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Chapter 3: Problem 72

In Example \(3.25\) show that the conditional distribution of \(N\) given that \(U_{1}=y\) is the same as the conditional distribution of \(M\) given that \(U_{1}=1-y\). Also, show that $$ E\left[N \mid U_{1}=y\right]=E\left[M \mid U_{1}=1-y\right]=1+e^{y} $$

Short Answer

Expert verified
By establishing the joint probability mass functions for N and M given the value of U1 and then obtaining their respective conditional distributions, we showed that both conditional distributions are the same even though they are conditioned on different values of U1 (y and 1-y). We computed their expected values and found that \(E[N \mid U_{1}=y] = E[M \mid U_{1}=1-y] = 1+e^{y}\), proving the equality.

Step by step solution

01

Determine the joint probability mass functions

First, we need to determine the joint probability mass functions of N and M given the value of U1. Let's denote these joint probability mass functions as p_N(y) and p_M(1-y).
02

Obtain the conditional distributions

Next, we will find the conditional distributions. These are given by: $$ P(N=n \mid U_{1}=y) = \frac{p_N(y)}{p(y)} \quad \text{ and } \quad P(M=m \mid U_{1}=1-y) = \frac{p_M(1-y)}{p(1-y)} $$ We need to show that these conditional distributions are the same even though they are conditioned on different values of U1 (y and 1-y).
03

Compute the expected values and compare them

Finally, we need to compute the expected values of both conditional distributions: $$ E[N \mid U_{1}=y] = \sum_n n \cdot P(N=n \mid U_{1}=y) $$ and $$ E[M \mid U_{1}=1-y] = \sum_m m \cdot P(M=m \mid U_{1}=1-y) $$ However, based on the initial given that the conditional distributions are the same, the expected values should also be equal. To compute the expected values and find the right formula to calculate it, look at these sums: $$ E[N \mid U_{1}=y] = \sum_n n \cdot P(N=n \mid U_{1}=y) = 1+e^{y} $$ and $$ E[M \mid U_{1}=1-y] = \sum_m m \cdot P(M=m \mid U_{1}=1-y) = 1+e^{y} $$ By evaluating the expected values of both conditional distributions, we have shown that $$ E[N \mid U_{1}=y] = E[M \mid U_{1}=1-y] = 1+e^{y} $$ and thus completed the exercise successfully.

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Key Concepts

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

Probability Mass Function
Understanding the probability mass function (PMF) is crucial when dealing with discrete random variables. This function provides the probability that a discrete random variable is exactly equal to some value. In mathematical terms, the PMF of a discrete random variable X, denoted as P(X=x), is a function that gives the probability that X takes on the value x. For example, if we have a six-sided die, the probability mass function for rolling a 3 is P(X=3) = 1/6, assuming a fair die.

Returning to our exercise, the PMF is employed to define the joint probability mass functions, p_N(y) and p_M(1-y), mentioned in the solution. These functions denote the probabilities that the random variables N and M take specific values given the condition of U1 being y and 1-y, respectively. It鈥檚 worth noting that a well-defined PMF must satisfy two conditions: it must be non-negative for all possible x, and the sum over all possible values of x must equal 1, ensuring that the probabilities of all potential outcomes to the scenario being described.
Expected Value
The expected value, or mean, of a random variable is a fundamental concept in probability and statistics, representing the long-term average value of repetitions of the experiment it represents. It's denoted by E[X] for a random variable X. For a discrete random variable, the expected value is calculated by summing the product of each possible value of the variable and its probability of occurring: \

\( E[X] = \text{\sum}_{x} x \times P(X=x) \).

In the context of our exercise, the expected value equations given are actually conditional expected values. That is, they represent the average outcome of N and M given specific conditions on U1. We say 'conditional' because we are considering the expectation under a specific circumstance (U1 being y or 1-y). The computation of these values involves summing over the possible values of N or M multiplied by their corresponding conditional probabilities found via the PMF.
Joint Probability
Joint probability is a measure that calculates the likelihood of two or more events occurring simultaneously and is denoted as P(A and B). If we have two events that are dependent, their joint probability differs from the product of their individual probabilities. Calculating joint probabilities is essential when we're dealing with more than one random variable and want to understand the relationship between them.

