Kitri's Variation, Act III

Yes Kitri can, using Markov's inequality: $P(\tau \geq 30) \leq \frac{E(\tau)}{30} = \frac{15}{30} = \frac{1}{2} < 1$ . So, the answer we seek is $\boxed{\text{Yes}}$ .

Here's a rough sketch of the proof. Assuming that $\tau$ is a random variable and that $\mu_\tau = E(\tau)$ is finite. Let the indicator variable $I$ be $I = I(\tau \geq 30)$ so that $I = 1$ if $\tau \geq 30$ and $I = 0$ if not. This implies two things: (i) $\tau \geq \tau\cdot I \geq 30\cdot I$ and (ii) $P(\tau \geq 30) = E(I)$ . Therefore, $E(\tau) \geq E(30\cdot I) = 30\cdot E(I) = 30\cdot P(\tau \geq 30)$ . But this is simply saying $P(\tau \geq 30) \leq \frac{E(\tau)}{30} = \frac{15}{30} = \frac{1}{2} < 1$ .