What is the moment generating function (MGF) of the Poisson distribution?

Study for the Society of Actuaries Exam P. Immerse in flashcards and multiple-choice questions, each with hints and explanations. Gear up for your exam success!

Multiple Choice

What is the moment generating function (MGF) of the Poisson distribution?

The moment generating function (MGF) of a Poisson distribution is derived from the definition of the MGF, which is given by the expected value of \( e^{tX} \) where \( X \) is a random variable representing the Poisson process. The Poisson distribution is characterized by a single parameter \( \lambda \), which indicates the average rate (or mean) of occurrences in a fixed interval of time or space. To find the MGF, we start with the definition: \[ M_X(t) = E[e^{tX}] = \sum_{k=0}^{\infty} e^{tk} \cdot P(X = k) \] For a Poisson random variable, the probability mass function is \( P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} \). Substituting this into our MGF formulation gives us: \[ M_X(t) = \sum_{k=0}^{\infty} e^{tk} \cdot \frac{e^{-\lambda}\lambda^k}{k!} \] This can be rearranged: \[ M_X(t) = e^{-\lambda} \sum

The moment generating function (MGF) of a Poisson distribution is derived from the definition of the MGF, which is given by the expected value of ( e^{tX} ) where ( X ) is a random variable representing the Poisson process. The Poisson distribution is characterized by a single parameter ( \lambda ), which indicates the average rate (or mean) of occurrences in a fixed interval of time or space.

To find the MGF, we start with the definition:

[

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

]

For a Poisson random variable, the probability mass function is ( P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!} ). Substituting this into our MGF formulation gives us:

[

M_X(t) = \sum_{k=0}^{\infty} e^{tk} \cdot \frac{e^{-\lambda}\lambda^k}{k!}

]

This can be rearranged:

[

M_X(t) = e^{-\lambda} \sum

More practice questions