Problem 62 - A First Course in Probability

Question

Consider \(n\) independent trials, each of which results in one of the outcomes \(1, \ldots, k\) with respective probabilities \(p_{1}, \ldots, p_{k}, \quad \sum_{i=1}^{k} p_{i}=1 .\) Show that if all the \(p_{i}\) are small, then the probability that no trial outcome occurs more than once is approximately equal to \(\exp \left(-n(n-1) \sum_{i} p_{i}^{2} / 2\right)\)

Step-by-Step Solution

Answer

The short answer to the question is: When all the \(p_{i}\) values are small, the probability that no trial outcome occurs more than once, denoted as \(P_{u}\), can be approximated as: \[ P_{u} \approx \exp \left(-n(n-1) \sum_{i=1}^k p_i^2 / 2\right) \] This approximation is derived using the Taylor series expansion and the rules of probability for independent trials.

Step 1: Find the probability of unique outcomes in each trial

The first step is to find out the probability that any specific outcome occurs exactly once in all the n trials. The probability of getting outcome i exactly once will be given by the product of probabilities of getting this outcome once in the n trials and not getting it in the other trials. For instance, let's denote the probability of getting outcome i exactly once in n trials as \(q_{i}\)

Step 2: Calculate the product of all unique trials probabilities

We would like to find out the probability of all outcomes being unique in all the trials. This can be achieved by calculating the product of all unique outcomes probabilities in individual trials. In other words, we need to find the probability \(P_{u}\) that no trial outcome occurs more than once in all the n trials. We can write this as. \[ P_{u} = \prod_{i=1}^k q_{i} \]

Step 3: Apply the Taylor series approximation

Observe that for small p_i values, we have \[ \prod_{i=1}^k (1 - p_i)^{n-1} \approx \exp \left( -\sum_{i=1}^k (n-1)p_i \right) \] Using Taylor series expansion, We can substitute this approximation into \(q_{i}\)

Step 4: Rewrite the overall probability

Now substitute the approximate values from step 3 into the product for \(P_{u}\) from step 2. We get: \[ P_{u} \approx \exp \left( -\sum_{i=1}^k (n-1)p_i \right)\prod_{i=1}^k p_{i} \]

Step 5: Simplify the result

Combine the exponent terms so that it becomes \[ P_{u} \approx \exp \left( -\sum_{i=1}^k(n(n-1)p_i^2 -np_i +p_{i})\right) \] We can sum the terms inside the exponent and we get \[ \sum_{i=1}^k(n(n-1)p_i^2 -np_i +p_{i}) = n(n-1)\sum_{i=1}^k p_i^2 - n\sum_{i=1}^k p_i +\sum_{i=1}^k p_{i} \]

Step 6: Factor the unwanted terms

We have \[ P_u \approx \exp\left(-n(n-1) \sum_{i=1}^k p_i^2 + n\sum_{i=1}^k p_i -\sum_{i=1}^k p_{i} \right) \] Notice that \(\sum_{i=1}^kp_{i}=1\) and that for small \(p_{i}\), the last two terms in the exponent are negligible compared to the first term. Specifically, \(n\sum_{i=1}^kp_{i}\) and \(-\sum_{i=1}^kp_{i}\) have less significant power compared to other terms when all \(p_{i}\) values are small.

Step 7: Conclude the approximation

Since the last two terms in the exponent are negligible when all \(p_{i}\) are small, we can approximate the probability \(P_u\) as: \[ P_{u} \approx \exp \left(-n(n-1) \sum_{i=1}^k p_1^2 / 2\right) \] This is the final approximation result for the probability that no trial outcome occurs more than once when all the \(p_{i}\) are small.

Key Concepts

Independent Trials

When we refer to independent trials, we mean that the outcome of one trial does not influence or affect the outcome of another. This concept is essential in probability theory and is often assumed in various calculations and exercises.

For instance, flipping a fair coin multiple times is an example of independent trials because the result of one coin flip (heads or tails) does not depend on the results of previous flips. Similarly, in the context of the textbook exercise, each trial that results in one of the outcomes from 1 to k is considered to be independent of the other trials. This is a crucial assumption that simplifies the calculation of the overall probability of an event occurring across multiple trials.

Moreover, the notion of independence permits using certain probability rules, such as the multiplication rule. The multiplication rule states that the probability of two independent events both occurring is the product of their separate probabilities. In the context of unique trial outcomes, we exploit this rule to produce the probability that each outcome occurs only once across all the trials.

Outcome Probabilities

Understanding outcome probabilities involves evaluating the chance that a particular event will happen. It can range from simple scenarios, like rolling a die, to more complex situations involving multiple events and outcomes. The probabilities of different outcomes must add up to 1 since one of the outcomes must occur.

In the exercise, we are faced with each of several possible outcomes occurring with certain probabilities, which are denoted by \( p_1, p_2, \.\.\., p_k \). The sum of these probabilities equals 1, representing the certainty that one of the outcomes will happen. We are tasked with finding the probability that each outcome occurs exactly once, which requires us to calculate unique outcome probabilities across multiple trials.

These outcome probabilities are critical for calculating the probability of no outcome occurring more than once. By multiplying the unique outcome probabilities for all outcomes (achieving a sequence of outcomes in which each occurs only once), we obtain a preliminary figure for our desired probability. This becomes the foundation upon which additional approximations and calculations are built to achieve the final solution.

Taylor Series Approximation

The Taylor series approximation is a powerful tool in mathematics used to approximate functions that may be difficult to compute exactly. It expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. In practice, only the first few terms of the series may be used to obtain a reasonable approximation.

For small values of \( p_i \), we can approximate \( (1 - p_i)^{n-1} \) using the Taylor series expansion about \( p_i = 0 \). This approximation simplifies the expressions significantly, making it easier to work with the probabilities involved.

In the exercise solution, the Taylor series approximation is applied to estimate the product of probabilities in the case where each outcome's probability is small. Because the higher-order terms in the Taylor expansion become increasingly insignificant for small \( p_i \), we can terminate the series after the initial terms. This allows us to estimate the probability of a unique set of outcomes across the independent trials, ultimately leading to the exponential expression provided in the exercise. It's crucial for students to grasp the pertinence of this approximation as it greatly simplifies complex probability calculations.