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I need to prove the relation between Brownian motion and diffusion equation. To do this we have to differentiate the density function of Brownian motion once with respect to t and twice with respect to y.

the density function of brownian motion is given by $p(t,x,y)=\frac{1}{\sqrt{2 \pi t}}e^{\frac{-(y-x)^2}{2t}}$

the diffusion equation which is given by $\frac{\partial p}{\partial t}=\frac{1}{2}\frac{\partial^2p}{\partial y^2}$,

However i couldn't verify this. there is something wrong in my way! any help please

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If you want somebody to spot your mistake, you must show your work. –  Did Aug 5 '12 at 20:04
ok , i considered the density function as product of two functions. so the derivative is equal to the first function times the derivative of the second, plus the second function times the derivative of the first \frac{1}{\sqrt{2\pi t}} (\frac{-1}{2t^2}e^{\frac{-(y-x)^2}{2t}}+\frac{-1}{2t\sqrt{2\pi t}}e^{\frac{(y-x)^2}{2t}} but i am not sure about it. it seems wrong ! –  Ruby Aug 5 '12 at 21:30
In the first part of the derivative, you missed a factor $(y-x)^2$ when differentiating $-(y-x)^2/(2t)$ in the exponential (and there is probably a sign error). Try to correct these and see if there is still a problem. –  Did Aug 5 '12 at 21:34
oh, thank you a lot. indeed i missed -(y-x)^2. however, the problem still there. is the derivative of the first term right ? –  Ruby Aug 5 '12 at 21:58
Once again: show your work, as completely as possible, including its full details in your post. –  Did Aug 5 '12 at 22:05

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