Some exercises have problems regarding direct use of the expected value. But one as follows asks to find the minimum expected value, how would I set the input values of integration for such a problem? Example.

Suppose $X$~$Exp(1)$. Find $E \ min (X,1)$

The general formula for the expected value of a continuous variable is $\int_{-\infty }^{\infty }xf(x)dx$ , but i can't see how to enter the values for the proper solution.

ps. Exp is the exponential distribution.


$$ E\min\{X,1\} = \int_{-\infty}^\infty \min\{x,1\} f(x) \, dx .$$ where $f(x)$ is the PDF.

  • $\begingroup$ But how do I know what's the minimum value between x and one? I'm integrating a function and suddenly there's the min{x,1} which i can't see how it fits in the integration process. $\endgroup$
    – matt_zarro
    Dec 17 '13 at 22:42
  • 2
    $\begingroup$ You know that $\min\{x,1\} = x$ if $x \le 1$, and $\min\{x,1\} = 1$ if $x \ge 1$. So the integral becomes $$\int_{-\infty}^1 x f(x) \, dx + \int_{1}^\infty f(x) \, dx .$$ $\endgroup$ Dec 17 '13 at 22:46
  • $\begingroup$ WOW! Thanks very much Stephen, that was brilliant!! $\endgroup$
    – matt_zarro
    Dec 17 '13 at 23:20

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