Jtdylktuy

I have to simulate the following game

Suppose that two players A and B each start with a stake of $5, and bet $0.5 on consecutive coin flips. The game ends when either one of the players has won all the money, that amounts to $10. Let $S_n$ be the fortune of player A at time n. Then ${S_n, n gt 0 }$ is a symmetric random walk with absorbing barriers at 0 and 10. Estimate $E[S_n]$ and $V[S_n]$ for $n = 50$.

I made a program and that's ok. Now I would to compare my results with the theoretical values. I don't know almost anything in probability, so that's just a curiosity. My question is

Which are the values of $E[S_n]$ and $V[S_n]$ in general? And for $n = 50$?

If there's no closed form, I will appreciate also an approximation (I got $E[S_n] approx 5$ and $V[S_{50}] approx 4.8$)

Thanks in advance.

Modeling the process

Define $mathcal{S}$ as the set of possible values for the Markov chain:
$$mathcal{S} = {0, 0.5, 1, 1.5, …, 9.5, 10}$$
Note that $S_0=5$ and $S_n in mathcal{S}$ for all $n in {0, 1, 2, …}$.
We have
$$S_{n+1} = S_n + A_n quad forall n in {0, 1, 2, ...} $$
where
$$ A_n = left{ begin{array}{ll}
(1/2)B_n &mbox{ if $S_n notin {0, 10}$} \
0 & mbox{ otherwise}
end{array}
right.$$
where ${B_n}_{n=0}^{infty}$ is an i.i.d. sequence with $P[B_n=1]=P[B_n=-1]=1/2$.
Then
$$boxed{E[A_n|S_n=s] = 0 quad, forall s in mathcal{S}} quad (Eq. 1) $$

Mean

So for each $n in {0, 1, 2, ...}$ we have
begin{align}
E[S_{n+1}] &overset{(a)}{=} sum_{s in mathcal{S}}E[S_{n+1}|S_n=s]P[S_n=s] \
&overset{(b)}{=} sum_{s in mathcal{S}}E[S_n + A_n|S_n=s]P[S_n=s] \
&= sum_{s in mathcal{S}}E[s + A_n|S_n=s]P[S_n=s] \
&= sum_{s in mathcal{S}}(s + E[A_n|S_n=s])P[S_n=s] \
&overset{(c)}{=} sum_{s in mathcal{S}}sP[S_n=s] \
&overset{(d)}{=} E[S_n]
end{align}
where (a) holds by the law of total expectation; (b) holds by the fact $S_{n+1}=S_n+A_n$; (c) holds by Eq. (1); (d) holds by definition of expectation.
Since $E[S_0]=5$ we conclude:
$$boxed{E[S_n]=5 quad forall n in {0, 1, 2, … }}$$

Limiting variance

We know $E[S_n]=5$ for all $n$ and so
$$Var(S_n) = E[(S_n-5)^2] = sum_{s in mathcal{S}}(s-5)^2P[S_n=s] $$
Since the process is equally likely to end up at state $0$ or $10$ we have
begin{align}
lim_{nrightarrowinfty} P[S_n=0] &= 1/2\
lim_{nrightarrowinfty} P[S_n=10] &= 1/2\
lim_{nrightarrowinfty} P[S_n=s] &= 0 quad forall s notin {0, 10}
end{align}
so
$$ boxed{lim_{nrightarrowinfty} Var(S_n) = (0-5)^2(1/2) + (10-5)^2(1/2) = 25} $$

Details on variance

Squaring the equation $S_{n+1} = S_n + A_n$ gives
$$S_{n+1}^2 = (S_n+A_n)^2 = S_n^2 + 2S_nA_n + A_n^2 $$
So
$$E[S_{n+1}^2|S_n] = S_n^2 + 2S_nE[A_n|S_n] + E[A_n^2|S_n] = S_n^2 + 0 + (1/4)1_{{S_n notin{0, 10}}}$$
where $1_{{S_n notin{0, 10}}}$ is an indicator function that is 1 if $S_n notin {0,10}$ and is 0 else. So
$$E[S_{n+1}^2] = E[S_n^2] + (1/4)P[S_n notin {0,10}]$$
Subtracting 25 from both sides gives
$$ Var(S_{n+1}) = Var(S_n) + (1/4)P[S_n notin {0,10}]$$
and $Var(S_0)=0$ so
$$ boxed{Var(S_n) = (1/4)sum_{i=0}^{n-1} P[S_i notin {0,10}] quad forall nin {1, 2, 3, ...} } $$
Since $P[S_i notin {0,10}] = 1$ for $i in {0, 1, 2, 3, ..., 9}$ we have $$boxed{Var(S_1)=1/4, Var(S_2)=2/4, Var(S_3) = 3/4, ..., Var(S_{10})= 10/4}$$
On the other hand:
$$ Var(S_{11}) = 10/4 + (1/4)underbrace{(1-2(1/2)^{10})}_{P[S_{10}notin{0,10}]}$$
In general, the variance increases as $nrightarrowinfty$ to approach a
limiting value of $25$. It is possible to compute $P[S_i notin {0,10}]$ for all $i$ (for example, by taking powers of a transition probability matrix), but this calculation is more involved.