Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why is this: $E(\tau \wedge T) = \int_0^Ttf_\tau(t)dt+T(1-P(\tau\leq T))$, where $f_\tau(t)$ is the pdf of $\tau$. I am thinking its because $E(\tau \wedge T) = \tau P(\tau\leq T)+T(1-P(\tau\leq T))$ since $\tau \wedge T =\min(\tau,T) = \tau$ if $\tau\leq T)$ and T if $\tau> T)$. But the first part is I cannot figure out.

share|cite|improve this question
up vote 1 down vote accepted

We have as you wrote \[ \tau \wedge T = \chi_{\{\tau \le T\}} \cdot \tau + T \cdot \chi_{\{\tau > T\}} \] where $\chi_A$ denotes the indicator function of $A$. We have $\chi_{\\{\tau \le T\\}} = \chi_{[0,T]} \circ \tau$, therefore \begin{align*} E(\tau \wedge T) &= E\bigl((\chi_{[0,T]}\cdot \mathrm{id}_{[0,\infty)}) \circ \tau\bigr) + T \cdot E(\chi_{\\{\tau > T\\}})\\ &= \int_0^\infty \chi_{[0,T]}(t) \cdot t \cdot f_\tau(t) \, dt + T \cdot P(\tau > T)\\ &= \int_0^T t f_\tau(t)\, dt + T \bigl(1 - P(\tau \le T)\bigr) \end{align*}

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.