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I was asked to show that the Chapman-Kolmogorov equation for Brownian motion $B(.)$, that is $$p(0,0,t,x)= \int_{-\infty}^{\infty} p(s,y,t,x)p(0,0,s,y) dy$$ where $p(s,y,t,x)$ is the transitional probability of Brownian motion.

Since $$p(s,y,t,x)=P(B(t)=x|B(s)=y)= \frac{1}{\sqrt{2\pi(t-s)}} e^{-\frac{(x-y)^2}{2(t-s)}}$$ In other words, I need to show $$\frac{1}{\sqrt{2\pi t}} e^{-\frac{x^2}{2t}} = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi (t-s)}} e^{-\frac{(x-y)^2}{2(t-s)}} \frac{1}{\sqrt{2\pi s}} e^{-\frac{y^2}{2s}} dy$$

I simplified the integral into $$\begin{align} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi (t-s)}} e^{-\frac{(x-y)^2}{2(t-s)}} \frac{1}{\sqrt{2\pi s}} e^{-\frac{y^2}{2s}} dy & = \frac{1}{2\pi \sqrt{s(t-s)}} \int_{-\infty}^{\infty} e^{-\frac{(x-y)^2}{2(t-s)}} e^{-\frac{y^2}{2s}} dy \\ & = \frac{1}{2\pi \sqrt{s(t-s)}} \int_{-\infty}^{\infty} e^{-\frac{s(x-y)^2+y^2(t-s)}{2s(t-s)}} dy \\ & = \frac{1}{2\pi \sqrt{s(t-s)}} \int_{-\infty}^{\infty} e^{-\frac{sx^2-2sxy+ty^2}{2s(t-s)}} dy \end{align}$$

But then I am stuck with the exponential in the integral. Any ideas to integrate that?

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  • $\begingroup$ Have you tried to write the argument of the exponential as a remarkable product? $\endgroup$ – Daddy Oct 1 '16 at 10:32
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We first write $$ \exp\left(-\frac{(x-y)^2}{2(t-s)}\right) \exp\left( -\frac{y^2}{2s}\right) = \exp \left(-\frac{x^2}{2t}\right) \exp \left(\frac{t}{2s(s-t)} \left( y-\frac{xs}{t} \right)^2\right) \, , $$ and then simply integrate the resulting integral to obtain $$ \frac{1}{\sqrt{2\pi t}} \, \exp \left(-\frac{x^2}{2t}\right) \, . $$

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