Estimating a mean from games of ruin

[StoneTemplePython]

Very interesting! So much to learn and explore here.I think I am missing something to get $S_n$ and $E[J]$ increase linearly together. I see $E[J]$ has to increase linearly with $E[Y_1...+Y_J]$. So does that rely on $E[Y_1...+Y_J]$ and $S_n$ increase linearly together, and is that obvious?

depending on how you interpret it, yes, because $S_J = \sum_{i=1}^J Y_i$

... it belatedly occurs to me that I should have been using $S_J$ or given the lack of homogeneity, $S_J^{(n)}$ or something like that to denote the $S_J$ on the nth "day". I think I merged notations in a not so great way.
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so, remember what lurks in the background for anything Wald Equality related is (a) a valid stopping rule and (b) $E\big[J\big] \lt \infty$. These are both satisfied here though the latter takes some work to show. (If you're interested in this sort of stuff, may I suggest https://ocw.mit.edu/courses/electri...62-discrete-stochastic-processes-spring-2011/ ? It's a lot of work, but not a lot of wasted work...)

Ignoring nuisances around boundary conditions with minimum bet size, for the time being, we stop when our cumulative bankroll = 0 or equivalently, when accumulated losses are equal to the bankroll we walked into the casino with on that day. Otherwise the game never stops. Hence we can say $E\big[S_J\big] =\text{bankroll at start of the day}$

The equation of Wald tells us $E\big[S_J\big] = E\big[J\big]\bar{Y}$ and the same game is being played over and over (and independently) so $\bar{Y}$ which is expected payoff per game, is constant.

Now let's have some fun with scaling. Choose some $\alpha \gt 0$ and scale

$\alpha E\big[S_J\big] = E\big[\alpha S_J\big] = E\big[(\alpha J)\big]\bar{Y} = \big(\alpha E\big[J\big]\big)\bar{Y} = \alpha E\big[J\big]\bar{Y}$

And then repeatedly choosing the right scalars along the way to homogenize your starting bankroll and hence homogenize expectation of J as well. Then back into expected value of Y after a lot of trials. That's really it.

 

[jim mcnamara]

@lalekl Please choose either A or B
A: accept suggestions
B: stop bringing up non-sequitur requirements and ad-hoc "spray", without complete counterexamples or proof.

There is no C:
We are volunteers here, in effect you abusing our time. Reddit is a nice place to have undirected arguments. PF is educational.

Thank you.

 

[Dale]

Now we’re cooking with whale oil! But how?

I would make a statistical model that could encompass a reasonable set of payout schedules, and have one of the parameters of the model be the EV. Then I would marginalize over all of the other parameters.

 

[lalekl]

depending on how you interpret it, yes, because $S_J = \sum_{i=1}^J Y_i$

... it belatedly occurs to me that I should have been using $S_J$ or given the lack of homogeneity, $S_J^{(n)}$ or something like that to denote the $S_J$ on the nth "day". I think I merged notations in a not so great way.
- - - -

so, remember what lurks in the background for anything Wald Equality related is (a) a valid stopping rule and (b) $E\big[J\big] \lt \infty$. These are both satisfied here though the latter takes some work to show. (If you're interested in this sort of stuff, may I suggest https://ocw.mit.edu/courses/electri...62-discrete-stochastic-processes-spring-2011/ ? It's a lot of work, but not a lot of wasted work...)

Ignoring nuisances around boundary conditions with minimum bet size, for the time being, we stop when our cumulative bankroll = 0 or equivalently, when accumulated losses are equal to the bankroll we walked into the casino with on that day. Otherwise the game never stops. Hence we can say $E\big[S_J\big] =\text{bankroll at start of the day}$

The equation of Wald tells us $E\big[S_J\big] = E\big[J\big]\bar{Y}$ and the same game is being played over and over (and independently) so $\bar{Y}$ which is expected payoff per game, is constant.

Now let's have some fun with scaling. Choose some $\alpha \gt 0$ and scale

$\alpha E\big[S_J\big] = E\big[\alpha S_J\big] = E\big[(\alpha J)\big]\bar{Y} = \big(\alpha E\big[J\big]\big)\bar{Y} = \alpha E\big[J\big]\bar{Y}$

And then repeatedly choosing the right scalars along the way to homogenize your starting bankroll and hence homogenize expectation of J as well. Then back into expected value of Y after a lot of trials. That's really it.

Thank you, between this and some of the inequalities I've mentioned before, I have a complete solution to the above problem. I will program it up and see how fast it converges.

 

[StoneTemplePython]

Thank you, between this and some of the inequalities I've mentioned before, I have a complete solution to the above problem. I will program it up and see how fast it converges.

Looking through some of my notes, it may turn out that some additional structure needs to be imposed.

1.) In particular I don't see the guaranty on finiteness of variance of your recurrence time that I was looking for (in essence $E\Big[\big(J^{(n)}\big)^2\Big]$ as we know $E\Big[\big(J^{(n)}\big)\Big]$ is finite)

My sense is no real problems, baring a pathological setup... and if the payoff mix of a gambling problem is finite (like say a real slot machine) or at least bounded, a lot of this should simplify. In this case you could directly show exponential decay in probabilities of being away from the mean with increasing numbers of bets (via chernoff bound) to assure $E\Big[\big(J^{(n)}\big)^2\Big] \lt \infty$, and not having mess around with more complex machinery like truncation arguments.

2.) it is possible that we may need to bound the range of starting bankrolls for the gambler, so $M\in [a, b]$ -- which after (1) then bounds $\sigma_{J^{(n)}}$. In general there are many different sufficient conditions for WLLN or SLLN for independent non identically distributed random variables like these $J^{(n)}$, but not clean cut necessary and sufficient recipes for the non iid case. My bias is to just pick a sufficient condition like the Kolmogorov Criterion for SLLN which is easy to apply when there is a $\text{max}\big(\sigma_{J^{(n)}}\big)$.

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Again, I wouldn't anticipate problems, but technically there's a few too many ambiguities / degrees of freedom still on the table.

 

[jim mcnamara]

Thanks for participating. The thread seems to have some good answers. Thread closed.