2.3 The reflection principle

Since $a+n-(t-n)=b$, $n$ is the number of upwards moves and $t-n$ is the number of downward moves needed for a path from $X_0=a$ to $X_t=b$. We can divide $t$ into $n$ upward moves and $t-n$ downward moves and this is same as the number of possible combinations. ∎

Consider a path from from $X_0=a$ to $X_t=b$ for which $X_s=0$ for some $s=1,\mathrm{\dots },t-1$. Let $s*$ be time when $X_s$ is first 0. Then we can map this path onto a path from $Y_0=-a$ to $Y_t=b$ by reflecting the path between time $0$ and $X_{s*}$. Formally we define $Y_s=-X_s$ for $s\le s*$ and $Y_s=X_s$ for $s>s*$.

For example consider a path with $X_0=1$, $X_3=0$ and $X_7=4$:

Now the key idea is that this mapping is a bijection between all paths from $Y_0=-a$ to $Y_t=b$ and all paths from $X_0=a$ to $X_t=b$ that hit the value $X_s=0$ for some $s=1,\mathrm{\dots },t-1$. Hence the number of the two sorts of paths are identical. ∎

To prove this result, first notice that for such a path $X_{2m}=1$. So we wish to count the number of paths from $X_0=1$ to $X_{2m}=1$ that do not have $X_s=0$ for any $s=1,\mathrm{\dots },2m-1$. We count the number of such paths as:

‘the number of paths $X_0=1$ to $X_{2m}=1$ ’ minus ‘the number of paths $X_0=1$ to $X_{2m}=1$ which hit 0 ’

Thus

Using that $N_{2m}^0(1,1)=N_{2m}(-1,1)$ we get

∎