Brownian bridge expectation

Question

Let $X_t=B_t-tB_1$, where $B_t$ represents standard Brownian motion for $0\leq t\leq 1$. For $0\leq s<t\leq 1$, find $$\mathbb{E}(X_s|X_t=1).$$

So, here's what I thought. We have $X_t=1$, so $B_t-tB_1=1$, which means that $B_1=\frac{1}{t}(B_t-1)$. Then, our expression becomes $$\mathbb{E}(B_s-sB_1|B_1=\frac{1}{t}(B_t-1))=\mathbb{E}(B_s|B_1=\frac{1}{t}(B_t-1))-s\mathbb{E}(B_1|B_1=\frac{1}{t}(B_t-1)).$$

The second term is easy to solve, it becomes $-\frac{s}{t}(B_t-1)$. What about the first term? I am suspecting that $\mathbb{E}(B_s|B_1=\frac{1}{t}(B_t-1))=\frac{s}{t}(B_t-1)$ using some kind of binomial, but I can't justify this. Is this approach correct?

Answer 1

Let's decompose $$X_s=B_s -sB_1 = \left( B_s - \frac{s}{t}B_t \right)+ \frac{s}{t}\left(B_t-tB_1 \right)$$

We have $\left( B_s - \frac{s}{t}B_t \right) = \left(1-\frac{s}{t}\right)B_s-\frac{s}{t}(B_t-B_s)$ and $(B_t-tB_1)=(1-t)B_t-t(B_1-B_t)$; so both follow normal distribution. Besides, \begin{align}Cov( B_s - \frac{s}{t}B_t,B_t-tB_1)& =s-ts-\frac{s}{t}t+\frac{s}{t}t\times t= 0 \end{align} Then $\left( B_s - \frac{s}{t}B_t \right)$and $\left(B_t-tB_1 \right)$ are independant. We deduce that $$E(B_s - \frac{s}{t}B_t |B_t-tB_1)=0$$

Hence, \begin{align} E(X_s|X_t) &= E\left( \left( B_s - \frac{s}{t}B_t \right)+ \frac{s}{t}\left(B_t-tB_1 \right) |B_t-tB_1\right)\\ &= E\left( B_s - \frac{s}{t}B_t |B_t-tB_1\right)+E\left(\frac{s}{t}\left(B_t-tB_1 \right)|B_t-tB_1\right)\\ &= \frac{s}{t}(B_t-tB_1) \\ &=\frac{s}{t}X_t \\ \end{align} We deduce the particular case $$E(X_s|X_t = 1) = \frac{s}{t}$$

Answer 2

Can you please check if the following works? I have left it in term of $B_1$

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We will assume a weiner process for the brownian motion with $B_{t+1} - B_t \sim \mathcal N(0,1) \forall t$ \begin{align*} X_t &= B_t-tB_1\\ X_{t-\Delta t} &= B_{t-\Delta t}-(t-\Delta t)B_1\\ X_t - X_{t-\Delta t} &= B_t - B_{t-\Delta t}-\Delta t B_1\\ X_{t-\Delta t} &= X_t + \Delta t B_1 - (B_t - B_{t-\Delta t})\\ \mathbb E [X_{t-\Delta t} |X_t = 1] &= \mathbb E [X_t + \Delta tB_1 - (B_t - B_{t-\Delta t}) |X_t = 1]\\ &= 1 + \Delta t B_1 + \mathbb E (B_t - B_{t-\Delta t}) |X_t = 1]\\ &= 1 + \Delta t B_1 \\ \end{align*} The last equality is by using the fact that the brownian motion is a weiner process with gaussian increments. We can rewrite this in terms of $s$ \begin{align*} \mathbb E [X_{s} |X_t = 1] &= 1+(t-s)B_1 \end{align*}

Answer 3

$(X_s)_{0\le s\le 1}$ is a Gaussian process with $\Bbb E[X_s] =0$ and $\Bbb E[X_sX_t]=s(1-t)$ for $0\le s\le t\le 1$. Consequently, if we choose $c=s/t$ then the covariance of $X_s-cX_t$ and $X_t$ is $0$, so $X_s-cX_t$ and $X_t$ are independent! With this choice, the conditional distribution of $X_s=(X_s-cX_t) +cX_t$, given that $X_t=x$, is normal with mean $cx$ and variance that of $X_s-cX_t$, namely $s(1-s)+c^2t(1-t)-2cs(1-t) = s(1-s/t)$.