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I want to see if I am doing this right:

Let $X_{1},\dots,X_{n}$ denote a random sample from the exponential distribution, with unknown location parameter $\theta$ and scale parameter $\lambda$, whose density is given by $$ p(x|\theta,\lambda)=\lambda\exp[-\lambda(x-\theta)],\ \theta\leq x<\infty, $$ where $-\infty<\theta<\infty$ and $0<\lambda<\infty$. Find the mean and the variance of $X-\theta$.

--For the mean, I computed $E(X-\theta)=E(X)-E(\theta)=E(X)-\theta$. After some integration by parts, I found that $E(X-\theta)=1/\lambda$.

--For the variance, I computed $Var(X-\theta)=Var(X)+Var(\theta)=Var(X)$. I computed $E(X^{2})=\theta^{2}+2\theta/\lambda+2/\lambda^{2}$, which yields a variance of $$ Var(X)=\left(\theta^{2}+\frac{2\theta}{\lambda}+\frac{2}{\lambda^{2}}\right)-\left(\theta+\frac{1}{\lambda}\right)^{2}.$$

Kirk Fogg
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    That is correct. The variance simplifies to $\frac{1}{\lambda^2}$. You could have made life simpler by noting that $X-\theta$ is an exponential that has location parameter $0$. It is standard that the mean is $\frac{1}{\lambda}$ and the variance is $\frac{1}{\lambda^2}$. And if you want mean, variance for $X$, mean is $\theta+\frac{1}{\lambda}$ and variance doesn't care about shift. – André Nicolas Jul 18 '13 at 22:46

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