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.


For the summing over m confusion, look again at the definition of p1,l(n) and how it is a summation. It appears that summing both sides over m, and factoring out the exponent gives the equality.

The second to last statement comes from the fact that when the Brownian particle starts at time t=0, it is at location x=0 with probability one, and the notation uses the kronecker delta to indicate this.

The last statement follows from the definition of p1(m,n) by integrating both sides of the solution.

  • $\begingroup$ 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 $\endgroup$ – strwars Feb 26 at 23:37

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