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Chapter 3: Problem 35

Consider \(n\) multinomial trials, where each trial independently results in outcome \(i\) with probability \(p_{i}, \sum_{i=1}^{k} p_{i}=1 .\) With \(X_{i}\) equal to the number of trials that result in outcome \(i\), find \(E\left[X_{1} \mid X_{2}>0\right]\)

Short Answer

Expert verified
The expected number of trials resulting in outcome 1 given that there's at least one trial resulting in outcome 2 is given by: $$E[X_1\mid X_2 >0] = \frac{\sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}}{1-p_1^n}$$

Step by step solution

01

In a multinomial distribution, the joint PMF of (\(X_1\), \(X_2\)) is given by: $$P(X_1=x_1, X_2=x_2) = \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}$$ Here, we consider only the cases where \(0\leq x_1 \leq n - 1\) and \(0\leq x_2 \leq n - x_1\), since \(X_1 + X_2 \leq n\). The rest of the trials will fall into outcomes 3,4,5,...,k. #Step 2: Find \(E[X_1 \mathbb{1}_{\{X_2 > 0\}}]\)#

In order to find \(E[X_1\mathbb{1}_{\{X_2>0\}}]\), we'll first compute \(x_1 P(X_1=x_1, X_2=x_2)\) for each possible value of \(x_1\) and \(x_2\) where \(x_2 > 0\), and then sum over all possibilities: $$E[X_1\mathbb{1}_{\{X_2>0\}}]= \sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 P(X_1=x_1, X_2=x_2) = \sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}$$ #Step 3: Find \(P(X_2 > 0)\)#
02

Next, we'll compute the probability \(P(X_2 > 0)\) which means at least one trial will result in outcome 2. We can calculate this as the complement: \(P(X_2 > 0) = 1 - P(X_2 = 0)\). Since all outcomes are independent, we have: $$P(X_2=0) = n!\frac{p_1^{n-x_2} p_3^{x_2}}{(n-x_2)!x_2!}$$ Substitute \(x_2=0\) to the above equation and subtract from 1: $$P(X_2 > 0) = 1 - n!\frac{p_1^n}{n!} = 1 - p_1^n$$ #Step 4: Calculate \(E[X_1 | X_2 > 0]\)#

Finally, we can use the definition of conditional expectation to find the desired expectation: $$E[X_1\mid X_2 >0] = \frac{E[X_1\mathbb{1}_{\{X_2>0\}}]}{P(X_2>0)}= \frac{\sum_{x_1=0}^{n-1} \sum_{x_2=1}^{n-x_1} x_1 \frac{n!}{x_1! x_2! (n-x_1-x_2)!} p_1^{x_1} p_2^{x_2} p_3^{n-x_1-x_2}}{1-p_1^n}$$ This expression gives us the expected number of trials resulting in outcome 1 given that there's at least one trial resulting in outcome 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multinomial Distribution
The multinomial distribution is a crucial statistical concept used when dealing with scenarios that have more than two possible outcomes. It extends from the binomial distribution to handle more than two categories. Imagine performing a series of experiments, where each trial can result in one of several category outcomes, such as rolling a dice. Each side of the dice is a possible outcome.
In a multinomial setup:
  • We have a fixed number of trials, denoted by \( n \).
  • Each trial is independent.
  • The probabilities for all possible outcomes add up to one, i.e., \( \sum_{i=1}^{k} p_{i} = 1 \), where \( k \) is the number of possible outcomes.
This forms the basis for more complex probability computations and allows us to model real-world scenarios where multiple outcomes are possible.
Probability Mass Function
In statistics, the Probability Mass Function (PMF) is essential for understanding how probabilities are distributed across discrete random variables. For a multinomial distribution, the PMF provides the probability of a particular combination of outcomes.
It is represented as:
  • \( P(X_1 = x_1, X_2 = x_2, ..., X_k = x_k) = \frac{n!}{x_1! x_2!...x_k!} p_1^{x_1} p_2^{x_2}...p_k^{x_k} \)
This formula shows the likelihood of obtaining the sequence \( (x_1, x_2, ..., x_k) \) from \( n \) trials. Each \( x_i \) represents the number of times outcome \( i \) is observed.
Understanding PMF helps predict the behavior of random variables, a useful tool in both simple and complex probability scenarios.
Joint Probability
Joint probability is the probability of multiple events occurring at the same time. In the context of our multinomial distribution, joint probability allows us to calculate the probability of specific outcomes happening together.
For two events, the joint probability \( P(A \cap B) \) can be found by multiplying the probabilities of each event under independence assumptions. However, in a multinomial distribution context, events are dependent on the same trial outcomes. It becomes more about working through a distribution formula like the PMF for multiple outcomes.
The joint probability can range from zero (no chance that both events occur) to one (certainty that both events occur together), enabling us to understand relationships between different outcomes within a single experiment or model.
Expectation Computation
Expectation or expected value is a vital statistical measure indicating the mean or average value of a random variable. In the context of conditional expectation, it is calculated given a certain condition, such as another variable taking a particular value.
The conditional expectation \( E[X | Y] \) provides insights into expected outcomes based on certain conditions.
To compute expectation for a multinomial distribution, you sum up the products of each possible outcome value and its probability. Conditional expectation specifically requires separating out the relevant probabilities, often involving calculating probabilities using available joint distributions or probability mass functions.
For example, finding \( E[X_1 | X_2 > 0] \), involves:
  • Computing \( E[X_1 \,\mathbb{1}_{\{X_2>0\}}] \), using only the values where \( X_2 \) is greater than zero.
  • Then, divide this by \( P(X_2 > 0) \) to get the conditional expectation.
This calculation is central in probabilistic models where events are connected, giving a deeper understanding of conditional scenarios.

