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Chapter 3: Problem 88

In Section 3.6.3, we saw that if \(U\) is a random variable that is uniform on \((0,1)\) and if, conditional on \(U=p, X\) is binomial with parameters \(n\) and \(p\), then $$ P\\{X=i\\}=\frac{1}{n+1}, \quad i=0,1, \ldots, n $$ For another way of showing this result, let \(U, X_{1}, X_{2}, \ldots, X_{n}\) be independent uniform \((0,1)\) random variables. Define \(X\) by \(X=\\# i: X_{i}

Short Answer

Expert verified
In summary, we found that the probability of X having exactly i values smaller than U is given by \( P\{X=i\}=\binom{n}{i}U^i(1-U)^{n-i} \), and the expected value of this probability is \( E[P\{X=i\}]=\frac{1}{n+1} \). This result is the same as the one provided in Section 3.6.3, which proves the relationship between this problem and the conditional binomial distribution based on a uniform random variable U.

Step by step solution

01

Finding the probability of X having a particular value

Since X is defined as the number of Xi values smaller than U, we want to calculate the probability of exactly i Xi values being smaller than U while the rest are larger than U. This event can be represented as follows: \(P\{X = i\} = P\{X_1 < U, X_2 < U, \ldots, X_i < U, X_{i+1} > U, \ldots, X_n > U\}\) We can first calculate the individual probabilities of each condition given that these random variables are uniform on (0,1): \( P\{X_j < U\} = P\{U - X_j > 0\} = U\) \(P\{X_k > U\} = P\{X_k - U > 0\} = 1 - U\)
02

Calculating the probability for all cases

Using the fact that all the random variables are independent, we can calculate the probability of all these conditions happening simultaneously by multiplying the probabilities of each condition: \(P\{X = i\} = \binom{n}{i} U^i (1 - U)^{n - i}\), where \(\binom{n}{i}\) is the number of possible combinations of choosing i elements from an n-element set. Now, we should find the expected value of this probability: \(E[P\{X = i\}] = \int_0^1 \binom{n}{i} U^i (1 - U)^{n - i} dU\)
03

Evaluate the integral

We will now evaluate the integral with respect to U: \(E[P\{X = i\}] = \binom{n}{i} \int_0^1 U^i (1 - U)^{n-i} dU\) This integral is equivalent to calculating the beta function: \(B(a, b) = \int_0^1 x^{a-1}(1-x)^{b-1}dx\), where a = i+1 and b = n-i+1. Using properties of the beta function, we know that \(B(a, b) = \frac{(a-1)!(b-1)!}{(a+b-1)!}\)
04

Calculate the expected value

Substitute the beta function into the expected value expression: \(E[P\{X = i\}] = \binom{n}{i} \frac{i!(n-i)!}{(n+1)!}\) Notice that this simplifies to: \(E[P\{X = i\}] = \frac{1}{n+1}\)
05

Relate to given result and conclude

Now that we have found \(E[P\{X = i\}] = \frac{1}{n+1}\), we see that it is equal to the provided result. This is consistent with the conditional binomial distribution given in the problem, as the probability \(P\{X = i\}\) also accounts for all possible values of p from the uniform random variable U.

