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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. \(m(t)=e^{2\left(e^{t}-1\right)}\)

Short Answer

Expert verified
a. Binomial (n=5, p=1/3); b. Geometric (p=1/2); c. Poisson (λ=2).

Step by step solution

01

Identifying the Type of Distribution (Part a)

The function given is \(m(t)=\left[(1 / 3) e^{t}+(2 / 3)\right]^{5}\). This resembles the form \([pe^t + (1-p)]^n\), which is the moment-generating function (MGF) for a Binomial distribution. Here, \(p = 1/3\) and \(n = 5\), meaning we have a Binomial distribution with parameters \(n = 5\) trials and probability of success \(p = 1/3\).
02

Identifying the Type of Distribution (Part b)

The MGF \(m(t)= \frac{e^{t}}{2- e^{t}}\) resembles the form of a geometric distribution MGF. The general form is \(\frac{pe^t}{1-(1-p)e^t}\) when simplified. Observing this function, it follows the geometric distribution MGF with \(p=1/2\). Thus, the distribution is geometric with success probability \(p = 1/2\).
03

Identifying the Type of Distribution (Part c)

The function \(m(t)= e^{2\left(e^{t}-1\right)}\) matches the moment-generating function of a Poisson distribution. The MGF of a Poisson distribution is \(e^{\lambda(e^{t}-1)}\). Here, \(\lambda = 2\), indicating that the random variable follows a Poisson distribution with a parameter \(\lambda = 2\).

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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 discrete probability distribution that describes the number of successes in a sequence of independent and identically distributed Bernoulli trials. Each trial can result in just two possible outcomes: success or failure. The distribution depends on two parameters:
  • n: number of trials
  • p: probability of success on each trial
The moment-generating function (MGF) of a Binomial distribution is crucial in identifying the distribution type. It is given by \( M(t) = [(1-p) + pe^t]^n \). In this expression, you observe how the probability of success and the number of trials shape the distribution.
In the provided exercise, the MGF \([(1/3)e^t + (2/3)]^5\) matches this formula, where the parameters are:
  • \(n = 5\) (number of trials)
  • \(p = 1/3\) (probability of success)
Thus, the random variable follows a Binomial distribution with these parameters. This distribution can be useful in real-world scenarios such as determining the probability of getting 3 heads in 5 coin flips if each coin has a 1/3 chance of landing heads.
Geometric Distribution
The Geometric distribution is another discrete probability distribution, but it measures the number of trials needed for the first success in a series of independent Bernoulli trials. It is defined by one parameter, \( p \), which is the probability of success on each trial. Unlike the Binomial distribution, it doesn't have a fixed number of trials.
The MGF of a Geometric distribution is given by \( M(t) = \frac{pe^t}{1-(1-p)e^t} \). This formula captures how the distribution behaves based on the probability of achieving success early.
In part b of the original exercise, the function \( \frac{e^t}{2-e^t} \) simplifies to match the MGF of a Geometric distribution with \( p = 1/2 \). Therefore, this identifies a Geometric distribution where the random variable measures the expected trials until the first success, given a 50% chance of success each time. This distribution is commonly used in scenarios like choosing a toll-free number which is immediately available.
Poisson Distribution
The Poisson distribution is a discrete distribution that arises in processes where events happen independently and at a constant average rate. It is commonly used for counting the number of events that occur in a fixed interval of time or space.
A Poisson distribution is characterized by a single parameter:
  • \( \lambda \): the average rate (mean number of events in the interval)
Its moment-generating function is \( M(t) = e^{\lambda(e^t-1)} \). This expression identifies how the expected rate influences the frequency and probability of the events in a given period.
In part c of the exercise, the MGF \( e^{2(e^t-1)} \) correlates directly to a Poisson distribution with \( \lambda = 2 \). This implies that on average, 2 events occur in the given time or space frame. The Poisson distribution is useful in scenarios such as counting the number of phone calls received by a call center in an hour, assuming calls arrive randomly at a known average rate.

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Most popular questions from this chapter

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. \(P(Y \leq 1)\)

In responding to a survey question on a sensitive topic (such as "Have you ever tried marijuana?"), many people prefer not to respond in the affirmative. Suppose that \(80 \%\) of the population have not tried marijuana and all of those individuals will truthfully answer no to your question. The remaining \(20 \%\) of the population have tried marijuana and \(70 \%\) of those individuals will lie. Derive the probability distribution of \(Y\), the number of people you would need to question in order to obtain a single affirmative response.

Let \(Y\) have a hypergeometric distribution $$p(y)=\frac{\left(\begin{array}{l} r \\ y \end{array}\right)\left(\begin{array}{l} N-r \\ n-y \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} \quad y=0,1,2, \ldots, n$$ a. Show that $$P(Y=n)=p(n)=\left(\frac{r}{N}\right)\left(\frac{r-1}{N-1}\right)\left(\frac{r-2}{N-2}\right) \cdots\left(\frac{r-n+1}{N-n+1}\right)$$ b. Write \(p(y)\) as \(p(y | r) .\) Show that if \(r_{1}\frac{p\left(y+1 | r_{1}\right)}{p\left(y+1 | r_{2}\right)}$$ c. Apply the binomial expansion to each factor in the following equation: $$(1+a)^{N_{1}}(1+a)^{N_{2}}=(1+a)^{N_{1}+N_{2}}$$ Now compare the coefficients of \(a^{n}\) on both sides to prove that $$\left(\begin{array}{c} N_{1} \\ 0 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n \end{array}\right)+\left(\begin{array}{c} N_{1} \\ 1 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n-1 \end{array}\right)+\cdots+\left(\begin{array}{c} N_{1} \\ n \end{array}\right)\left(\begin{array}{c} N_{2} \\ 0 \end{array}\right)=\left(\begin{array}{c} N_{1}+N_{2} \\ n \end{array}\right)$$ d. Using the result of part ( \(\underline{\mathrm{c}}\) ), conclude that $$\sum_{y=0}^{n} p(y)=1$$

If \(Y\) has a geometric distribution with success probability \(p,\) show that $$P(Y=\text { an odd integer })=\frac{p}{1-q^{2}}$$.

The number of cars driving past a parking area in a one-minute time interval has a Poisson distribution with mean \(\lambda\). The probability that any individual driver actually wants to park his or her car is \(p .\) Assume that individuals decide whether to park independently of one another. a. If one parking place is available and it will take you one minute to reach the parking area, what is the probability that a space will still be available when you reach the lot? (Assume that no one leaves the lot during the one- minute interval.) b. Let \(W\) denote the number of drivers who wish to park during a one-minute interval. Derive the probability distribution of \(W\)

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