Time-dependent Gambler's ruin (coefficient probabilities taken from a probability distribution)
$begingroup$
Consider modifying the standard gambler's ruin recurrence relation (i.e. probability of winning given starting wealth of $i$, maximum exit wealth of $c$, and probability of winning $p$):
$
begin{equation}
u(i) = pu(i+1) + (1-p)u(i-1)
end{equation}
$
I would like to modify the expression above to replace the constant $p$ with with the pmf of some particular distribution (of special interest is the binomial distribution, but happy to look at other distributions as well). Therefore, the recurrence relation now becomes:
$
begin{equation}
u(i) = f(i)u(i+1) + (1-f(i))u(i-1)
end{equation}
$
where $f(i)$ computes the pmf of the binomial distribution over $c$ trials, with exactly $i$ successes, where the probability of a success is $i/c$. In other words,
$
begin{equation}
f(i) = {c choose i}(frac{i}{c})^i(1-frac{i}{c})^{c-i}
end{equation}
$
I've looked around extensively, and very surprisingly I have not seen anyone treat this. Is there a closed-form expression?
probability probability-distributions recurrence-relations markov-chains
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
$endgroup$
add a comment |
$begingroup$
Consider modifying the standard gambler's ruin recurrence relation (i.e. probability of winning given starting wealth of $i$, maximum exit wealth of $c$, and probability of winning $p$):
$
begin{equation}
u(i) = pu(i+1) + (1-p)u(i-1)
end{equation}
$
I would like to modify the expression above to replace the constant $p$ with with the pmf of some particular distribution (of special interest is the binomial distribution, but happy to look at other distributions as well). Therefore, the recurrence relation now becomes:
$
begin{equation}
u(i) = f(i)u(i+1) + (1-f(i))u(i-1)
end{equation}
$
where $f(i)$ computes the pmf of the binomial distribution over $c$ trials, with exactly $i$ successes, where the probability of a success is $i/c$. In other words,
$
begin{equation}
f(i) = {c choose i}(frac{i}{c})^i(1-frac{i}{c})^{c-i}
end{equation}
$
I've looked around extensively, and very surprisingly I have not seen anyone treat this. Is there a closed-form expression?
probability probability-distributions recurrence-relations markov-chains
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
$endgroup$
1
$begingroup$
This comment may not be directly related - but some observations. The question reminds me of time dependent random walk and of Wright-Fisher models. Also, just as an observation, it will likely require dealing with law of total total prob/expectation on a case by case depending on the marginal distribution of $p$.
$endgroup$
– Just_to_Answer
Jan 5 at 21:29
add a comment |
$begingroup$
Consider modifying the standard gambler's ruin recurrence relation (i.e. probability of winning given starting wealth of $i$, maximum exit wealth of $c$, and probability of winning $p$):
$
begin{equation}
u(i) = pu(i+1) + (1-p)u(i-1)
end{equation}
$
I would like to modify the expression above to replace the constant $p$ with with the pmf of some particular distribution (of special interest is the binomial distribution, but happy to look at other distributions as well). Therefore, the recurrence relation now becomes:
$
begin{equation}
u(i) = f(i)u(i+1) + (1-f(i))u(i-1)
end{equation}
$
where $f(i)$ computes the pmf of the binomial distribution over $c$ trials, with exactly $i$ successes, where the probability of a success is $i/c$. In other words,
$
begin{equation}
f(i) = {c choose i}(frac{i}{c})^i(1-frac{i}{c})^{c-i}
end{equation}
$
I've looked around extensively, and very surprisingly I have not seen anyone treat this. Is there a closed-form expression?
probability probability-distributions recurrence-relations markov-chains
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
$endgroup$
Consider modifying the standard gambler's ruin recurrence relation (i.e. probability of winning given starting wealth of $i$, maximum exit wealth of $c$, and probability of winning $p$):
$
begin{equation}
u(i) = pu(i+1) + (1-p)u(i-1)
end{equation}
$
I would like to modify the expression above to replace the constant $p$ with with the pmf of some particular distribution (of special interest is the binomial distribution, but happy to look at other distributions as well). Therefore, the recurrence relation now becomes:
$
begin{equation}
u(i) = f(i)u(i+1) + (1-f(i))u(i-1)
end{equation}
$
where $f(i)$ computes the pmf of the binomial distribution over $c$ trials, with exactly $i$ successes, where the probability of a success is $i/c$. In other words,
$
begin{equation}
f(i) = {c choose i}(frac{i}{c})^i(1-frac{i}{c})^{c-i}
end{equation}
$
I've looked around extensively, and very surprisingly I have not seen anyone treat this. Is there a closed-form expression?
probability probability-distributions recurrence-relations markov-chains
probability probability-distributions recurrence-relations markov-chains
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
edited Jan 8 at 17:53
ux74bn1
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
asked Jan 5 at 18:43
ux74bn1ux74bn1
163
163
1
$begingroup$
This comment may not be directly related - but some observations. The question reminds me of time dependent random walk and of Wright-Fisher models. Also, just as an observation, it will likely require dealing with law of total total prob/expectation on a case by case depending on the marginal distribution of $p$.
$endgroup$
– Just_to_Answer
Jan 5 at 21:29
add a comment |
1
$begingroup$
This comment may not be directly related - but some observations. The question reminds me of time dependent random walk and of Wright-Fisher models. Also, just as an observation, it will likely require dealing with law of total total prob/expectation on a case by case depending on the marginal distribution of $p$.
$endgroup$
– Just_to_Answer
Jan 5 at 21:29
1
1
$begingroup$
This comment may not be directly related - but some observations. The question reminds me of time dependent random walk and of Wright-Fisher models. Also, just as an observation, it will likely require dealing with law of total total prob/expectation on a case by case depending on the marginal distribution of $p$.
$endgroup$
– Just_to_Answer
Jan 5 at 21:29
$begingroup$
This comment may not be directly related - but some observations. The question reminds me of time dependent random walk and of Wright-Fisher models. Also, just as an observation, it will likely require dealing with law of total total prob/expectation on a case by case depending on the marginal distribution of $p$.
$endgroup$
– Just_to_Answer
Jan 5 at 21:29
add a comment |
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$begingroup$
This comment may not be directly related - but some observations. The question reminds me of time dependent random walk and of Wright-Fisher models. Also, just as an observation, it will likely require dealing with law of total total prob/expectation on a case by case depending on the marginal distribution of $p$.
$endgroup$
– Just_to_Answer
Jan 5 at 21:29