What is the moment-generating function (MGF) of a geometric distribution?

What is the moment-generating function (MGF) of a geometric distribution?

• A. ((1 - qe^t)/p)
• B. ((1 - qe^t)/p)^-1
• C. (pe^t)/(1-qe^t)
• D. (qe^t)/(1-pe^t)

Answer: (B)

The moment-generating function (MGF) of a geometric distribution is crucial for understanding the properties of the distribution, particularly its moments. For a geometric distribution with success probability \( p \) (where \( q = 1 - p \) is the probability of failure), the MGF is derived from the power series expansion. For a geometric random variable \( X \) representing the number of trials until the first success, the MGF \( M_X(t) \) is given by: \[ M_X(t) = E[e^{tX}] = \sum_{k=0}^{\infty} e^{tk} P(X = k) = \sum_{k=0}^{\infty} e^{tk} (q^k p) \] This series can be simplified using the geometric series sum formula. Given \( P(X = k) = p q^k \), the MGF becomes: \[ = p \sum_{k=0}^{\infty} (qe^t)^k = p \frac{1}{1 - qe^t} \] for \( |qe^t| < 1 \). Rearranging yields: \[ M_X(t) = \frac

The moment-generating function (MGF) of a geometric distribution is crucial for understanding the properties of the distribution, particularly its moments. For a geometric distribution with success probability ( p ) (where ( q = 1 - p ) is the probability of failure), the MGF is derived from the power series expansion.

For a geometric random variable ( X ) representing the number of trials until the first success, the MGF ( M_X(t) ) is given by:

[

M_X(t) = E[e^{tX}] = \sum_{k=0}^{\infty} e^{tk} P(X = k) = \sum_{k=0}^{\infty} e^{tk} (q^k p)

]

This series can be simplified using the geometric series sum formula. Given ( P(X = k) = p q^k ), the MGF becomes:

[

= p \sum_{k=0}^{\infty} (qe^t)^k = p \frac{1}{1 - qe^t}

]

for ( |qe^t| < 1 ). Rearranging yields:

[

M_X(t) = \frac