The distribution of the sum of independent Poisson variables is which of the following?

• A. Normal(sum lambda, sum lambda)
• B. Binomial(sum lambda, p)
• C. Poisson(sum lambda)
• D. Gamma(sum lambda, 1)

Answer: (C)

The counts add up to another Poisson variable because Poisson distributions are closed under independent addition. If you have independent X1, X2, ..., Xk with X_i ~ Poisson(lambda_i), then the sum X = X1 + X2 + ... + Xk follows Poisson with parameter equal to the sum of the lambdas.

This comes from the probability generating function: for Poisson(lambda) the pgf is exp(lambda(z-1).) For independent variables, the pgf of the sum is the product of the individual pgfs, which gives exp((sum lambda_i)(z-1)), the pgf of Poisson(sum lambda_i). The same idea using moment generating functions yields M_X(t) = exp((sum lambda_i)(e^t - 1)).

The other options don’t fit because a normal distribution would be an approximation for large sums, a binomial distribution requires a fixed number of trials with a constant success probability, and a gamma distribution is continuous and arises from sums of certain continuous variables, not a discrete count.