Covariance of Brownian bridge increments

Question

$\text{I need to prove the following:}\\$
$$\text{Cov}[\left(W(t_{i+1})-W(t_i)\right),\left(W(t_{j+1})-W(t_j)\right)|Z]= \begin{cases} \left(t_{i+1}-t_i\right)-\frac{\displaystyle\left(t_{i+1}-t_i\right)^2}{\displaystyle T} &\text{if } i=j\\ -\frac{\displaystyle\left(t_{i+1}-t_i\right)\left(t_{j+1}-t_j\right)}{\displaystyle T} &\text{otherwise} \end{cases}\\ $$ where $ W(t)$ is a standard Brownian motion and $Z=\frac{W_T}{\sqrt{T}}$ is an $\mathcal{F}_T$-measurable standard Gaussian random variable. My guess is that this could be done using some results for the Brownian bridge. Also, I was searching for the answer to this question on this website and there are some hints you can find, but somehow I cannot come to the endpoint. I would appreciate any help.

Answer 1

Let $\widetilde{W}(t)=W(t)-\frac{t}TW(T)$, since $\{\widetilde{W}(t),\,0\le t\le T\}$ is a gaussian process and uncorrelated with $Z=\frac{W(T)}{\sqrt{T}}$, therefore $\{\widetilde{W}(t),\,0\le t\le T\}$ is independent with $Z$ too and \begin{gather} \mathsf{E}[\widetilde{W}(t)|Z]=0,\quad \mathsf{E}[\Delta\widetilde{W}(t_i)|Z]\stackrel{\text{def}}=\mathsf{E}[\widetilde{W}(t_{i+1})-\widetilde{W}(t_i)|Z]=0, \quad \mathsf{E}[\Delta W(t_i)|Z]=\frac{\Delta t_i}{\sqrt{T}}Z.\\ \mathsf{E}[\widetilde{W}(s)\widetilde{W}(t)]=\mathsf{E}[W(s)W(t)]-\frac{st}{T^2}\mathsf{E}[W^2(T)]=s\wedge t-\frac{st}T.\\ \begin{aligned} \mathsf{cov}[\Delta W(t_i),\Delta W(t_j)|Z] &=\mathsf{cov}[\Delta\widetilde{W}(t_i),\Delta\widetilde{W}(t_j)|Z] =\mathsf{E}[\Delta\widetilde{W}(t_i)\Delta\widetilde{W}(t_j)]\\ &=\begin{cases} \Delta t_i-\dfrac{(\Delta t_i)^2}{T}, & i=j,\\ -\dfrac{\Delta t_i\Delta t_j}{T}, &i\ne j. \end{cases} \end{aligned} \end{gather}

Some complements of above expressions: 1. Since $\mathsf{E}[W(T)]=0$ and $$\mathsf{E}[\widetilde{W}(t)]=\mathsf{E}\Bigl[W(t)-\frac{t}TW(T)\Bigl]=\mathsf{E}[W(t)]-\frac{t}T\mathsf{E}[W(T)]=0,$$ then $$\mathsf{cov}[\widetilde{W}(t),Z]=\mathsf{E}[\widetilde{W}(t)Z]-\mathsf{E}[\widetilde{W}(t)]\mathsf{E}{Z}=\mathsf{E}[W(t)Z]-\frac{t}{\sqrt{T}}\mathsf{E}[Z^2] =\frac{t}{\sqrt{T}}-\frac{t}{\sqrt{T}}=0.$$ Therefore, $\{\widetilde{W}(t),0\le t\le T\}$ and $Z$ are uncorrelated. Meanwhile, $\{\widetilde{W}(t),0\le t\le T,Z=\frac{W(T)}{\sqrt{T}}\}$ is Gaussian, So $\{\widetilde{W}(t),0\le t\le T\}$ and $Z$ are independent too.

4. Since $\Delta W(t_i)-\mathsf{E}(\Delta W(t_i)|Z)=\Delta W(t_i)-\frac{\Delta t_i}{\sqrt{T}}Z=\Delta\widetilde{W}(t_i)$, therefore \begin{align} \mathsf{cov}[\Delta W(t_i),\Delta W(t_j)|Z] &=\mathsf{E}[(\Delta W(t_i)-\mathsf{E}(\Delta W(t_i)|Z))(\Delta W(t_j)-\mathsf{E}(\Delta W(t_j)|Z))|Z]\\ &=\mathsf{E}\Bigl[\Bigl(\Delta W(t_i)-\frac{\Delta t_i}{\sqrt{T}}Z\Bigr)\Bigl(\Delta W(t_j)-\frac{\Delta t_j}{\sqrt{T}}Z\Bigr)\Bigr] \\ &=\mathsf{E}[\Delta\widetilde{W}(t_i)\Delta\widetilde{W}(t_j)|Z]=\mathsf{E}[\Delta\widetilde{W}(t_i)\Delta\widetilde{W}(t_j)]. \end{align} Similarly, since $\mathsf{E}[\Delta\widetilde{W}(t_i)|Z]=0$ and $$ \mathsf{cov}[\Delta\widetilde W(t_i),\Delta\widetilde W(t_j)|Z]=\mathsf{E}[\Delta\widetilde W(t_i) \Delta\widetilde W(t_j)|Z]=\mathsf{E}[\Delta\widetilde W(t_i) \Delta\widetilde W(t_j)]. $$ 5. For $i<j$ (i.e. $i+1\le j$), \begin{align} \mathsf{E}[&\Delta\widetilde{W}(t_i)\Delta\widetilde{W}(t_j)] =\mathsf{E}[(\widetilde{W}(t_{i+1})-\widetilde{W}(t_i))(\widetilde{W}(t_{j+1})-\widetilde{W}(t_j)]\\ &=(t_{i+1}\wedge t_{j+1}-t_i\wedge t_{j+1})-\frac{(t_{i+1}-t_i)t_{j+1}}{T} -(t_{i+1}\wedge t_j -t_i\wedge t_j)-\frac{(t_{i+1}-t_i)t_{j}}{T}\\ &=(t_{i+1}-t_i)-(t_{i+1}-t_i)-\frac{(t_{i+1}-t_i)(t_{j+1}-t_j)}{T}\\ &=-\frac{\Delta t_i\Delta t_j}T. \end{align}