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Let \(X_{1}, X_{2}, X_{3}\) be iid with common mgf \(M(t)=\left((3 / 4)+(1 / 4) e^{t}\right)^{2}\), for all \(t \in R\) (a) Determine the probabilities, \(P\left(X_{1}=k\right), k=0,1,2\). (b) Find the mgf of \(Y=X_{1}+X_{2}+X_{3}\) and then determine the probabilities, \(P(Y=k), k=0,1,2, \ldots, 6\)

Short Answer

Expert verified
For part a, the probabilities \(P(X_{1}=k)\) for \(k=0,1,2\) are proved to be 0.5625, 0.375, and 0.0625 respectively. In part b, the MGF of \(Y\) is found to be \((3 / 4 + 1 / 4e^t)^6\), which indicates that Y is a binomial distributed variable with parameters \(n=6\) and \(p=1/4\). After determination, the probabilities \(P(Y=k)\) for \(k=0,1,2,...,6\) can be calculated using the binomial distribution formula.

Step by step solution

01

Identify the Distribution

The function \(M(t)=\left((3 / 4)+(1 / 4)e^{t}\right)^{2}\) is the moment generating function of a binomial distribution with n=2 and probability of success p = 1/4. This is because the mgf for a binomial distribution is given by \((1-p+pe^t)^n\) where \(n\) is number of trials and \(p\) is probability of success.
02

Determine Probabilities P(X1=k)

Using the pdf of binomial distribution \(P(X=k)=C(n, k)p^k(1-p)^{n-k}\), the probabilities are calculated as \(P(X_{1}=0)=((2 choose 0)*(1/4)^0*(3/4)^2)= 9/16=0.5625\), \(P(X_{1}=1)=((2 choose 1)*(1/4)^1*(3/4)^1)=3/8=0.375\), \(P(X_{1}=2)=((2 choose 2)*(1/4)^2*(3/4)^0)= 1/16=0.0625\)
03

Find the MGF of Y

The MGF of \(Y=X_{1}+X_{2}+X_{3}\) can be found by multiplying the MGFs of \(X_1, X_2, X_3\) as they are independent and identical. Therefore, \(M_Y(t)=(M(t))^3=((3 / 4 + 1 / 4e^t)^2)^3 = ((3 / 4 + 1 / 4e^t)^6)\)
04

Determine Probabilities P(Y=k)

After finding \(M_Y(t)\), we can deduce that Y is binomial with parameters \(n=6\) and \(p=1/4\). Therefore, using the pdf of binomial distribution \(P(Y=k)=C(6, k)*(1/4)^k*(3/4)^{6-k}\), probabilities \(P(Y=k)\) for \(k=0,1,2,...,6\) can be calculated accordingly.

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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 fixed number of independent trials of a binary experiment. In each trial, there are only two outcomes: success or failure. It is ideal for situations where you are counting the number of successes over multiple trials, like flipping a coin multiple times.

For a binomial distribution, we need two parameters:
  • **Number of trials (n):** This indicates how many times the experiment is conducted.
  • **Probability of success (p):** This is the likelihood of achieving a success in a single trial.
One of the key properties of a binomial distribution is its moment generating function (mgf). The mgf is particularly useful as it can quickly identify a distribution and also helps in finding distribution-related moments, like mean and variance. For a binomial distribution, the mgf is given by:
  • \((1-p+pe^t)^n\)
This expression helps in computing the probabilities for different numbers of successes.
Independent Identically Distributed (iid) Variables
Independent Identically Distributed (iid) variables are a set of random variables that have two key properties: independence and identical distribution. Let's break them down.

**Independence:** This means that the outcome of one variable does not affect the others. In other words, knowing the value of one variable provides no information about the others.

**Identically Distributed:** Each variable comes from the same probability distribution and has the same probability mass function (PMF), if it's discrete, or probability density function (PDF), if it's continuous. This ensures that each variable behaves the same way statistically.

