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{\bf Question}

Suppose that a random variable $X$ has the pdf

$$f(x)=e^{-(x-\mu)},\ x>\mu.$$

Obtain the mean and variance of $X$.

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{\bf Answer}

Let $Y=X-\mu$.

Then the transformation is one-to-one and increasing.  Using the
transformation technique, the pdf of $Y$ is

$$g(y)=e^{-y},\ \ y>0$$

Hence $Y \sim \mathrm{exponential} (\beta=1)$

Therefore $E(Y)=1=\mathrm{var}(Y)$

Therefore $E(X)=E(Y+\mu)=1+\mu$

$\mathrm{var}(X)=\mathrm{var}(Y)=1$

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