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Chapter 5: Problem 40

Suppose that \(Y_{1}\) and \(Y_{2}\) are independent binomial distributed random variables based on samples of sizes \(n_{1}\) and \(n_{2},\) respectively. Suppose that \(p_{1}=p_{2}=p .\) That is, the probability of "success" is the same for the two random variables. Let \(W=Y_{1}+Y_{2} .\) In Chapter 6 you will prove that \(W\) has a binomial distribution with success probability \(p\) and sample size \(n_{1}+n_{2}\). Use this result to show that the conditional distribution of \(Y_{1}\), given that \(W=w\), is a hypergeometric distribution with \(N=n_{1}+n_{2},\) and \(n=w,\) and \(r=n_{1}\)

Short Answer

Expert verified
The conditional distribution of \( Y_1 \), given \( W = w \), is hypergeometric with \( N=n_1+n_2 \), \( n=w \), \( r=n_1 \).

Step by step solution

01

Understand the Problem

We need to show that the conditional distribution of \( Y_1 \), given \( W = w \), follows a hypergeometric distribution with parameters \( N = n_1 + n_2 \), \( n = w \), and \( r = n_1 \). The total successes \( W \) follows a binomial distribution based on the combined sample size \( n_1 + n_2 \) and the same probability of success \( p \).
02

Identify the Binomial Setup

Recognize that both \( Y_1 \sim \text{Binomial}(n_1, p) \) and \( Y_2 \sim \text{Binomial}(n_2, p) \). The random variable \( W = Y_1 + Y_2 \) is the sum of two independent binomial variables, each with the same probability of success \( p \). Thus, \( W \sim \text{Binomial}(n_1 + n_2, p) \).
03

Recall Hypergeometric Distribution

The hypergeometric distribution models the number of successes in a sequence of draws without replacement. Its parameters are the population size \( N \), the number of success states in the population \( n \), and the number of draws \( r \).
04

Relate Conditional Distribution to Hypergeometric

The conditional distribution of \( Y_1 \) given \( W = w \) can be seen as selecting \( w \) successes (from \( Y_1 + Y_2 \)) distributed over two groups \( n_1 \) and \( n_2 \). This mirrors the setup of drawing \( w \) successes without replacement from \( n_1 + n_2 \) total opportunities, where \( n_1 \) corresponds to one subgroup. Thus, \( Y_1|W=w \sim \text{Hypergeometric}(n_1 + n_2, w, n_1) \).
05

Conclusion

By recognizing the combinatorial nature of the problem and comparing it with the properties of the hypergeometric distribution, it confirms that \( Y_1 \), given \( W = w \), follows a hypergeometric distribution. Specifically, \( Y_1|W=w \) selects \( w \) success states from \( n_1 \) success opportunities out of \( N = n_1 + n_2 \).

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Key Concepts

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

conditional distribution
In simple terms, conditional distribution is the probability distribution of a random variable, provided that another random variable takes on a specific value. In this case, we are looking at the distribution of \( Y_1 \) given that \( W = w \). This means we are interested in the probability characteristics of \( Y_1 \) when the total number of successes \( W \) across both trials is exactly \( w \).
This concept is pivotal because it allows us to determine the likelihood of a particular outcome within a subset of a broader condition. In our exercise, the conditional setup transforms into a hypergeometric distribution. This happens because we are essentially picking a subset from a larger set in one trial context, without replacement. This resembles drawing \( w \) items from a combined dataset of \( n_1 + n_2 \), where our focus is on the subgroup of size \( n_1 \).
binomial distribution
The binomial distribution is central to understanding many probability problems. It models the number of successes in a series of independent and identical trials, where each trial has only two possible outcomes: success or failure.
  • \( Y_1 \sim \text{Binomial}(n_1, p) \) means \( Y_1 \) represents the number of successes in \( n_1 \) trials with success probability \( p \).
  • Similarly, \( Y_2 \sim \text{Binomial}(n_2, p) \).
  • The total successes \( W = Y_1 + Y_2 \) is a binomial distribution because it sums samples from two independent binomial processes.
Understanding the binomial distribution helps us see why \( W \) can also be represented as \( \text{Binomial}(n_1 + n_2, p) \): it鈥檚 a large single trial combining all the smaller ones. Recognizing \( Y_1 \) and \( Y_2 \) contributions in this context makes the transition to the conditional hypergeometric more intuitive.
probability distribution
A probability distribution provides a model for how a random variable behaves. It tells us the likelihood of every possible outcome of the variable. Different situations utilize different probability distributions depending on their nature.
In this exercise, we see two primary types of probability distributions:
  • Binomial Distribution: As described earlier, it applies when each trial is independent, and the probability of success remains constant across trials.
  • Hypergeometric Distribution: It contrasts with binomial because it involves trials without replacement, which changes the probabilities as 'draws' occur sequentially.
The exercise transitions from a binomial setup to a conditional hypergeometric distribution due to known constraints \( W=w \). This shift helps us validate our approach to problems involving combined experiments (adding \( Y_1 \) and \( Y_2 \)). It emphasizes how knowledge of specific distributions can simplify problem-solving by matching real-world conditions to theoretical models effectively.

