Deriving the Moment Generating Function for the Sum of Geometric Variables

This solution explains how to find the moment generating function (MGF) for the sum of two independent and identically distributed (i.i.d.) geometric random variables.

Understanding Geometric Distribution and MGF

A geometric random variable represents the number of trials needed to achieve the first success in a sequence of independent Bernoulli trials, or alternatively, the number of failures before the first success. Let's consider the definition where the random variable counts the number of failures before the first success. If $ X $ follows a geometric distribution with success probability $ p $, its probability mass function (PMF) is given by:

$ P(X=k) = (1-p)^k p, \quad k = 0, 1, 2, \dots $

Here, $ p $ is the parameter, and $ q = 1-p $ is the probability of failure.

The moment generating function (MGF) of a random variable $ X $, denoted as $ M_X(t) $, is defined as $ M_X(t) = E[e^{tX}] $, where $ t $ is a real number such that the expectation exists. For a geometric random variable $ X $ (counting failures before success) with parameter $ p $, the MGF is:

$ M_X(t) = \sum_{k=0}^{\infty} e^{tk} P(X=k) $

$ M_X(t) = \sum_{k=0}^{\infty} e^{tk} (1-p)^k p $

$ M_X(t) = p \sum_{k=0}^{\infty} (e^t (1-p))^k $

This is a geometric series with first term 1 and common ratio $ r = e^t(1-p) $. The sum converges if $ |r| < 1 $. Thus, the MGF is:

$ M_X(t) = p \left( \frac{1}{1 - e^t(1-p)} \right) = \frac{p}{1 - qe^t} $

So, for a geometric variable $ X $ with parameter $ p $ (counting failures before success), $ M_X(t) = \frac{p}{1 - qe^t} $.

MGF of the Sum of Independent Variables

We are given two random variables, $ X $ and $ Y $, that are independent and identically distributed (i.i.d.) geometric variables with parameter $ p $. This means:

• $ M_X(t) = \frac{p}{1 - qe^t} $
• $ M_Y(t) = \frac{p}{1 - qe^t} $
• $ X $ and $ Y $ are independent.

A key property of moment generating functions is that for independent random variables $ X $ and $ Y $, the MGF of their sum $ (X+Y) $ is the product of their individual MGFs:

$ M_{X+Y}(t) = M_X(t) M_Y(t) $

Calculating the MGF of (X+Y)

Using the property mentioned above and the MGF for a single geometric variable:

$ M_{X+Y}(t) = \left( \frac{p}{1 - qe^t} \right) \times \left( \frac{p}{1 - qe^t} \right) $

$ M_{X+Y}(t) = \left( \frac{p}{1 - qe^t} \right)^2 $

Conclusion

Therefore, the moment generating function of $ (X+Y) $, where $ X $ and $ Y $ are i.i.d. geometric random variables with parameter $ p $ (counting failures before the first success), is $ \left( \frac{p}{1 - qe^t} \right)^2 $.