[ QUOTE ]
Via a roundabout route, I've been looking at
this post by BruceZ containing his own derivation of a risk of ruin formula, using coin-flip games as the basis.
The game is an even-money bet of size b with a biased coin: you win $b with probability p and lose $b with probability q = 1-p.
BruceZ writes this formula for the variance of the game:
variance = p*b^2 + q*(-b)^2 – EV^2
Is that right? I don't know much stats, but I thought variance was computed as the average of the squares of the deviations of actual results from the mean, i.e.
p(b - EV)^2 + q(-b - EV)^2
which doesn't seem to be the same thing.
Guy.
[/ QUOTE ]
Sure it is. Expand it out.
pb^2 + q(-b)^2 + EV^2(p + q) - 2bEV(p - q)
p + q = 1
EV = pb + q(-b) = b(p - q)
-2bEV(p - q) = -2EV^2
so we have
pb^2 + q(-b)^2 + EV^2 - 2EV^2
= pb^2 + q(-b)^2 - EV^2
Actually this form is just
variance = E(x^2) - [E(x)]^2
where E means expected value.