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Chapter 2: Problem 63

Calculate the moment generating function of a geometric random variable.

Short Answer

Expert verified
The moment generating function (MGF) of a geometric random variable X, with probability p of success, is given by: M_X(t) = \(p*\frac{e^t}{1 - (1-p)e^t}\).

Step by step solution

01

Define the probability mass function for a geometric random variable

The probability mass function for a geometric random variable X is given by: P(X=k) = (1-p)^(k-1)p, where k is the number of trials and p is the probability of success on each trial.
02

Define the moment generating function for a random variable

The moment generating function for a random variable X is given by the formula: M_X(t) = E(e^(tX)), where E denotes the expected value, and t is a real number.
03

Calculate the expected value of e^(tX) using the given probability mass function

To calculate the expected value of e^(tX), we will use the formula: E(e^(tX)) = 危 [P(X=k) * e^(tk)], where the summation is over all positive integers k. Now, we have: E(e^(tX)) = 危 [(1-p)^(k-1)p * e^(tk)].
04

Factor out the common terms from the summation

We can factor out the p from the summation: E(e^(tX)) = p*(危 ((1-p)^(k-1) * e^(tk)). Now we need to find a closed-form expression for the summation.
05

Simplify the summation and calculate the MGF

Notice that the summation is a geometric series: 危 ((1-p)^(k-1) * e^(tk)), with common ratio (1-p)*e^t. So we have: 危 ((1-p)^(k-1) * e^(tk))= (e^t) / (1 - (1-p)e^t), provided that the denominator is nonzero. Therefore, the MGF for a geometric random variable will be: M_X(t) = p*(e^t) / (1 - (1-p)e^t). This is the moment generating function for a geometric random variable.

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Most popular questions from this chapter

Show that when \(r=2\) the multinomial reduces to the binomial.

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their policy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger who shows up?

Let \(X\) and \(Y\) each take on either the value 1 or \(-1\). Let $$ \begin{aligned} p(1,1) &=P\\{X=1, Y=1\\} \\ p(1,-1) &=P[X=1, Y=-1\\} \\ p(-1,1) &=P[X=-1, Y=1\\} \\ p(-1,-1) &=P\\{X=-1, Y=-1\\} \end{aligned} $$ Suppose that \(E[X]=E[Y]=0\). Show that (a) \(p(1,1)=p(-1,-1) ;\) (b) \(p(1,-1)=p(-1,1)\). Let \(p=2 p(1,1) .\) Find (c) \(\operatorname{Var}(X)\); (d) \(\operatorname{Var}(Y)\) (e) \(\operatorname{Cov}(X, Y)\).

If \(X\) is a nonnegative integer valued random variable, show that (a) $$ E[X]=\sum_{n=1}^{\infty} P[X \geq n\\}=\sum_{n=0}^{\infty} P(X>n\\} $$ Hint: Define the sequence of random variables \(I_{n}, n \geq 1\), by $$ I_{n}=\left\\{\begin{array}{ll} 1, & \text { if } n \leq X \\ 0, & \text { if } n>X \end{array}\right. $$ Now express \(X\) in terms of the \(I_{n}\). (b) If \(X\) and \(Y\) are both nonnegative integer valued random variables, show that $$ E[X Y]=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} P(X \geq n, Y \geq m) $$

The random variable \(X\) has the following probability mass function: $$ p(1)=\frac{1}{2}, \quad p(2)=\frac{1}{3}, \quad p(24)=\frac{1}{6} $$ Calculate \(E[X]\)
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