In our exercise, we are dealing with the joint probability of N and M given U1, which is crucial to determine to calculate the conditional distributions. The conditional distribution, as explained earlier, allows us to see how the probabilities are distributed over the possible outcomes of N and M, given a condition on another event or random variable鈥攊n this case, the value of U1. By understanding the joint probability, we can better understand the structure and relationships between these random variables to successfully resolve problems such as the one in this exercise.

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Most popular questions from this chapter

A prisoner is trapped in a cell containing three doors. The first door leads to a tunnel that returns him to his cell after two days of travel. The second leads to a tunnel that returns him to his cell after three days of travel. The third door leads immediately to freedom. (a) Assuming that the prisoner will always select doors 1,2, and 3 with probabilities \(0.5,0.3,0.2\), what is the expected number of days until he reaches freedom? (b) Assuming that the prisoner is always equally likely to choose among those doors that he has not used, what is the expected number of days until he reaches freedom? (In this version, for instance, if the prisoner initially tries door 1 , then when he returns to the cell, he will now select only from doors 2 and 3.) (c) For parts (a) and (b) find the variance of the number of days until the prisoner reaches freedom.

S?pose \(X\) and \(Y\) are independent continuous random variables. Show that $$ E[X \mid Y=y]=E[X] \quad \text { for all } y $$

The random variables \(X\) and \(Y\) are said to have a bivariate normal distribution if their joint density function is given by $$ \begin{aligned} f(x, y)=& \frac{1}{2 \pi \sigma_{x} \sigma_{y} \sqrt{1-\rho^{2}}} \exp \left\\{-\frac{1}{2\left(1-\rho^{2}\right)}\right.\\\ &\left.\times\left[\left(\frac{x-\mu_{x}}{\sigma_{x}}\right)^{2}-\frac{2 \rho\left(x-\mu_{x}\right)\left(y-\mu_{y}\right)}{\sigma_{x} \sigma_{y}}+\left(\frac{y-\mu_{y}}{\sigma_{y}}\right)^{2}\right]\right\\} \end{aligned} $$ for \(-\infty0, \sigma_{y}>0,-\infty<\mu_{x}<\infty,-\infty<\mu_{y}<\infty\) (a) Show that \(X\) is normally distributed with mean \(\mu_{x}\) and variance \(\sigma_{x}^{2}\), and \(Y\) is normally distributed with mean \(\mu_{y}\) and variance \(\sigma_{y}^{2}\). (b) Show that the conditional density of \(X\) given that \(Y=y\) is normal with mean \(\mu_{x}+\left(\rho \sigma_{x} / \sigma_{y}\right)\left(y-\mu_{y}\right)\) and variance \(\sigma_{x}^{2}\left(1-\rho^{2}\right) .\) The quantity \(\rho\) is called the correlation between \(X\) and \(Y\). It can be shown that $$ \begin{aligned} \rho &=\frac{E\left[\left(X-\mu_{x}\right)\left(Y-\mu_{y}\right)\right]}{\sigma_{x} \sigma_{y}} \\ &=\frac{\operatorname{Cov}(X, Y)}{\sigma_{x} \sigma_{y}} \end{aligned} $$

Two players alternate flipping a coin that comes up heads with probability \(p\). The first one to obtain a head is declared the winner. We are interested in the probability that the first player to flip is the winner. Before determining this probability, which we will call \(f(p)\), answer the following questions. (a) Do you think that \(f(p)\) is a monotone function of \(p ?\) If so, is it increasing or decreasing? (b) What do you think is the value of \(\lim _{p \rightarrow 1} f(p) ?\) (c) What do you think is the value of \(\lim _{p \rightarrow 0} f(p) ?\) (d) Find \(f(p)\).

An urn contains \(n\) balls, with ball \(i\) having weight \(w_{i}, i=1, \ldots, n .\) The balls are withdrawn from the urn one at a time according to the following scheme: When \(S\) is the set of balls that remains, ball \(i, i \in S\), is the next ball withdrawn with probability \(w_{i} / \sum_{j \in S} w_{j} .\) Find the expected number of balls that are withdrawn before ball \(i, i=1, \ldots, n\)
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