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Let $\{W_t,t \geq 0\}$ be a standard Brownian motion under $\mathbb P$. Let $T_a$ be the hitting time of level $a$, that is: $$T_a= \text{inf}\{t \geq 0:W_t=a\}.$$ From a proposition, we know that $$\mathbb E[\exp(-\theta T_a)] = \exp(-a \sqrt{2\theta}).$$

How do we make use of the above preposition to calculate $\mathbb E[T_a]$?

I am supposed to obtain $\infty$ as the expectation.

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  • $\begingroup$ Are you sure this isn't the characteristic function, and so should be $E[\exp(-i\theta T_a)]$ where $i$ is the imaginary unit? because the mgf only exists if all moments exist, and as @vinux's answer shows, using this to calculate the first moment doesn't work... $\endgroup$
    – probabilityislogic
    Feb 21, 2012 at 12:33
  • $\begingroup$ @prob: $T_a$ is nonnegative almost surely, so for any $\theta \geq 0$, we have $\exp(-\theta T_a) \leq 1$ almost surely, so the expectation definitely exists (for such $\theta$)! $\endgroup$
    – cardinal
    Feb 21, 2012 at 12:41

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This is similar to calculating expectation from M.G.F. Since $ e^x = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots. $ use differentiation technique for deriving expectation.

$$ -\dfrac{\partial}{\partial \theta} E[\exp(-\theta T_a)]_{\theta=0} =-\dfrac{\partial}{\partial \theta} \exp(-\sqrt{2\theta}) \Big \vert_{\theta=0} = E[T_a]$$

This works when the moments are finite. It seems the expectation is not finite.

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  • $\begingroup$ Thank you for the respone. I do not understand the first part though. Could you kindly elaborate. $\endgroup$
    – Probabilityman
    Feb 21, 2012 at 5:42
  • $\begingroup$ Also How will i compute the partial differentiation with E(exp(-theta*Ta)) with E being a part of the equation. If you could provide me a head start on how to differentiate I would be able to proceed. $\endgroup$
    – Probabilityman
    Feb 21, 2012 at 5:44
  • $\begingroup$ I will modify my solution. $\endgroup$
    – vinux
    Feb 21, 2012 at 5:46
  • $\begingroup$ Hmm. Do you not need $\mathbb E(e^{−\theta T_a})$ to exist in an open interval around zero to apply such a result? I think you can get at it by simpler means. $\endgroup$
    – cardinal
    Feb 21, 2012 at 15:54
  • $\begingroup$ I agree cardinal. I initially wrote the approach. later I added the actual function and realised it is not that simple. $\endgroup$
    – vinux
    Feb 21, 2012 at 16:20

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