Gambler's ruin: $\mathbb{P}\left\{ X_\tau=n \mid X_1 = k+1\right\}=\mathbb{P}\left\{ X_\tau = n\mid X_0 = k+1\right\}\,?$

Question

I don't understand the last part. How do we get $p_{k+1}$ and $p_{k-1}$? I understand it intuitively but I'm have trouble writing things down. It seems like we define $\tau$ as

$$\tau :=\min\left\{t\ge 0:X_t=0\text{ or } X_t = n\right\}$$

And I think $p_k$ is

$$\mathbb{P}\left\{X_\tau=n \mid X_0=k \right\}$$

Let $\Delta_i$'s be Bernoulli with $p=0.5$. I have

\begin{align*} \mathbb{P}\left\{ X_\tau = n\mid X_0=k\right\}&= \frac{1}{2} \mathbb{P}\left\{ X_\tau=n \mid X_1 = k+1\right\} +\frac{1}{2} \mathbb{P}\left\{ X_\tau=n \mid X_1 = k-1\right\} \end{align*}

But not the quesiton is, why is it true that

$$\mathbb{P}\left\{ X_\tau=n \mid X_1 = k+1\right\}=\mathbb{P}\left\{ X_\tau = n\mid X_0 = k+1\right\}\,?$$

Answer 1

Note that at a certain step $p_k$, by definition, represent the probability that the gambler reach fortune $n$ with $k$ dollars, thus since at the next step we have only two possibilities:

• the gambler win and earns $k+1$ dollars
• or the gambler loose and earns $k-1$ dollars

for the law of total probability we have

$$p_k=\frac12 p_{k-1}+\frac12 p_{k+1}$$