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let [$]\tau = \inf\{t>0, W_t = a\}[$] and [$]g[$] a continuous function, what is the value of [$]E\left[\int_0^{\tau} g(W_s) ds \right][$]?

[$]g(a)\tau[$].

Is this do-able?

I think it's doable.EBal, do you mean [$]g(a)E[\tau][$]?I believe [$]E[\tau]=a^2[$] which leads to [$]g(a)a^2[$]. This is consistent for [$]g[$] constant for example.Anyone with a detailed proof?

Yes, I meant [$]E(\tau)[$], but now that I start thinking about this the expectation of [$]\tau[$] is infinite.

QuoteOriginally posted by: PenielI think it's doable.EBal, do you mean [$]g(a)E[\tau][$]?I believe [$]E[\tau]=a^2[$] which leads to [$]g(a)a^2[$]. This is consistent for [$]g[$] constant for example.Anyone with a detailed proof? [$]E[\tau]=\infty[$] So you don't know the answer?

The answer is [$]\infty[$]. If for example [$]g[$] has a minimum it is pretty obvious.

OK, got it, thanks.

but why don't we slightly change the stopping time:[$]\tau_\mu=\inf\{t>0, W_t+\mu t=a\}[$] or [$]\tau_b=\inf\{t>0, W_t=a~\text{or}~W_t=-b\}[$],then let either [$]\mu\to0^+[$] or [$]b\to+\infty[$]. not that this changes the conclusion of the original problem, but it sheds light on how in general the approach might be.we need [$]E\left[F_\tau\right][$] for [$]dF_t=g(W_t)dt,~F(0)=0[$]. the strategy is to find some process [$]G(W_t)[$] so that [$](F-G)[$] is a martingale. note by introducing either nonzero mu or finite b, the stopping time is almost surely bounded so we can apply the optional stopping theorem and take advantage of the fact that [$]W_\tau[$] is given above.applying Ito's lemma, we easily get [$]G''=2g[$] and can choose initial conditions, e.g., [$]G(0)=0,~G'(0)=0[$]. the rest is to calculate [$]E[G][$] given an explicit form of g. it's helpful to note that the p.d.f. of [$]\tau_\mu[$] is inverse Gaussian, while [$]P(W_{\tau_b}=a)[$] and [$]P(W_{\tau_b}=b)[$] are well-known.just for reference, e.g., if g(x)=x, then G(x)=x^3/3, one gets the expectation of -a/mu^2 or ab(a-b)/3, both tend to infinity quadratically