# Probability of a Large Brownian Particle being at a Certain Position at a Certain Time

I am currently trying to follow along with my notes from a lecture and I am getting very lost in my professor's solution for determining the probability a Brownian particle sits at a specific position $$x=ma$$ at time $$t=n\tau$$.

To do this, he states that in order for the particle to be at $$x=ma$$ at time $$t=n\tau$$, it must have been at either $$x=(m-1)a$$ or $$x=(m+1)a$$ at previous time $$t=(n-1)\tau$$. With this in mind, and specifying that the particle has equal probability of going one way or the other, he comes to

$$p_1(m,n)=\frac{1}{2}p_1(m-1,n-1)+\frac{1}{2}p_1(m+1,n-1)$$

So far I understand this. He then seeks a solution to the above formula through using the Fourier transform method on the probability such that

$$p_{1,l}(n)=\sum^{\infty}_{m=-\infty}p_1(m,n)*e^{-ilm}$$

for $$-\pi \leq l \leq \pi$$ such that

$$p_1(m,n)=\frac{1}{2\pi}\int^{+\pi}_{-\pi}p_{1,l}(n)*e^{ilm}*dl$$

where $$\frac{1}{2\pi}\int^{+\pi}_{-\pi}e^{il(m-m')}dl=\delta_{m,m'}$$. I can follow along with this as well.

Where I get lost is he states that by multiplying the formula for $$p_{1}(m,n)$$ by $$e^{-ilm}$$ and summing over $$m$$ we achieve

\begin{align} p_{1,l}(n)=\sum_{m=-\infty}^\infty p_{1}(m,n)*e^{-ilm}&=\sum_{m=-\infty}^\infty \Bigg[\frac{1}{2}p_1(m-1,n-1)+\frac{1}{2}p_1(m+1,n-1)\Bigg]e^{-ilm}\\ &=\frac{e^{-il}+e^{il}}{2}p_{1,l}(n-1)\\ &=cos(l)*p_{1,l}(n-1) \end{align}

I've been trying to replicate this, but to no avail. I've currently done the following:

\begin{align} p_1(n)=\sum_{m=-\infty}^\infty p_1(m,n)*e^{-ilm} &= \sum_{m=-\infty}^\infty \Bigg[\frac{1}{2}p_1(m-1,n-1)+\frac{1}{2}p_1(m+1,n-1)\Bigg]*e^{-ilm} \\ &= \sum_{m=-\infty}^\infty \frac{1}{2}\frac{1}{2\pi}\Bigg[ \int^{+\pi}_{-\pi}p_{1,l}(n-1)*e^{il(m-1)}*dl +\int^{+\pi}_{-\pi}p_{1,l}(n-1)*e^{il(m+1)}*dl \Bigg]*e^{-ilm} \\ &= \frac{1}{2}\frac{1}{2\pi}\sum_{m=-\infty}^\infty \Biggr[ \int^{+\pi}_{-\pi} \bigg[\big(e^{ilm}e^{-il} + e^{ilm}e^{il}\big) p_{1,l}(n-1)*dl\bigg] \Biggr]*e^{-ilm} \\ &= \frac{1}{2}\frac{1}{2\pi} \sum_{m=-\infty}^\infty \Biggr[ \int^{+\pi}_{-\pi} \bigg[\big(e^{-il} + e^{il}\big) p_{1,l}(n-1)*dl\bigg] \Biggr] \end{align}

And that's where I get stuck. I know that $$\frac{e^{-il} + e^{il}}{2}=cos(l)$$, but am unsure how I'd be able to pull this out.

Taking $$p_{1,l}(n)=cos(l)*p_{1,l}(n-1)$$ as a fact, it is said that iterating this relationship yields

$$p_{1,l}(n)=\Big[cos(l)\Big]^{n}*p_{1,l}(0)$$

Which makes sense. From here, and using initial conditions that at $$t=0$$ the Brownian particle was at position $$x=0$$, we're told

$$p_{l,0}(0)=\sum^{+\infty}_{m=-\infty}\delta_{m,0}*e^{-ilm}=1$$

and that by using this we can come to the solution

$$p_{l}(m,n)=\frac{1}{2\pi}\int^{+\pi}_{-\pi}\big[cos(l)\big]^n*cos(lm)*dl$$

Neither of these last 2 statements make sense to me either, but that could be because I am stuck at the spot I mentioned previously. I just included it in case anyone had more insight on the matter. Apologies if this is obvious and thank you for any assistance.

• Thank you for the reply @RichardG , I'm just confused by how that exponential would come out of the integral and how that summation to get $p_1(n)$ can occur when the $e^{ilm}$ and ${e^-ilm}$ would cancel each other out I would think – strwars Feb 26 at 23:37