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I've recently become interested in a type of biased random walk. In the original context, the simplest problem I cannot answer is as follows:

Suppose you enter a simple but lousy betting game. The price to enter is \$3, and a fair coin is flipped. If it lands heads you win \$4, otherwise you win nothing. So, each game either increases your wealth by \$1 or decreases it by \$3. If you begin with just \$3 and repeatedly enter the game until you can no longer pay the fee, what is the expected number of games you will play?

Actually, I'd like to be able to answer this question for any game where the cost is $m$, the prize is $m+1$, and we have any given amount of starting wealth $w$. This similar question+answer shows that the expectation diverges when $m=1$ for any starting wealth.

Another way to phrase the question would be as a random walk. If you consider our position on the number line to be our total wealth after paying the fee, the case $m=w=1$ is a random walk from the origin that ends whenever the position becomes negative. The general case $m$ is a sort of random walk, but really it's more of a "stumble", since we either advance one step or reverse $m$ steps. I am interested in the expected length of these $m$-stumbles. Here is what I know so far:

The expected number of games, $g_{m,w}$, satisfies an easy recurrence relating your starting wealth:

$$ g_{m,w} = 1 + \frac{1}{2}g_{m,w+1} + \frac{1}{2}g_{m,w-m}$$

Which I find has generating function:

$$\sum_{w \ge 0} g_{m,w}z^n = \left(\alpha_m - \frac{2z}{1-z}\right)\left(\frac{1}{1 - 2z + z^{m+1}}\right)z^{m}$$

where $\alpha_m = g_{m,m}$ represents the average length of an $m$-stumble from the origin. The $g$ terms are related to the Fibonacci n-step numbers by $A_{w}\alpha_m - 2B_{w}$ where A is a partial sum of the first $w$ n-step numbers $\sum_{k \le w}F_{k}^{(n)}$, and B is a double convolution, i.e. the sum of the first $w-1$ partial sums $\sum\sum_{k < w}F^{(n)}_{k}$. The easiest way to find $g_{m,w}$ given some $g_{m,s}$ that I know of is to use the recurrence directly.

I have found the following double sum formula for general $\alpha_m$:

$$\alpha_m = \sum_{n = 0}^{\infty} \sum_{k=1}^{m} k\binom{mn + n + k}{n} / 2^{mn + n + k}$$

which more-or-less follows from the definition of the Fuss-Catalan numbers. Each term represents the contribution to the expectation of games ending at flip number $(m+1)n+k$. The inner sum does not include the term $k=0$ because it is not possible to lose after flip numbers divisible by $m+1$. (After flip $n$ your total wealth must be congruent to $n+m \mod{(m+1)}$, since it increases by exactly $1$ or $-m \equiv +1 \mod{(m+1)}$ each flip, and it begins at $w = m \equiv -1 \mod{(m+1)}$, so if you haven't already lost your wealth must be at least $m$ meaning you are guaranteed to have enough to pay the fee for these games.)

I don't know how to evaluate this sum, but I do know at least one value and have good guesses for others. We know from the linked question $\alpha_1 = \infty$, but for $m > 1$, $\alpha_m$ converges. The one value I know is:

$$\alpha_2 = \sum_{n=0}^{\infty} \binom{3n+1}{n}/2^{3n+1} + 2\sum_{n=0}^{\infty}\binom{3n+2}{n}/2^{3n+2} = 1 + \sqrt{5}$$

Which is the value Mathematica gives me. I don't how how it arrives at this value, but it appears correct and matches my simulations. Unfortunately, Mathematica chokes on $\alpha_3$ and higher. I guessed that $\alpha_m$ is the root of an $m$'th order polynomial, so $\alpha_2 = 1 + \sqrt{5} = (x^2 - 2x - 4)_{+}$, meaning $\alpha_2$ is the unique positive real root of $x^2 - 2x - 4 = 0$.

