Covariance of Brownian bridge increments $\text{I need to prove the following:}\\$
$$\text{Cov}[\left(W(t_{i+1})-W(t_i)\right),\left(W(t_{j+1})-W(t_j)\right)|Z]=    
\begin{cases}
\left(t_{i+1}-t_i\right)-\frac{\displaystyle\left(t_{i+1}-t_i\right)^2}{\displaystyle T} &\text{if  } i=j\\
-\frac{\displaystyle\left(t_{i+1}-t_i\right)\left(t_{j+1}-t_j\right)}{\displaystyle T} &\text{otherwise}
\end{cases}\\
$$ 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. 
 A: 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}