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Most popular questions from this chapter

Let \(Y_{1}\) and \(Y_{2}\) be jointly distributed random variables with finite variances. a. Show that \(\left[E\left(Y_{1} Y_{2}\right)\right]^{2} \leq E\left(Y_{1}^{2}\right) E\left(Y_{2}^{2}\right) .\) [Hint: Observe that \(E\left[\left(t Y_{1}-Y_{2}\right)^{2}\right] \geq 0\) for any real number t or, equivalently, $$t^{2} E\left(Y_{1}^{2}\right)-2 t E\left(Y_{1} Y_{2}\right)+E\left(Y_{2}^{2}\right) \geq 0$$ This is a quadratic expression of the form \(A t^{2}+B t+C\); and because it is nonnegative, we must have \(B^{2}-4 A C \leq 0 .\) The preceding inequality follows directly.] b. Let \(\rho\) denote the correlation coefficient of \(Y_{1}\) and \(Y_{2} .\) Using the inequality in part (a), show that \(\rho^{2} \leq 1\)

Let \(Z\) be a standard normal random variable and let \(Y_{1}=Z\) and \(Y_{2}=Z^{2}\). a. What are \(E\left(Y_{1}\right)\) and \(E\left(Y_{2}\right) ?\) b. What is \(E\left(Y_{1} Y_{2}\right) ?\left[\text { Hint: } E\left(Y_{1} Y_{2}\right)=E\left(Z^{3}\right), \text { recall Exercise 4.199. }\right]\) c. What is \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right) ?\) d. Notice that \(P\left(Y_{2}>1 | Y_{1}>1\right)=1 .\) Are \(Y_{1}\) and \(Y_{2}\) independent?

Suppose that a company has determined that the the number of jobs per week, \(N\), varies from week to week and has a Poisson distribution with mean \(\lambda\). The number of hours to complete each job, \(Y_{i},\) is gamma distributed with parameters \(\alpha\) and \(\beta\). The total time to complete all jobs in a week is \(T=\sum_{i=1}^{N} Y_{i} .\) Note that \(T\) is the sum of a random number of random variables. What is a. \(E(T | N=n) ?\) b. \(E(T)\), the expected total time to complete all jobs?

A retail grocery merchant figures that her daily gain \(X\) from sales is a normally distributed random variable with \(\mu=50\) and \(\sigma=3\) (measurements in dollars). \(X\) can be negative if she is forced to dispose of enough perishable goods. Also, she figures daily overhead costs \(Y\) to have gamma distribution with \(\alpha=4\) and \(\beta=2\). If \(X\) and \(Y\) are independent, find the expected value and variance of her net daily gain. Would you expect her net gain for tomorrow to rise above \(\$ 70 ?\)

In Exercise 5.12 we were given the following joint probability density function for the random variables \(Y_{1}\) and \(Y_{2},\) which were the proportions of two components in a sample from a mixture of insecticide: $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 2, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1,0 \leq y_{1}+y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find \(P\left(Y_{1} \geq 1 / 2 | Y_{2} \leq 1 / 4\right)\) b. Find \(P\left(Y_{1} \geq 1 / 2 | Y_{2}=1 / 4\right)\)
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