A value for $\alpha_3$ would answer the initial problem statement in this question. I find the following guesses match the value of my sum with very high accuracy, to at least 100 digits:

$$ \begin{align} \alpha_3 &= (x^3 - 4x - 4)_{+} \\ \alpha_4 &= (3x^4 + 4x^3 - 8x^2 - 24x - 16)_{+} \\ \alpha_5 &= (2x^5 + 5x^4 - 20x^2 - 32x - 16)_{+} \end{align} $$

but even knowing these values I have no idea how to prove them, even for the apparently simple cases of $\alpha_2$ and $\alpha_3$. Does anyone know how to find $\alpha_3$ or general $\alpha_m$?

Brocellous
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1 Answers1

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I'm gonna show that $\alpha_m$ is the (unique) solution of $\alpha^{m+1}=(\alpha-2)(\alpha+2)^m$ with $\alpha>2$.

My point of departure is your double sum, rewritten as $$\alpha_{m-1}=\sum_{k=1}^{m-1}k\cdot 2^{-k}G_{m,k}(2^{-m}),\qquad G_{m,k}(z)=\sum_{n=0}^\infty\binom{mn+k}{n}z^n.$$

Since you're familiar with the Fuss–Catalan numbers, let's recall that if $$F_{m,k}(z)=\sum_{n=0}^\infty\frac{k}{mn+k}\binom{mn+k}{n}z^n$$ and $F_m(z)=F_{m,1}(z)$, then $F_{m,k}(z)=\big(F_m(z)\big)^k$ and $F_m(z)=1+z\big(F_m(z)\big)^m$.

Now $G_{m,k}(z)$ can be expressed in terms of $F_m(z)$. We have $$G_{m,k}(z)=F_{m,k}(z)+\frac{mz}{k}F_{m,k}'(z)=\big(F_{m}(z)\big)^k R_m(z),$$ where $R_m(z)=1+mzF_m'(z)/F_m(z)$. On the other hand, $F_m'(z)=\big(F_m(z)\big)^m R_m(z)$ by taking the derivative of $F_m(z)=1+z\big(F_m(z)\big)^m$. Hence $$R_m(z)=1+\frac{mz}{F_m(z)}\frac{F_m(z)-1}{z}R_m(z);$$ solving this, we get $R_m(z)=F_m(z)/\big(m-(m-1)F_m(z)\big)$ and $$G_{m,k}(z)=\frac{\big(F_m(z)\big)^{k+1}}{m-(m-1)F_m(z)}.$$

Thus, $2^{-k}G_{m,k}(2^{-m})=2x_m^{k+1}/\big(m-2(m-1)x_m\big)$, where $x_m=F_m(2^{-m})/2\color{blue}{\in(1/2,1)}$ implied by $m-2(m-1)x_m>0$. And $x_m$ is the unique solution $x\in(1/2,1)$ of $$0=x^m-2x+1=(x-1)(x^{m-1}+\ldots+x-1).$$

Then $\beta_m=x_{m+1}$ is the unique solution $\beta>1/2$ of $\beta^m+\ldots+\beta=1$, and $$\alpha_m=\frac{2\beta_m\gamma_m}{1-m(2\beta_m-1)},\qquad\gamma_m=\sum_{k=1}^m k\beta_m^k.$$ Now $\beta_m\gamma_m=m\beta_m^{m+1}+\sum_{k=1}^m(k-1)\beta_m^k=m(2\beta_m-1)+\gamma_m-1$ gives $$\gamma_m=\frac{1-m(2\beta_m-1)}{1-\beta_m}\implies\alpha_m=\frac{2\beta_m}{1-\beta_m}\implies\beta_m=\frac{\alpha_m}{\alpha_m+2},$$ and if we put $\beta=\alpha/(\alpha+2)$ into $\beta^{m+1}=2\beta-1$, we get the claimed equation for $\alpha=\alpha_m$.

metamorphy
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