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Most popular questions from this chapter

\(A, B\), and \(C\) are evenly matched tennis players. Initially \(A\) and \(B\) play a set, and the winner then plays \(C\). This continues, with the winner always playing the waiting player, until one of the players has won two sets in a row. That player is then declared the overall winner. Find the probability that \(A\) is the overall winner.

A deck of \(n\) cards, numbered 1 through \(n\), is randomly shuffled so that all \(n !\) possible permutations are equally likely. The cards are then turned over one at a time until card number 1 appears. These upturned cards constitute the first cycle. We now determine (by looking at the upturned cards) the lowest numbered card that has not yet appeared, and we continue to turn the cards face up until that card appears. This new set of cards represents the second cycle. We again determine the lowest numbered of the remaining cards and turn the cards until it appears, and so on until all cards have been turned over. Let \(m_{n}\) denote the mean number of cycles. (a) Derive a recursive formula for \(m_{n}\) in terms of \(m_{k}, k=1, \ldots, n-1\). (b) Starting with \(m_{0}=0\), use the recursion to find \(m_{1}, m_{2}, m_{3}\), and \(m_{4}\). (c) Conjecture a general formula for \(m_{n}\). (d) Prove your formula by induction on \(n\). That is, show it is valid for \(n=1\), then assume it is true for any of the values \(1, \ldots, n-1\) and show that this implies it is true for \(n\). (e) Let \(X_{i}\) equal 1 if one of the cycles ends with card \(i\), and let it equal 0 otherwise, \(i=1, \ldots, n\). Express the number of cycles in terms of these \(X_{i}\). (f) Use the representation in part (e) to determine \(m_{n}\). (g) Are the random variables \(X_{1}, \ldots, X_{n}\) independent? Explain. (h) Find the variance of the number of cycles.

Data indicate that the number of traffic accidents in Berkeley on a rainy day is a Poisson random variable with mean 9 , whereas on a dry day it is a Poisson random variable with mean \(3 .\) Let \(X\) denote the number of traffic accidents tomorrow. If it will rain tomorrow with probability \(0.6\), find (a) \(E[X]\); (b) \(P[X=0\\}\) (c) \(\operatorname{Var}(X)\)

Suppose there are \(n\) types of coupons, and that the type of each new coupon obtained is independent of past selections and is equally likely to be any of the \(n\) types. Suppose one continues collecting until a complete set of at least one of each type is obtained. (a) Find the probability that there is exactly one type \(i\) coupon in the final collection. Hint: Condition on \(T\), the number of types that are collected before the first type \(i\) appears. (b) Find the expected number of types that appear exactly once in the final collection.

There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities \(0.3,0.5,0.7 .\) A coin is randomly selected from among these three and is then flipped ten times. Let \(N\) be the number of heads obtained on the ten flips. (a) Find \(P[N=0\\}\). (b) Find \(P[N=n\\}, n=0,1, \ldots, 10\) (c) Does \(N\) have a binomial distribution? (d) If you win \(\$ 1\) each time a head appears and you lose \(\$ 1\) each time a tail appears, is this a fair game? Explain.
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