Sessions of random length and variance

[marv]

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What are E(N) and Var(N) equal to here?

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E[N] = infty.

Let N be the first time |Z_n| > c, where c is at least 1. Let T(n) = min(n,N). Since (S_n)^2 - n is a martingale and T(n) is a bounded stopping time, we may apply the optional sampling theorem to get

E[(S_T(n))^2] = E[T(n)].

Hence,

E[T(n)] >= E[I*(S_N)^2],

where I = 1 if N < n and I = 0 otherwise. Let n go to infinity. By monotone convergence, E[N] >= E[S_N^2]. But S_N^2 > (c^2)N. If E[N] is finite, this is a contradiction.

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Nice.

[PairTheBoard]

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So it appears that the random session lengths can create bias in either direction. I wonder if one direction is more common than the other and, if so, what would explain that?

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I think one of the most common tendencies is for players to keep playing until they are ahead, or exhasted, or way behind and give up on the session. I would have thought this would produce a higher variance for Z_N, so I was suprised by your Example 1 which is a version of this method - exhastion comes after 2 rounds.

However, I wonder if this is just a peculiarity of the session's short length - max rounds = 2. Suppose you let N be the first time S_n = 1. Of course N will have infinite expected value. And you won't get the same kind of convergence of the truncated series like in your other example. Still, it seems to me that this is the kind of thing to look at. You get a lot of small wins and a few large losses. Seems like that's got to pump up the variance.

PairTheBoard

[jason1990]

If N is the first time S_n = 1, then (Z_N)^2 = 1/N, so Var(Z_N) < E[(Z_N)^2] < 1.

We could try letting N be the first time S_n = k. Then E[(Z_N)^2] = k^2 E[1/N]. Now,

E[1/N] = E[\int_0^\infty e^{-bN} db]
= \int_0^\infty E[e^{-bN}] db.

If b > 0 and a = cosh^{-1}(e^b) > 0, then M_n = e^{aS_n - bn} is a martingale. This means that for T = min(n,N)

1 = E[M_T] = E[e^{aS_T - bT}].

Letting n go to infinity, dominated convergence gives E[e^{-bN}] = e^{-ak}. Hence,

E[1/N] = \int_0^\infty \exp{-k cosh^{-1}(e^b)} db.

Make the change of variables e^b = cosh(u) to get

E[1/N] = \int_0^\infty e^{-ku}tanh(u) du.

For k = 2, the integrand has antiderivative

e^{-2x}/2 - ln(1 + e^{-2x}).

This gives k^2 E[1/N] = 4(ln(2) - 1/2) = 0.7726.

With the help of Mathematica, I have calculated k^2 E[1/N] for several values of k, and it appears that this quantity increases monotonically to 1. This would imply Var(Z_N) < E[(Z_N)^2] < 1 for all k.

To compute Var(Z_N) exactly, we would need to compute E[1/sqrt{N}]. I do not have a method for this, but we can compute it in the Brownian motion case. If N is the first time a Brownian motion reaches level k, then E[1/N] = 1/k^2 and E[1/sqrt{N}] = sqrt{2/pi}/k. This suggests that, for large k,

Var(Z_N) = 1 - 2/pi = 0.3634.

[PairTheBoard]

That's interesting. I can't think of any other common ways players might bias their sessions. At least for this type of method it looks like you've shown what you were thinking was true about the miscalculation of variance based on sessions rather than hands.

I guess what Z_n gives is a normalized measure of how far your session is from its EV of zero. Your example of stopping when |Z_n|>2 translates to someone playing until they hit a really big lucky or unlucky streak. A big enough streak to throw their session off by 2 standard deviations for the session. This doesn't sound like it would be a very common tendency for most players. I think most players who have a stopping criteria - possibly unconscious - base it on a goal for S_n, which your work above indicates reduces the variance for Z_N.

PairTheBoard