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)



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 \( K_Y(t) \) is given as:

\[
K_Y(t) = e^t - 1
\]

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

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

Thus, the distribution of \( Y \) 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 \( t = 0 \) give cumulants directly, simplifying calculations for higher-order moments!

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).

This problem asks to determine the distribution of a random variable Y given its cumulant generating function (CGF) K_Y(t) = e^t − 1. Learn how to identify that Y follows a Poisson(1) distribution using key probability theory concepts and formulas, including the CGF for Poisson distributions. Suitable for undergraduate students studying probability and statistics.

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K Y ( t )

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