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**heathrowjohnny** Find the mean and variance for $\displaystyle X $ given the pdf **$\displaystyle f(x) = .15e^{-0.15(x-5)}, \ x \geq .5 $**

If $\displaystyle Y = X-0.5 $ find the mean and variance of $\displaystyle Y $.

So $\displaystyle M_{X}(t) = \int_{0.5}^{\infty} .15e^{-0.15(x-5)}e^{tx} \ dx = .15e^{0.75} \int_{0.5}^{\infty} e^{x(t-0.15)} \ dx$

$\displaystyle M_{X}(t) = \frac{.15e^{0.5}}{t-0.15}e^{x(t-0.15)} $ evaluated from $\displaystyle x = 0.5 $ to $\displaystyle x = \infty $.

I have a feeling that this is to messy just to find the mean and variance.

And to find the mean and variance for $\displaystyle Y, $ I wrote $\displaystyle X = Y+0.5 $, and so $\displaystyle f_{Y}(y) = .15e^{-0.15(y-4.5)}, \ y \geq 0.5 $. Then do the same thing.