Question

An event independently occurs on each day with probability \(p .\) Let \(N(n)\) denote the total number of events that occur on the first \(n\) days, and let \(T_{r}\) denote the day on which the \(r\) th event occurs. (a) What is the distribution of \(\mathrm{N}(n)\) ? (b) What is the distribution of \(T_{1}\) ? (c) What is the distribution of \(T_{r}\) ? (d) Given that \(N(n)=r\), show that the set of \(r\) days on which events occurred has the same distribution as a random selection (without replacement) of \(r\) of the values \(1,2, \ldots, n\)

Short answer

In summary: (a) The distribution of N(n) is Binomial(n, p): \[P(N(n) = k) = \binom{n}{k}p^k(1-p)^{n-k}\] (b) The distribution of T_1 is a geometric series: \[P(T_1 = k) = (1-p)^{k-1}p\] (c) The distribution of T_r is given by the conditional probability formula: \[P(T_r = m) = \sum_{k=1}^{m-1} P(T_{r-1} = k) (1-p)^{m-k-1}p\] (d) The set of r days with events has the same distribution as a random selection of r values from the set {1,2,...,n} when N(n) = r, as the probabilities of specific sets of r days are equal and correspond to the uniform distribution of choosing r values.

Step-by-step solution

(a) Finding the Distribution of N(n)

The total number of events that occur on the first n days is N(n). Since an event occurs on each day with probability p independent of other days, N(n) follows a binomial distribution. Therefore, the distribution of N(n) is Binomial(n, p), which can be represented as follows: \[P(N(n) = k) = \binom{n}{k}p^k(1-p)^{n-k}\]

(b) Finding the Distribution of T_1

We are looking for the distribution of T_1, which is the day the first event occurs. The first event occurs on day k with probability p (the independent probability of the event occurring per day) times the probability that the event will not occur in the first k-1 days (which is (1-p)^(k-1)). Therefore, the distribution of T_1 is given by the following geometric series: \[P(T_1 = k) = (1-p)^{k-1}p\]

(c) Finding the Distribution of T_r

To find the distribution of T_r, the day on which the rth event occurs, we will use the concept of conditional probability. Given T_{r-1} = k, the rth event will occur on day m (where m > k) with probability p * (1-p)^(m - k - 1), since the rth event occurs with probability p and there should be no event between day k and m. Then, we sum these conditional probabilities over all possible values of k, ranging from 1 to m-1: \[P(T_r = m) = \sum_{k=1}^{m-1} P(T_{r-1} = k) (1-p)^{m-k-1}p\]

(d) Proving the Distribution

We are given that N(n)=r. This means that there are r events in the first n days. We will show that the set of r days with events has the same distribution as a random selection of r values from the set {1,2,...,n}. Notice that the favorable outcomes correspond to the ways of choosing r events from n days, which are counted by the binomial coefficient \(\binom{n}{r}\). Each of these \(\binom{n}{r}\) outcomes is equally likely. Furthermore, the probability of any specific set of r days having events under the Binomial(n, p) distribution is also equal, as: \[P(\text{Specific r days with events}) = p^r(1-p)^{n-r}\] Thus, the uniform distribution of choosing r values from the set {1,2,...,n} corresponds to the distribution of the set of r event days when N(n) = r.

Key concepts

Binomial Distribution

Understanding the binomial distribution is essential when discussing probability models where there are two potential outcomes for an event, often termed success and failure. Imagine you're flipping a coin several times; how would you calculate the likelihood of getting a specific number of heads? That's where the binomial distribution comes in.

Its formula is represented as \[P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\] where \(n\) is the number of trials, \(k\) is the number of desired successes, \(p\) is the probability of success on a single trial, and \(1-p\) is the probability of failure. As shown in the exercise, since events occur independently each day with probability \(p\), the total number of events in \(n\) days (\(N(n)\)) follows a binomial distribution, Binomial(\(n\), \(p\)).

For instance, if a student wants to calculate the likelihood of getting exactly four heads in six coin tosses where each toss has a 50% chance of resulting in heads, they would apply the binomial formula with \(n\) equaling 6, \(k\) equaling 4, and \(p\) being 0.5. These calculations demonstrate binomial distribution's power to predict outcomes over several trials.

Geometric Distribution

Moving on to the geometric distribution, this model is applicable in scenarios where you are interested in the number of trials needed for a first success in a series of independent trials, each with the same probability of success. Unlike the binomial distribution that handles multiple successes, the geometric distribution focuses on the first one.

Its formula is \[P(X = k) = (1-p)^{k-1}p\] where \(X\) is the random variable representing the trial on which the first success occurs, \(p\) is the success probability, and \(k\) is the number of trials. This distribution was demonstrated in part (b) of the exercise for \(T_1\), which is the day the first event happens. It illustrates that the likelihood of the first event occurring on day \(k\) is solely dependent on whether events did not occur in the preceding \(k-1\) days and then finally occurring on the \(k\)th day.

Conditional Probability

Conditional probability comes into play when evaluating the likelihood of an event given the occurrence of another event. It's the probability of event A happening provided that event B has already occurred, often expressed as \(P(A|B)\).

In our exercise, conditional probability was used to determine the distribution of the \(T_r\), the day on which the \(r\)th event occurs. We looked at the likelihood of this happening given the day the previous event happened. This involves summing up probabilities of all possible prior days' events to get the total probability for \(T_r\): \[P(T_r = m) = \sum_{k=1}^{m-1} P(T_{r-1} = k) (1-p)^{m-k-1}p\]
The calculation of conditional probabilities often involves complex operations but is fundamental in understanding sequences of dependent events.

Combinatorics

Combinatorics is a field of mathematics primarily concerned with counting, sequencing, and arranging groups of objects. In probability, it frequently deals with calculating the number of ways certain outcomes can occur. Part (d) of our exercise required understanding of combinatorial principles.

When given that \(N(n)=r\), the query was, 'How many ways can these \(r\) events occur over the span of \(n\) days?' The answer resides in the binomial coefficient \[\binom{n}{r}\] which tells us the number of ways to choose \(r\) successes (event days) out of \(n\) trials (days), without regard to order. As the exercise postulated, this combinatorial count equates to how a random selection of \(r\) days from \(1\) to \(n\) without replacement would distribute, proving the powerful link between combinatorics and probability.