The problem asks to find the distribution of $U = X + Y$, where both $X$ and $Y$ are independent random variables, each following a Poisson distribution with parameter $\lambda$, i.e.,

• $X \sim \text{Poisson}(\lambda)$
• $Y \sim \text{Poisson}(\lambda)$

To solve this, we will use the fact that the sum of two independent Poisson-distributed random variables also follows a Poisson distribution. The parameters of the resulting distribution are the sum of the individual parameters.

Solution:

1. Step 1: Identify the distribution of $X$ and $Y$
   The probability mass function (PMF) of a Poisson random variable is: $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots$ for both $X$ and $Y$, as they are both Poisson-distributed with parameter $\lambda$.

2. Step 2: Distribution of $U = X + Y$
   Since $X$ and $Y$ are independent, the sum of two independent Poisson random variables is also Poisson-distributed. Specifically, if $X \sim \text{Poisson}(\lambda_1)$ and $Y \sim \text{Poisson}(\lambda_2)$, then: $U = X + Y \sim \text{Poisson}(\lambda_1 + \lambda_2)$ In this case, both $X$ and $Y$ have parameter $\lambda$, so: $U = X + Y \sim \text{Poisson}(2\lambda)$

Conclusion:

The distribution of $U = X + Y$ is: $U \sim \text{Poisson}(2\lambda)$

Let me know if you need further clarification or details on the steps!

Related Questions:

1. How would the distribution change if $X$ and $Y$ were not independent?
2. What is the expected value and variance of the distribution $U$?
3. How do we calculate the probability $P(U = k)$ for the Poisson distribution?
4. What happens if $X$ and $Y$ have different Poisson parameters, say $\lambda_1$ and $\lambda_2$?
5. How would you derive the moment-generating function (MGF) of the sum of two independent Poisson random variables?

Tip:

When summing independent Poisson random variables, you can directly add their rate parameters to find the rate of the resulting Poisson distribution.