Step 1: Concept Overview:
The problem requires finding the moment generating function (MGF) for the sum of two independent and identically distributed (i.i.d.) geometric random variables. A crucial MGF property is that the MGF of the sum of independent variables equals the product of their individual MGFs.
Step 2: Approach:
1. Determine the MGF of a single geometric random variable.
2. Apply the rule: If \(X\) and \(Y\) are independent, then \( M_{X+Y}(t) = M_X(t) . M_Y(t) \).
3. Recognize that the sum of \(r\) i.i.d. geometric variables follows a Negative Binomial distribution, with MGF \( \left(\frac{p}{1-qe^t}\right)^r \).
Step 3: Detailed Solution:
First, find the MGF of a single geometric random variable \(X\). Based on the answer choices, we consider the geometric distribution that counts failures (\(k=0, 1, 2, \dots\)) before the first success.
The MGF is:
\[ M_X(t) = \frac{p}{1-qe^t} \]
where \(q = 1-p\).
Since X and Y are i.i.d., their MGFs are identical:
\[ M_X(t) = M_Y(t) = \frac{p}{1-qe^t} \]
Because X and Y are independent, the MGF of their sum, \(Z = X+Y\), is the product of their MGFs:
\[ M_{X+Y}(t) = M_X(t) . M_Y(t) \]
\[ M_{X+Y}(t) = \left(\frac{p}{1-qe^t}\right) . \left(\frac{p}{1-qe^t}\right) \]
\[ M_{X+Y}(t) = \left(\frac{p}{1-qe^t}\right)^2 \]
This matches the MGF of a Negative Binomial distribution with parameters \(r=2\) and \(p\), as expected for the sum of two i.i.d. geometric variables.
Step 4: Answer:
The moment generating function of (X+Y) is \( \left(\frac{p}{1-qe^t}\right)^2 \).