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Key Concepts

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

Uniform Distribution
The Uniform Distribution is a type of probability distribution where all outcomes are equally likely. Imagine spinning a fair wheel. No matter where you stop, each section of the wheel has the same chance of being landed on.
  • In mathematical terms, if we consider a uniform distribution over an interval \(a, b\), each point within this interval is equally probable.
  • For a Uniform Distribution on (0, 1), the probability of observing any value, say U, is constant between 0 and 1.
When applied to a random variable U, such as in our exercise, it means that the random numbers generated from U are evenly spread across the interval (0, 1).
Thus, there isn't any favoritism towards any region within this interval.
Binomial Distribution
The Binomial Distribution is used to model the number of successful outcomes in a fixed number of experiments. Consider flipping a coin multiple times; you're interested in how many times it lands on heads.
  • It is described by two main parameters: the number of trials (n) and the probability of success in each trial (p).
  • The formula to determine the exact probability of seeing a certain number of successes is given by:\[P(X = i) = \binom{n}{i} p^i (1-p)^{n-i}\]where \(\binom{n}{i}\) denotes "n choose i," or the number of ways to select i successes in n trials.
In our exercise, the Binomial Distribution comes into play by examining how many of the uniform random variables fall below a certain threshold U, which itself follows a Uniform Distribution.
The integration of these distributions reveals the uniform probability result.
Beta Function
The Beta Function is a key tool in integration problems, especially those arising in probability. It is integral to deriving various distribution properties.
  • The Beta Function, denoted as \(B(a, b)\), is defined as:\[B(a, b) = \int_0^1 x^{a-1}(1-x)^{b-1}dx\]
  • It serves as a normalization constant in problems pertaining to probability distributions like the Beta Distribution, which is a generalization of the concept of a binomial trial.
Returning to our exercise, the Beta Function bridges the gap between integrating random variables over the uniform distribution interval and simplifying the derived expression by applying known beta properties.
These properties streamline the evaluation process, revealing that certain integrals simplify to ratios of factorials, aligning with well-known distribution results.
Expected Value
The Expected Value is the theoretical mean of a random variable's probability distribution. It tells us what to expect on average if we could repeat a random process infinite times.
  • Formally, it is represented as the sum of all possible values of a random variable, each multiplied by its probability of occurrence.
  • For discrete random variables like in a binomial distribution, it is calculated using:\[E[X] = \sum_{i=0}^{n} i \cdot P(X=i)\]
In continuous cases, expected value uses integration:\[E[X] = \int_{-\infty}^{\infty} x f(x) dx\]where \(f(x)\) is the probability density function.
In the context of our step-by-step solution, the expected value helps us affirm that the probability result \(\frac{1}{n+1}\) aligns perfectly with the continuous integration of the Beta Function, thereby reaffirming the uniform distribution outcome.
This powerful concept unifies discrete and continuous probability analyses, offering insights into results derived from complex problems.

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Most popular questions from this chapter

Independent trials, resulting in one of the outcomes \(1,2,3\) with respective probabilities \(p_{1}, p_{2}, p_{3}, \sum_{i=1}^{3} p_{i}=1\), are performed. (a) Let \(N\) denote the number of trials needed until the initial outcome has occurred exactly 3 times. For instance, if the trial results are \(3,2,1,2,3,2,3\) then \(N=7\) Find \(E[N]\). (b) Find the expected number of trials needed until both outcome 1 and outcome 2 have occurred.

A coin, having probability \(p\) of landing heads, is continually flipped until at least one head and one tail have been flipped. (a) Find the expected number of flips needed. (b) Find the expected number of flips that land on heads. (c) Find the expected number of flips that land on tails. (d) Repeat part (a) in the case where flipping is continued until a total of at least two heads and one tail have been flipped.

The opponents of soccer team \(\mathrm{A}\) are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores against a class \(i\) opponent is a Poisson random variable with mean \(\lambda_{i}\), where \(\lambda_{1}=2, \lambda_{2}=3\). This weekend the team has two games against teams they are not very familiar with. Assuming that the first team they play is a class 1 team with probability \(0.6\) and the second is, independently of the class of the first team, a class 1 team with probability \(0.3\), determine (a) the expected number of goals team A will score this weekend. (b) the probability that team \(\mathrm{A}\) will score a total of five goals.

Let \(X_{1}, X_{2}, \ldots\) be independent continuous random variables with a common distribution function \(F\) and density \(f=F^{\prime}\), and for \(k \geqslant 1\) let $$ N_{k}=\min \left\\{n \geqslant k: X_{n}=k \text { th largest of } X_{1}, \ldots, X_{n}\right\\} $$ (a) Show that \(P\left\\{N_{k}=n\right\\}=\frac{k-1}{n(n-1)}, n \geqslant k\). (b) Argue that $$ f_{X_{N_{k}}}(x)=f(x)(\bar{F}(x))^{k-1} \sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(F(x))^{i} $$ (c) Prove the following identity: $$ a^{1-k}=\sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(1-a)^{i}, \quad 0

Let \(X\) be exponential with mean \(1 / \lambda ;\) that is, $$ f_{X}(x)=\lambda e^{-\lambda x}, \quad 01]\)
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