Let $B$ be a standard Brownian motion. Define a Brownian bridge $b$ by $b_t=B_t-tB_1$. Let $\mathbb{W'}$ be the law of this process.

According to Wikipedia,

A Brownian bridge is a continuous-time stochastic process B(t) whose probability distribution is the conditional probability distribution of a Wiener process W(t) (a mathematical model of Brownian motion) given the condition that B(0) = B(1) = 0.

Surely it makes no sense to condition on a probability 0 event? So I'm trying to show that $\mathbb{W'}$ is the weak limit as $\epsilon\to 0$ of Brownian motion conditioned upon the event $\{|B_1|\leq \epsilon\}$. How do we prove this?

Thank you.


2 Answers 2


Brownian motion $B_t$ over the interval $[0,1]$ can be decomposed into two independent terms. That is, the process $X_t=B_t-tB_1$ and the random variable $Y=B_1$. As these are joint normal, to prove that they are independent, it is enough to show that the covariance ${\rm Cov}(X_t,Y)={\rm Cov}(B_t,B_1)-t{\rm Var}(B_1)=t-t$ vanishes.

The distribution of $B$ conditional on $\vert B_1\vert < \epsilon$ is just the same as that of $X$ plus the independent process $tY$ (conditioned on $\vert Y\vert < \epsilon$). As $\epsilon$ goes to zero, this converges to the distribution of $X$, which is a Brownian bridge.

  • $\begingroup$ Didn't you mean $Y=tB_1$ (and the equation then yields $t^2-t^2$)? $\endgroup$
    – Mr_3_7
    Jul 17, 2016 at 21:47
  • $\begingroup$ Well, you are correct that we should decompose $B_t=X_t+tB_1$ so I updated the argument to do this. $\endgroup$ Jul 31, 2016 at 19:26

The question says: "Surely it makes no sense to condition on a probability 0 event?"

Suppose $X,Y$ are jointly normally distributed random variables. "Jointly" means every linear combination of them is a normally distributed random variable. Suppose their means and variances are $\mu_X$, $\mu_Y$, $\sigma_X^2$, and $\sigma_Y^2$, and their correlation is $\rho$. It is commonplace to read in textbooks that the conditional distribution of $X$ given the probability-0 event that $Y=y$ is normal with expected value $$E(X\mid Y=y) = \mu_X + \rho\sigma_X\left(\frac{y-\mu_Y}{\sigma_Y}\right)$$ and variance $$(1-\rho^2)\sigma_X^2.$$ So you're conditioning on a probability-$0$ event.

Going a small step further, one can condition on a random variable rather than on an event, and get $$E(X\mid Y) = \mu_X + \rho\sigma_X\left(\frac{Y-\mu_Y}{\sigma_Y}\right),$$ and that is a random variable in its own right. It is the random variable whose expecation one finds in the "law of total expectation" $$ E(E(X\mid Y))=E(X), $$ and the "law of total variance": $$ \operatorname{var}(X) = \operatorname{var}(E(X\mid Y)) + E(\operatorname{var}(X \mid Y)). $$

  • $\begingroup$ What's your definition of $\mathbb{E}(X|Y=y)$? $\endgroup$ Sep 9, 2011 at 1:58
  • 2
    $\begingroup$ hi, considering your argument " So you're conditioning on a probability-0 event.",i think you are not putting very well. In fact, $E(X\vert Y)$ (by Doob-Dynkin's lemma) is a function of $Y$, say $E(X\vert Y) = g(Y)$ then if $Y = y$, then we have the abused notation $E\big( X\vert Y = y \big) = g(y)$ while it does not really mean " conditioning on a 0-probability event"... best regards.. $\endgroup$
    – Chival
    Oct 13, 2014 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.