Variance of Brownian Bridge

Question

With $W(t)$ denoting the standard Wiener process, Brownian Bridge $B(t)$ can be defined on $t \in [0,T]$ as:

$$B(t):=W(t)-\frac{t}{T}W(T)$$

Computing the expectation gives $\mathbb{E}[B(t)]=\mathbb{E}[W(t)]-\frac{t}{T}\mathbb{E}[W(T)]=0-\frac{t}{T}0=0.$ From this we can deduce that $Var(B(t))=\mathbb{E}[B(t)^2]$. Here is my attempt:

$$\mathbb{E}[B(t)^2]=\mathbb{E}\left[\left(\sqrt{t}W(1)-\frac{t}{\sqrt{T}}W(1)\right)^2\right]=\mathbb{E}[W(1)^2]\left(\sqrt{t}\left(1-\sqrt{\frac{t}{T}}\right)\right)^2=\\=t\left(1-2\frac{\sqrt{t}}{\sqrt{T}}+\frac{t}{T}\right)=t\left(\frac{T-2\sqrt{t}\sqrt{T}+t}{T}\right)=\frac{t}{T}\left(\sqrt{T}-\sqrt{t}\right)^2$$

Now wikipedia gives the variance as $\frac{t}{T}(T-t)$, so I must be making a mistake somewhere, but I cannot figure out where? Any hints would be greatly appreciated,

Answer 1

Although $W(t)$ and $\sqrt{t}W(1)$ have the same distribution, they are not equal. The correct computation is $$\mathbb{E}[B(t)^2] = \mathbb{E}[(W(t) - (t/T)W(T))^2]$$ $$= \mathbb{E}[W(t)^2] - 2 \frac{t}{T} \mathbb{E}[W(t) W(T)] + \left(\frac{t}{T}\right)^2 \mathbb{E}[W(T)^2]$$

We know $\text{Var}(W(t)) = \mathbb{E}[W(t)]^2 = t$ and $$\text{Cov}(W(t), W(T)) = \mathbb{E}[W(t)W(T)] = \min(t,T) = t$$ Thus, we conclude $$\mathbb{E}[B(t)^2] = t - 2 \frac{t}{T} t + \left(\frac{t}{T}\right)^2 T = \frac{t}{T} (T-t)$$ as on Wikipedia.

Answer 2

Notice that $\forall t \in [0,T]$ we have that: $min\left(t,\frac{t^2}{T}\right)=\frac{t^2}{T}$. Furthermore, I will make use of the following three lemmas (which I leave to prove as exercise):

• Lemma 1: $cov(W_t,W_s)=min(s,t)$

• Lemma 2: $Var(W_t - W_s)=Var(W_t)+Var(W_s)-2Cov(W_s,W_t)$

• Lemma 3: $\frac{t}{T}W(T)=W\left(\frac{t^2}{T}\right)$ in Distribution

We then have:

$$Var(B(t))=Var\left(W_t-\frac{t}{T}W_T\right)=Var(W_t)+Var\left(W\left(\frac{t^2}{T}\right)\right)-2Cov\left(W_t,W_{\frac{t^2}{T}}\right)=\\=t+\frac{t^2}{T}-2min\left(t,\frac{t^2}{T}\right)=t+\frac{t^2}{T}-2\frac{t^2}{T}=\\=\frac{t(T-t)}{T}$$