Ruin probability

The aim of this note is to study the behaviour of the gambler’s ruin, a finite state Markov chain with absorting state at its ends.

Def Gambler’s ruin chain. Let $(X_n{)}_{n\ge 0}$ be the MC with state space $S=\{0,1,\dots ,N\}$ and transition probabilities $P(X_{n+1}=k+1|X_n=k)=p$ and $P(X_{n+1}=k-1|X_n=k)=q$, with the boundary conditions $P(X_{n+1}=0|X_n=0)=P(X_{n+1}=N|X_n=N)=1$.

The value $X_n$ is the fortune of the player, if it reaches the state $0$ he is ruined!

The following are interesting questions:

1. What is the probability of ruin starting from $X_0=i$.
2. What is the average lenght of the game?

Let’s define the event ruin:

$$R:=\bigcup _{n\ge 0}\{X_n=0\}$$

which is the same as $h_i:=P(H^0<\infty |X_0=i)$ using the notation in Hitting probability and mean hitting times.

Using Markov property we obtain the following system of equations:

$$\{\begin{array}{l}{h}_0=1\\ h_1=q{h}_0+p{h}_2\\ \dots \\ h_n=q{h}_{n-1}+p{h}_{n+1}\\ \dots \\ h_N=0\end{array}$$

Solving RHEs via ansatz §

we solve this systems of linear equations from an ansatz:

$$h_k=C{a}^k$$

substituting we get

$$C{a}^n=qC{a}^{n-1}+pC{a}^{n+1}$$

assuming $C\ne 0$

$$a=q+p{a}^2{a}^2-\frac{1}{p}a+\frac{q}{p}=0$$

solving for $a$

$$\frac{1}{2p}\pm \frac{1}{2}\sqrt{\frac{1}{p^2}-4\frac{q}{p}}$$ $$\frac{1}{2p}(1\pm \sqrt{1-4qp})$$

we get two values for $a=\{1,\frac{q}{p}\}$. Since we are solving linear equations the general solution has the form

$$h_k=C_1{a}_1^k+C_2{a}_2^k=C_1+C_2{\left(\frac{q}{p}\right)}^k$$

let’s find $C_1$ and $C_2$ using the boundary conditions

$$\{\begin{array}{l}{h}_0=C_1+C_2=1\\ h_k=C_1+C_2{\left(\frac{q}{p}\right)}^N=0\end{array}$$ $$(1-C_2)+C_2{\left(\frac{q}{p}\right)}^N=0$$ $$C_2=\frac{1}{1-{\left(\frac{q}{p}\right)}^N}{C}_1=1-\frac{1}{1-{\left(\frac{q}{p}\right)}^N}=\frac{-{\left(\frac{q}{p}\right)}^N}{1-{\left(\frac{q}{p}\right)}^N}$$

and the solution we were looking for is

$$h_k=\frac{{\left(\frac{q}{p}\right)}^k-{\left(\frac{q}{p}\right)}^N}{1-{\left(\frac{q}{p}\right)}^N}$$

Remark In the limit $N\to \infty$ the ruin probability is simply a geometric random variable.

For the symmetric case $p=q=1/2$ this is not enough, since we only get $a=1$. First note that in the symmetric case, our difference equations is a discretization of the laplace equation:

$$h_n=\frac{h_{n+1}+h_{n-1}}{2}$$

rearranging

$$h_{n+1}-2h_n+h_{n-1}=0$$

which is the finite difference formula for the second derivative. Linear functions solve laplace equation in $1$d, so an ansatz is

$$g(k)=C_{2}k$$

and a general solution is

$$h_k=C_1+C_{2}k$$

using the boundary conditions we get

$$h_k=1-\frac{k}{N}$$