Variance of the minimum of two identical exponential random variables

Question

Let X and Y be two independent and identically distributed exponential random variables, with mean 2. What is the variance of Z = min(X;Y) ?

E(X) = 2; E(X) = 1/lambda, then lambda = 1/2; V(X) = 1/(lambda^2) = 4;

The answer however is 1.

Answer 1

Let $G$ be the CDF of $Z$. We have: \begin{align} 1-G(z)=\Pr(Z> z)=\Pr(X>z)\Pr(Y>z)=[\exp(-0.5z)]^2\implies G(z)=1-\exp(-z). \end{align} Thus, $Z$ is exponential with mean parameter $\lambda_Z=1$ with mean and variance: $$ E(Z)=\lambda_Z^{-1}=1,\quad\operatorname{Var}(Z)=\lambda_Z^{-2}=1. $$