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I learned that Brownian bridge can be used to find the conditional expectation of a Brownian motion conditioning on the end point. For example, suppose $W(t)$ is a standard Wiener process starting from $0$. I want to calculate the conditional expectation $E[W(t)|W(T)=w]$ where $t<T$. Consider Brownian bridge $W(t)-\frac{t}{T}W(T)$, it is independent with $W(T)$, thus $$ E[W(t)|W(T)=w]=E[\frac{t}{T}W(T)|W(T)=w] = \frac{t}{T}w $$. And since the conditional distribution of multivariate normal distribution is also normal distribution, $(W(t)|W(T)=w)$ is normal. The expectation is already calculated, which is $\frac{t}{T}w$, but in order to calculate the conditional variance, I need to calculate $E[W^2(t)|W(T)=w]$. Is there a way to compute it using Brownian bridge? Or any other easy way, without computing the joint density?

Thank you!

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  • $\begingroup$ Oh yes! Thank you so much! $\endgroup$ – Edward Wang Jan 27 '18 at 19:01
  • $\begingroup$ Exactly the same approach works: decompose $$W(t)^2=(t/T)^2W(T)^2+2(t/T)W(T)U+U^2$$ with $$U=W(t)-(t/T)W(T)$$ and compute $$E(W(T)^2\mid W(T))\qquad E(W(T)U\mid W(T))\qquad E(U^2\mid W(T))$$ $\endgroup$ – Did Jan 27 '18 at 19:01
  • $\begingroup$ Yeah you don't need to edit. I understand your idea actually. $\endgroup$ – Edward Wang Jan 27 '18 at 19:02

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