I am having trouble finding the variance of this distribution. the mean is 2$\lambda$, but the variance is $\lambda^2$ ? In the solution, they say $\operatorname{Var}(\bar Y) =\lambda^2 $, I'm really puzzled... Shifting a distribution will not really cause its variance to change right?
Really appreciate any help! Thanks
You are right, variance is invariant under shifting. Basically, your random variable can be expressed as $Y=X+\lambda$, where $X\sim Exp(1/\lambda)$, hence the expected value is $E[Y]=\lambda + E[X] = 2\lambda$, and the variance is $$ Var(Y) = Var(X) = \lambda^2 $$