91Ó°ÊÓ

Let's consider the given strategy: stopping at the first event to occur after time \(s\). We'll win if no event occurs between time \(s\) and \(T\). The number of events occurring in an interval follows a Poisson distribution. So, the number of events happening between \(s\) and \(T\) is Poisson distributed with parameter \(\lambda (T-s)\). We'll win if there are no events in this interval, which means the probability of winning is the probability of having zero events in the interval \((s, T]\). Using the formula for the Poisson distribution we have: $$P(\text{win}) = P(X=0) = \frac{e^{-\lambda(T-s)} (\lambda(T-s))^0}{0!} = e^{-\lambda(T-s)}$$

Now we need to find the value of \(s\) that maximizes the winning probability. To achieve this, we should take the derivative of the winning probability function with respect to s and set it equal to zero. $$\frac{dP(\text{win})}{ds} = \frac{d}{ds} e^{-\lambda(T-s)} = \lambda e^{-\lambda(T-s)}$$ Next, set the derivative equal to zero and solve for \(s\): $$\lambda e^{-\lambda(T-s)} = 0$$ The exponential term cannot be zero so the derivative is nonzero for any \(s\). Therefore, we have to analyze the function at its boundary. Let's evaluate the winning probability function at the boundaries \(s = 0\) and \(s = T\): $$P(\text{win})|_{s=0} = e^{-\lambda (T-0)} = e^{-\lambda T}$$ $$P(\text{win})|_{s=T} = e^{-\lambda(T-T)} = e^0 = 1$$ As \(T > 1/\lambda\), we have \(e^{-\lambda T} < 1\). Thus, the maximum winning probability occurs at \(s = T\).

Now that we have the optimal value of \(s\), we can plug it back into the winning probability equation to find the overall winning probability with the best strategy: $$P(\text{win})|_{s=T} = e^{-\lambda(T-T)} = e^0 = 1$$ However, note that in order for us to stop at the first event after time \(T\), there must be at least one event between time \(0\) and \(T\). The probability of having at least one event between time \(0\) and \(T\) is \(1-e^{-\lambda T}\). So, our winning probability should be multiplied by this factor: $$P(\text{win}) = 1 \times (1-e^{-\lambda T}) = 1-e^{-\lambda T}$$ Now we are asked to show that this probability is \(1/e\). To do so, we remember that \(T > 1/\lambda\), which means that \(T\lambda > 1\). Then, the inequality can be written as: $$1 - e^{1-T\lambda} < P(\text{win}) = 1 - e^{-\lambda T} < 1 - e^{T\lambda -1}$$ Since \(\lim_{n\to\infty}(1-\frac{1}{n})^{n}=1/e\), we see that the winning probability gets closer to \(1/e\) as \(T\) increases with respect to \(\lambda\).