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I'm just brushing up my math skills and I came across the following problem.

A lifetime X of a certain device is exponential with parameter $\mu$ years. What is the expected value of $\max\{\mu/2 , X\}$?

I have no clue on how to start solving this. Any ideas?

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Your edit is truly mystifying. The new version does not make sense and also makes the answer appear disjointed. –  cardinal Oct 16 '11 at 1:34
    
I have rolled back your last edit and cleaned it up a bit. I hope you don't mind. Please check that it reads as intended. –  cardinal Oct 16 '11 at 12:01

1 Answer 1

By the definition of an expectation trough the density, $$ \mathsf E g(X) = \int\limits_{-\infty}^\infty g(t)f_X(t)\,dt $$ where $f_X(t)$ is a density function of r.v. $X$. In your case $X\sim\mathcal E(\mu)$ so $f_X(t) = 0$ for $t<0$ and $f_X(t) = \mu\mathrm e^{-\mu t}$ for $t\geq 0$. As a result $$ \mathsf E\max\{\mu/2,X\}= \int\limits_{0}^\infty \max\{\mu/2,t\}\mu\mathrm e^{-\mu t}\,dt = \int\limits_{0}^{\mu/2} \frac12\mu^2\mathrm e^{-\mu t}\,dt+\int\limits_{\mu/2}^\infty t\mu\mathrm e^{-\mu t}\,dt. $$

Do you know how to find value of these integrals?

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@ManiKrish: yes, $\mu$ is a constant, so first term integrates directly and second - by parts. –  Ilya Oct 15 '11 at 19:51
    
Is "trough" a typo for "through"? or is there really such a thing as "an expectation trough"? –  Gerry Myerson Oct 15 '11 at 22:32

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