a_i$, the linear combination of coordinates isThe Gaussianity of deterministic Brownian stochastic integrals makes this a centered normal random variable. Indeed step-function approximations give linear combinations of independent normal increments, and the Itô isometry gives convergence in $L^2$; their characteristic functions converge to that of a normal variable with the limiting variance. Since every linear combination is normal, the coordinate vector is a Gaussian random vector. Appending the deterministic coordinate $X_1=0$ preserves that property, so $X$ is a centered Gaussian process on $[0,1]$.

For $0\le s\le t<1$, the covariance version of the Itô isometry givesAt $t=1$ the covariance is zero because $X_1=0$, agreeing with the formula. Thus the Brownian bridge covariance kernel isIts finite-dimensional distributions identify it as a Brownian bridge; continuity at the terminal endpoint is established next.

The endpoint value printed in this part is erroneous. The correct conclusion isAlready $\mathbb{E}{X}_t^2=t(1-t)\to 0$ gives convergence in probability to zero and rules out an almost sure limit of one.

To prove the stronger endpoint convergence, use time reversal of a Brownian bridge. Set $Y_t=X_{1-t}$, including $Y_0=X_1=0$. Both $X$ and $Y$ are centered Gaussian processes, and for $s\le t$,They therefore have the same finite-dimensional distributions. The path of $X$ is continuous at zero, since its stochastic integral is continuous there. In particular its values at positive rational times tend to zero at zero with probability one. Equality of the laws on the countable rational coordinates transfers this event to $Y$. The path of $Y$ is continuous at every time in $(0,1]$, so its supremum on any interval bounded away from zero is determined by rational times. The rational endpoint condition consequently implies $Y_t\to 0$ as $t\downarrow 0$ along all real times. Hence $X_t\to 0$ as $t\uparrow 1$, and setting $X_1=0$ gives a continuous version on the whole closed interval. This proof uses equality of coordinate laws, without assuming terminal continuity in advance.