In practice, assuming variables are iid allows statisticians and mathematicians to make inferences about populations based on sample data. In our example, the variables \(X_1, X_2, X_3\) are iid, meaning they share a common distribution with the same mgf, making calculations more straightforward. This set-up is crucial for applying functions like mgf or pmf across multiple trials.
Probability Mass Function
The Probability Mass Function (PMF) is a function that gives the probability that a discrete random variable is exactly equal to some value. It is fundamental in understanding how the probabilities are distributed across different outcomes of a random experiment.

For a binomial distribution, the PMF is denoted as:
\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]
  • \(\binom{n}{k}\) is the binomial coefficient, representing the number of ways to choose \(k\) successes in \(n\) trials.
  • \(p^k\) is the probability of having \(k\) successes.
  • \((1-p)^{n-k}\) is the probability of having \(n-k\) failures.
This equation provides a complete picture of the likelihood of each number of successes across the given trials. For example, calculating \(P(X=0)\) for \(n=2\) and \(p=1/4\) gives the probability of having zero successes, while \(P(X=1)\) and \(P(X=2)\) give the probabilities for one and two successes, respectively. The PMF is a crucial concept for anyone working with or studying discrete distributions and aids in calculating expected values and variances.

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Most popular questions from this chapter

Let \(f\left(x_{1}, x_{2}\right)=21 x_{1}^{2} x_{2}^{3}, 0

Determine the mean and variance of the sample mean \(\bar{X}=5^{-1} \sum_{i=1}^{5} X_{i}\), where \(X_{1}, \ldots, X_{5}\) is a random sample from a distribution having pdf \(f(x)=\) \(4 x^{3}, 0

Let the random variables \(X_{1}\) and \(X_{2}\) have the joint pmf described as follows: $$ \begin{array}{c|cccccc} \left(x_{1}, x_{2}\right) & (0,0) & (0,1) & (0,2) & (1,0) & (1,1) & (1,2) \\ \hline p\left(x_{1}, x_{2}\right) & \frac{2}{12} & \frac{3}{12} & \frac{2}{12} & \frac{2}{12} & \frac{2}{12} & \frac{1}{12} \end{array} $$ and \(p\left(x_{1}, x_{2}\right)\) is equal to zero elsewhere. (a) Write these probabilities in a rectangular array as in Example 2.1.3, recording each marginal pdf in the "margins." (b) What is \(P\left(X_{1}+X_{2}=1\right)\) ?

Suppose \(X_{1}\) and \(X_{2}\) have the joint pdf $$ f\left(x_{1}, x_{2}\right)=\left\\{\begin{array}{ll} e^{-x_{1}} e^{-x_{2}} & x_{1}>0, x_{2}>0 \\ 0 & \text { elsewhere } \end{array}\right. $$ For constants \(w_{1}>0\) and \(w_{2}>0\), let \(W=w_{1} X_{1}+w_{2} X_{2}\) (a) Show that the pdf of \(W\) is $$ f_{W}(w)=\left\\{\begin{array}{ll} \frac{1}{w_{1}-w_{2}}\left(e^{-w / w_{1}}-e^{-w / w_{2}}\right) & w>0 \\ 0 & \text { elsewhere } \end{array}\right. $$ (b) Verify that \(f_{W}(w)>0\) for \(w>0\). (c) Note that the pdf \(f_{W}(w)\) has an indeterminate form when \(w_{1}=w_{2}\). Rewrite \(f_{W}(w)\) using \(h\) defined as \(w_{1}-w_{2}=h\). Then use l'Hôpital's rule to show that when \(w_{1}=w_{2}\), the pdf is given by \(f_{W}(w)=\left(w / w_{1}^{2}\right) \exp \left\\{-w / w_{1}\right\\}\) for \(w>0\) and zero elsewhere.

A fair die is cast at random three independent times. Let the random variable \(X_{i}\) be equal to the number of spots that appear on the \(i\) th trial, \(i=1,2,3\). Let the random variable \(Y\) be equal to \(\max \left(X_{i}\right)\). Find the cdf and the pmf of \(Y\). Hint: \(P(Y \leq y)=P\left(X_{i} \leq y, i=1,2,3\right)\).

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