Hi,If W(t) is a standard Wiener process, the first passage time T(x) is defined as:T(x) = inf{t: W(t) = x}T(x) has density function |x|Exp[-x^2/(2t)]/Sqrt[2 Pi t^3]I have a problem understanding why E[T(x)] is Infinity for all x > 0.In my opinion this is rather disheartening. Could someone kindly explain why this is so ? Thanks.

Since this is the brainteaser forum, it does suggest the puzzle:You're playing a fair game in a casino with unlimited capital, making constant bets at constant time intervals.Take blackjack, for example, with a modest amount of card counting.Suppose, at some point you're down $X. Since the game is fair, you decide to keep playing until you break even.Which is smaller: your expected time to break-even or your expected time 'till you die at the table?

Well, I know the simple solution that requires a lot so i don't think that it will help.If E[T(x)] were finite then according to Wald identities E[W(T(x))] were equal to 0, but it is not as W(T(x)) = x. Or you can estimate the integral, as you know the distribution of T it's straightforward method, but I still prefer the former.

Another method would be to compute its Laplace transform E(exp(a T)) using Wald's martingale, then differentiate this with respect to 'a' which leads something like E(T e(a T)) and compute the limit when 'a' tends to zero, using dominated convergence theorem everywhere.But I am not sure this is of great help to understand qualitatively what happens.

Qualitatively one can say that as you cannot (statistically) win in a fair game something should be wrong with the strategy of waiting until you get x and stop. That "something" in this case is infinite mean of the waiting time.

QuoteOriginally posted by: amiralievQualitatively one can say that as you cannot (statistically) win in a fair game something should be wrong with the strategy of waiting until you get x and stop. That "something" in this case is infinite mean of the waiting time.Martingale are not necessarly a fair game:x_n can take two values: x_n={2^n/(2^n-1);-2^n}P(x_n=2^n/(2^n-1))=(2^n-1)/2^nP(x_n=-2^n)=1/2^nand let S_n=x_1+...+x_nwe have that Ex_n=0 and E(S_n|S_n-1)=S_n, S_n is f_n measurable.So S_n is a martingale.But S_n goes to infinity with probability 1.....Do you want to play with me on that "fair" game?

QuoteOriginally posted by: meteorMartingale are not necessarly a fair game:let x_n iid and x_n can take two values: x_n={2^n/(2^n-1);-2^n}P(x_n=2^n/(2^n-1))=(2^n-1)/2^nP(x_n=-2^n)=1/2^nand let S_n=x_1+...+x_nwe have that Ex_n=0 and E(S_n|S_n-1)=S_n, S_n is f_n measurable.So S_n is a martingale.But S_n goes to infinity with probability 1.....Do you want to play with me on that "fair" game?Well, if you want a short answer - yes.Let me explain. First of all there are some minor details (typos I think, like x_n being iid) that could be misinterpreted, but in general you are correct: this is well-known example of non-uniformly integrable martingale. The problem is it has nothing to do with what I said about a fair game. Althought as this martingale indeed goes to infinity a.s. we won't play till infinity, and on any finite time horizon (and on any reasonable stopping time) expectation is still zero. Considering stopping times, in this case optional sampling will fail with finite expectation of "bad" stopping time - if that is what you mean. But in this case "something wrong" with this strategy is having infinite mean of the drawdown in your capital, so I don't think that using this strategy would be a good idea Do you still want to play?

QuoteOriginally posted by: amiralievDo you still want to play?I have plenty of time .... I just wanted to point out that martingale are not always fair.... and you are right x_n are not iid (I edited my post now).

Well, it depends on what you call "fair" but you are right that behaviour on infinity could be "bad".