Anyone has an idea how to constract a Master equation for the Brownian bridge?
In the form of: W[i+1]=W[i]+Pr*DeltaX-Pl*DeltaX where Pr (Pl) is the prob. to jump right (left) and DeltaX is the length of the jump (i.e Constant).
Pl=1-Pr.
What is Pr ?
Anyone has an idea how to constract a Master equation for the Brownian bridge?
In the form of: W[i+1]=W[i]+Pr*DeltaX-Pl*DeltaX where Pr (Pl) is the prob. to jump right (left) and DeltaX is the length of the jump (i.e Constant).
Pl=1-Pr.
What is Pr ?
For each $N\geqslant1$, the discrete Brownian bridge of length $2N$ starting and ending at $0$ is a process $(X_n)_{0\leqslant n\leqslant 2N}$ with integer values such that $X_0=X_{2N}=0$ and, for every $0\leqslant n\leqslant 2N-1$, $|X_{n+1}-X_n|=1$. Thus, for each $0\leqslant n\leqslant 2N$, the support of the distribution of $X_n$ is $S_n=\{-s_n,-s_n+2,\ldots,s_n-2,s_n\}$, where $$ s_n=\min\{n,2N-n\}. $$ In other words, the path $(X_n)_{0\leqslant n\leqslant 2N}$ is confined to the square of vertices $(0,0)$, $(N,N)$, $(2N,0)$ and $(N,-N)$. The transition probabilities at time $n$ are $$ \mathbb P(X_{n+1}=x+1\mid X_n=x)=p_n(x), $$ and $$ \mathbb P(X_{n+1}=x-1\mid X_n=x)=1-p_n(x), $$ where, for every $|x|\leqslant2N-n$, $$ p_n(x)=\frac12\left(1-\frac{x}{2N-n}\right). $$ The process $(X_n)_{0\leqslant n\leqslant 2N}$ uses the transition probabilities $p_n(x)$ only for $x$ in $S_n$. For each $n\geqslant N$, $p_n(x_n)=0$ and $p_n(-x_n)=1$, as the constraint that the path stays in the square mentioned above indicates (the constraint on the part $n\leqslant N$ being automatically satisfied by the paths of $(Y_n)_{n}$).
To prove the formula for $p_n(x)$, recall that $(X_n)_{0\leqslant n\leqslant2N}$ is distributed like a standard random walk $(Y_n)_{0\leqslant n\leqslant2N}$ conditionally on $Y_{2N}=0$, hence, for every $x$ in $S_n$, one is looking for $$ p_n(x)=\mathbb P(Y_{n+1}=x+1\mid Y_n=x,Y_{2N}=0)=\frac{\mathbb P(Y_{2N}=0\mid Y_{n+1}=x+1)}{2\cdot\mathbb P(Y_{2N}=0\mid Y_{n}=x)}. $$ This is $$ p_n(x)=\frac{\mathbb P(Y_{2N-n-1}=x+1)}{2\cdot\mathbb P(Y_{2N-n}=x)}=\frac{{2N-n-1\choose N+\frac12(x-n)}}{{2N-n\choose N+\frac12(x-n)}}=\frac{N-\frac12(x+n)}{2N-n}, $$ which is equivalent to the formula above.