= Joint Laplace transform of Brownian square and its time integral
{c}
{title2=$\mathbb Ee^{-\alpha B_t^2-(\beta^2/2)\int_0^tB_s^2ds}=[\cosh(\beta t)+(2\alpha/\beta)\sinh(\beta t)]^{-1/2}$}
For standard <Brownian motion> and $\alpha,t\geq0$, $\beta>0$, the title gives the joint <Laplace transform> of the terminal square and the <integrated square of Brownian motion>. At $\beta=0$ the answer is $(1+2\alpha t)^{-1/2}$. To prove the formula, use the <stochastic exponential>
$$
Z_t=\exp\!\left[-\beta\int_0^tB_s\,dB_s-\frac{\beta^2}{2}\int_0^tB_s^2ds\right]
=\exp\!\left[\frac{\beta t}{2}-\frac\beta2B_t^2-\frac{\beta^2}{2}\int_0^tB_s^2ds\right].
$$
On every finite interval $[0,T]$, $0<Z_t\leq e^{\beta T/2}$, so the <bounded local martingale criterion> makes it a <martingale>. The <Girsanov theorem> changes $B$ into the zero-started <Ornstein-Uhlenbeck process> $dB=dW-\beta Bdt$ under the terminal density $Z_t$. Its terminal <variance> is $v=(1-e^{-2\beta t})/(2\beta)$. Multiplying $Z_t$ by $\exp[(\beta/2-\alpha)B_t^2-\beta t/2]$ recovers the integrand on the left of the title. The <Gaussian quadratic exponential moment> then gives $e^{-\beta t/2}[1-(\beta-2\alpha)v]^{-1/2}$, which simplifies to the title formula. At $\alpha=0$ this reduces to the <Laplace transform of the integrated square of Brownian motion> after <Brownian scaling>.
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