Math Problem Statement

Suppose Y is a random variable with cumulant generating function (cgf)

K Y ( t )

e t − 1.

What is the distribution of Y ?

N(0,1)

exponential (1)

Poisson(1)

U(0,1)

Solution

The cumulant generating function (CGF) is related to the distribution of a random variable by the fact that the cumulants (which are related to moments) can be derived from it.

The cumulant generating function KY(t)K_Y(t) is given as:

KY(t)=et1K_Y(t) = e^t - 1

This form matches the CGF of a Poisson(1) distribution, because:

  • For a Poisson(λ) distribution, the CGF is KY(t)=λ(et1)K_Y(t) = \lambda(e^t - 1).
  • When λ=1\lambda = 1, the CGF simplifies to KY(t)=et1K_Y(t) = e^t - 1, which is exactly the given function.

Thus, the distribution of YY is Poisson(1).

Answer:

Poisson(1)

Would you like more details or have any questions?

Here are 5 related questions:

  1. What is the cumulant generating function for a normal distribution?
  2. How are cumulants related to moments in probability theory?
  3. What are the first four cumulants for the Poisson distribution?
  4. How can the CGF be used to find the mean and variance of a distribution?
  5. What is the CGF of an exponential distribution?

Tip: The cumulant generating function is especially useful because the derivatives of the CGF at t=0t = 0 give cumulants directly, simplifying calculations for higher-order moments!

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Cumulant Generating Function
Poisson Distribution

Formulas

K_Y(t) = e^t − 1
K_Y(t) for Poisson(λ) = λ(e^t − 1)

Theorems

Poisson Distribution CGF Theorem

Suitable Grade Level

Undergraduate Probability and Statistics