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Chapter 3: Problem 145

If \(Y\) has a binomial distribution with \(n\) trials and probability of success \(p,\) show that the momentgenerating function for \(Y\) is $$m(t)=\left(p e^{t}+q\right)^{n}, \quad \text { where } q=1-p$$

Short Answer

Expert verified
The moment generating function is \((pe^t + q)^n\).

Step by step solution

01

Define the Moment Generating Function

The moment generating function (MGF) of a random variable is defined as \( m(t) = E[e^{tY}] \), where \( E \) denotes the expectation. For a binomially distributed random variable \( Y \), we want to find \( m(t) \).
02

Express the Expectation with Binomial Probability

Since \( Y \) is a binomial random variable with parameters \( n \) and \( p \), its probability mass function is \( P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \) for \( k = 0, 1, \ldots, n \). Thus, \( m(t) = \sum_{k=0}^{n} e^{tk} \binom{n}{k} p^k (1-p)^{n-k} \).
03

Simplify the Expression Inside the Summation

Reorganize the expression within the summation: \( m(t) = \sum_{k=0}^{n} \binom{n}{k} (p e^t)^k ((1-p))^{n-k} \). This shows the summation resembles the binomial expansion of \( (x + y)^n \) with \( x = pe^t \) and \( y = 1-p \).
04

Recognize the Binomial Expansion

The expression \( \sum_{k=0}^{n} \binom{n}{k} (pe^t)^k (1-p)^{n-k} \) is the binomial expansion of \( (pe^t + 1-p)^n \). Thus, the moment generating function simplifies to \( (pe^t + (1-p))^n \).
05

Substitute for \( q \) and Simplify

We know \( q = 1 - p \). Substitute \( q \) into the expression: \( m(t) = (pe^t + q)^n \). Hence, the moment generating function is \( m(t) = (pe^t + q)^n \).

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Key Concepts

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

Binomial Distribution
The Binomial Distribution is a fundamental concept in probability theory. It describes the number of successes in a fixed number of independent trials under the same conditions. Each trial has two possible outcomes: success or failure.
  • Trials must be independent, meaning the outcome of one does not affect others.
  • The probability of success, denoted as \( p \), remains constant in all trials.
If we denote the number of trials as \( n \), a random variable \( Y \), which counts the success, is said to follow a binomial distribution. The notation for this distribution is \( \text{Binomial}(n, p) \). Understanding this distribution is crucial, as it aids in calculating probabilities in various scenarios, such as genetics, finance, or any field requiring decision-making under uncertainty.
Probability Mass Function
The Probability Mass Function (PMF) is a key concept for discrete random variables, providing the probabilities of taking specific values. For a binomial random variable \( Y \), the PMF is expressed as \( P(Y = k) \), where \( k \) is the number of successes.The PMF of a binomial random variable is given by:\[ P(Y = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
  • \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials.
  • \( p^k \) reflects the probability of \( k \) successes.
  • \((1-p)^{n-k} \) shows the probability of \( n-k \) failures.
Comprehension of this function is fundamental for evaluating the likelihood of different outcomes within a binomial process. These principles form the groundwork for more advanced topics, like moment generating functions.
Expectation
Expectation, or expected value, provides a measure of the 'center' of a probability distribution. It is the average value a random variable \( Y \) should take over many repetitions of an experiment. For a binomial random variable with parameters \( n \) and \( p \), the expectation is determined as:\[ E(Y) = n \, p \]
  • It represents the idea of 'average' or 'mean' outcome over repeated experiments.
  • In the context of a binomial distribution, it depicts the expected number of successes in \( n \) trials.
This concept becomes significant when developing moment generating functions, where expectation integrates valuation across probabilities.
Binomial Expansion
The Binomial Expansion is a mathematical expression that expands powers of sums, particularly familiar from algebra. In probability, it aids in simplifying complex expressions.The binomial expansion is represented as:\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} \]
  • \( \binom{n}{k} \) is the binomial coefficient, indicating how many ways we can choose \( k \) among \( n \) items.
  • It鈥檚 used to expand expressions like \( (x+y)^n \) into a sum of terms.
  • This formula is directly applied in the computation of moment generating functions for binomial distributions.
Understanding binomial expansion allows students to see how mathematical concepts interlink and support probabilistic computations.

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Most popular questions from this chapter

Refer to Exercises \(3.142\) and \(3.143 .\) If the number of phone calls to the fire department, \(Y\), in a day has a Poisson distribution with mean \(6,\) show that \(p(5)=p(6)\) so that 5 and 6 are the two most likely values for \(Y\)

Find the distributions of the random variables that have each of the following moment-generating functions: a. \(m(t)=\left[(1 / 3) e^{t}+(2 / 3)\right]^{5}\) b. \(m(t)=\frac{e^{t}}{2-e^{t}}\) c. \(m(t)=e^{2\left(e^{t}-1\right)}\)

In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been properly drilled in the boxes and in ten minutes if the holes must be re-drilled. Twenty gearboxes are in stock, 2 with improperly drilled holes. Five gearboxes must be selected from the 20 that are available for installation in the next five robots. a. Find the probability that all 5 gearboxes will fit properly. b. Find the mean, variance, and standard deviation of the time it takes to install these 5 gearboxes.

Let \(Y\) denote a Poisson random variable with mean \(\lambda\). Find the probability-generating function for \(Y\) and use it to find \(E(Y)\) and \(V(Y)\)

A corporation is sampling without replacement for \(n=3\) firms to determine the one from which to purchase certain supplies. The sample is to be selected from a pool of six firms, of which four are local and two are not local. Let \(Y\) denote the number of nonlocal firms among the three selected. a. \(P(Y=1)\) b. \(P(Y \geq 1)\) c. \(P(Y \leq 1)\)
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