Question

Events occur according to a Poisson process with rate \(\lambda .\) Each time an event occurs, we must decide whether or not to stop, with our objective being to stop at the last event to occur prior to some specified time \(T\), where \(T>1 / \lambda\). That is, if an event occurs at time \(t, 0 \leqslant t \leqslant T\), and we decide to stop, then we win if there are no additional events by time \(T\), and we lose otherwise. If we do not stop when an event occurs and no additional events occur by time \(T\), then we lose. Also, if no events occur by time \(T\), then we lose. Consider the strategy that stops at the first event to occur after some fixed time \(s, 0 \leqslant s \leqslant T\). (a) Using this strategy, what is the probability of winning? (b) What value of \(s\) maximizes the probability of winning? (c) Show that one's probability of winning when using the preceding strategy with the value of \(s\) specified in part (b) is \(1 / e\).

Short answer

The probability of winning using the given stopping strategy is \(P(W) = e^{-\lambda T}\), which doesn't depend on the choice of the stopping time s. Therefore, there is no specific value of s that maximizes the winning probability. It turns out that the optimal stopping strategy is to stop after 1 event has occurred, and the probability of winning when using this optimal strategy is \(\frac{1}{e}\).

Step-by-step solution

Find the probability of winning using the given strategy

Given that we stop at the first event happening after time s, we want to find the probability of no additional events happening between time s and time T. The number of events happening in a Poisson process with rate \(\lambda\) in the interval \(ss)\). This is the same as the probability of 0 events happening in the interval \(0s)\) and \(P(0)\): \(P(W) = P(S>s) * P(0) = e^{-\lambda s} * e^{-\lambda(T-s)} = e^{-\lambda T}\)

Determine the value of s to maximize the probability of winning

To find the value of s that maximizes the probability of winning, we can take the derivative of the probability function with respect to s and set it equal to 0: \(\frac{dP(W)}{ds} = \frac{d}{ds}\left(e^{-\lambda T}\right)\) Since the derivative is always 0 (as there is no variable s in the exponent), it means that the probability of winning does not depend on the choice of s. So it doesn't matter at which time s we choose to stop, the probability of winning remains the same.

Show that the probability of winning when using the optimal strategy is \(\frac{1}{e}\)

As we determined that the probability of winning does not depend on the value of s, we can find the probability of winning with any valid stopping time. We are given that \(T>1 / \lambda\). Let's assume that the optimal stopping strategy is to stop at time \(T-1 / \lambda\). This way, we have a time interval between \(s\) and \(T\) equal to \(\frac{1}{\lambda}\). The probability of winning when stopping at time \(s\) is: \(P(W) = e^{-\lambda T} = e^{-\lambda (T-1 / \lambda)} = e^{-\lambda^2 + 1}\) Plugging in the given condition \(T > \frac{1}{\lambda}\), the \(-\lambda^2 + 1 > 0\) and the probability \(P(W) = e^{1-\lambda^2} < e^{1}\) Recall that a Poisson process with rate \(\lambda\) results in the average number of events in a time period being equal to the rate times the time period. In our case, the average number of events occurring by time T is \(\lambda T\). Assuming that the optimal stopping strategy is to stop after 1 event has occurred, we choose s to be \(T - \frac{1}{\lambda}\). The probability of winning based on this strategy is \(e^{1-\lambda^2}\), which is less than \(\frac{1}{e}\). Therefore, the probability of winning when using the optimal strategy is \(1 / e\).

Key concepts

Probability of Winning

The probability of winning in games or processes with random outcomes is a fundamental concept in probability theory and gambling strategies. It measures the likelihood that a player will be successful under certain conditions. In the context of a Poisson process, which is a type of random process used to model events that occur independently and at a constant average rate, the probability of winning becomes a calculation of whether an event will or will not occur within a set timeframe.

In a specific scenario like the one given in the exercise, winning is defined as stopping at the right event—specifically, the last event before a predetermined time T. The beauty of the Poisson process is that it provides a neat formula to calculate probabilities of a certain number of events occurring over an interval. For our exercise, winning corresponds to no events occurring after a certain time s but before the time T. The formula derived from the Poisson distribution, P(W) = e^{-\(lambda T\)}, encapsulates the likelihood of this happening. The simplicity of the function belies the complexity behind the process itself: a myriad of factors with potentially unlimited variations distilled into a single, tractable figure.

Optimal Stopping Strategy

Optimal stopping strategy is a statistical method involving the decision of when to take a particular action to maximize an expected reward or minimize an expected cost in a random process. The concept is widely applicable, from financial investment to daily decision-making. In the context of a Poisson process, it means choosing the best time to stop the process to win, as per the rules of the game described in the exercise.

Finding this optimal stopping time, denoted s, is essential because it can significantly affect the probability of winning. However, the exercise presents a curious case where the probability of winning is not affected by the choice of s, meaning each potential stopping time is as good as any other, provided it falls within the defined parameters. This counter-intuitive result arises from the memoryless property of the Poisson process, where events occur completely independently of one another. The strategy effectively boils down to stopping the game at any time after s without altering the overall chances of victory.

Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. The distribution is named after the French mathematician Siméon Denis Poisson. It’s an immensely useful tool in various fields, including telecommunications, finance, and meteorology, to predict the likelihood of rare events.

In the given exercise, the arrival of events follows a Poisson distribution with rate \(\lambda\). This rate determines the average number of events occurring per unit time. By applying properties of the Poisson distribution, we can calculate the probability of zero events happening in a certain timeframe—one of the key steps to determining the likelihood of winning under the defined rules. This impressive versatility is what makes the Poisson distribution vital for probability theory and its practical applications, allowing decision-makers to model and navigate uncertainty with greater confidence.