Question on exponential distribution

Let {X1,...,Xn} be a random sample from the exponential distribution with probability density function f(x) = (mu)^-1 exp(-x/mu) for x > 0, and 0 otherwise, where mu>0 is an unknown parameter.

a) Compute the mean and the variance of Xi.

Any help would be greatly appreciated :)

Re: Question on exponential distribution

Just try the usual integration $\displaystyle \int_{0}^{\infty} xf(x) dx$ for mean and $\displaystyle \int_{0}^{\infty} x^2f(x) dx$ for variance.

Re: Question on exponential distribution

Hi, I did this and found that E(X)=mu, and Var(X)=mu^2. So, would E(Xi) be the sum from i=1 to k of mu i, and would Var(Xi) be the sum from i to k of (mu i)^2?

Re: Question on exponential distribution

$\displaystyle \mu$ and $\displaystyle \mu^2$ are $\displaystyle E(X_i)$ and $\displaystyle Var(X_i)$ respectively. The summation will come in the answer if you want $\displaystyle E(\sum_{i=1}^{n} X_i)$ and $\displaystyle Var(\sum_{i=1}^{n} X_i)$