For the homework, to turn in:
This graph was produced with the Maple command plot(subs(T0 = 10, a = 20, p = 1-q, Ruin), q = 0 .. 1) where the expression Ruin was defined with Ruin := ((q/p)^a-(q/p)^T0)/((q/p)^a-1)
This graph was produced with the Maple command plot(subs(a = 20, p = .45, q = .55, Ruin), T0 = 0 .. 20)
This graph was produced with the Maple command plot(subs(a = 20, p = .55, q = .45, Ruin), T0 = 0 .. 20)
Use “first step analysis” to write three equations in three unknowns (with two additional boundary conditions) that give the probability that the gambler will be ruined. Solve the equations to find the ruin probability.
Solution: The first step equations for the ruin are:
$$\begin{array}{llll}\hfill {q}_{4}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{3}& =\left(3\u22154\right){q}_{2}+\left(1\u22154\right){q}_{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{2}& =\left(1\u22152\right){q}_{1}+\left(1\u22152\right){q}_{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{1}& =\left(1\u22154\right){q}_{0}+\left(3\u22154\right){q}_{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{0}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$The solution of the equations is: ${q}_{1}=5\u22158$, ${q}_{2}=1\u22152$, ${q}_{3}=3\u22158$.
Solution: Writing the set of first-step equations:
$$\begin{array}{llll}\hfill {q}_{16}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{12}& =0.6{q}_{8}+0.4{q}_{16}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{10}& =0.6{q}_{8}+0.4{q}_{12}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{8}& =0.6{q}_{6}+0.4{q}_{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{6}& =0.6{q}_{4}+0.4{q}_{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{4}& =0.6{q}_{0}+0.4{q}_{8}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {q}_{0}& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}& \hfill \end{array}$$Rewrite the equations as:
$$\begin{array}{llll}\hfill 0.6{q}_{8}-{q}_{12}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 0.6{q}_{8}-{q}_{10}+0.4{q}_{12}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 0.6{q}_{6}-{q}_{8}+0.4{q}_{10}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 0.6{q}_{4}-{q}_{6}+0.4{q}_{8}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -{q}_{4}+0.4{q}_{8}& =-0.6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$In matrix form this is
The solution is ${q}_{4}=0.90857$, ${q}_{6}=0.85371$, ${q}_{8}=0.77143$, ${q}_{10}=0.64800$, ${q}_{12}=0.46286$.
Solution: The random walk starts at the origin ${T}_{0}=0$. Consider the case that the first step is to the left, occurring with probability $q$, so that ${T}_{1}=-1$. Then if the walk is ever to reach the goal $a>0$, then the walk must necessarily pass through the origin again. Hence for the walk to reach the goal $a>0$ the walk must first step to the right with probability $p$. Then the walk starts from state one, and we seek the subsequent probability that the walk reaches $a$ before reaching $0$. In our terminology, that is the probability of victory from $1$ before being ruined. That is ${p}_{1}$ or equivalently $1-{q}_{1}$. Hence the combined probability of these two independent events is $p\left(1-{q}_{1}\right)$.