# Expected value of a minimum

Under a group insurance policy, an insurer agrees to pay 100% of the medical bills incurred during the year by employees of a company, up to a max of $1$ million dollars. The total bills incurred, $X$, has pdf $$f_X(x) = \begin{cases} \frac{x(4-x)}{9}, & \text{for } 0 < x < 3\\ 0& \text{else} \end{cases}$$ where $x$ is measured in millions. Calculate the total amount, in millions of dollars, the insurer would expect to pay under this policy.

So I was able to obtain part of the solution, which was $$E(\min(X,1)) = \int_0^1 x\cdot \frac{x(4-x)}{9} dx \tag1$$

However, the solution has $(1)$, plus $$E(\min(X,1)) = \int_1^3 1\cdot \frac{x(4-x)}{9} dx \tag2$$

What I don't understand is if the problem explicitly states that they agree to pay up to $1$ million, why would you even have to bother with $(2)$?

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The minimum function is $x$ for $x\lt 1$ and $1$ for $1\lt x\le 3$. – André Nicolas Mar 19 '13 at 6:04

Let $x$ be the total bills in millions. For $x \in [0,1]$ the payout is $x$, for $x \in [1,\infty)$, the payout is $1$. Hence the payout as a function of $x$ is $p(x) = \min(x,1)$, and you wish to compute $Ep$. \begin{eqnarray} Ep &=& \int_0^\infty p(x) f_X(x) dx \\ &=& \int_0^1 x f_X(x) dx + \int_1^3 1 f_X(x)dx + \int_3^\infty 1 f_X(x)dx \\ &=& \frac{13}{108} + \frac{22}{27}+ 0 \\ &=& \frac{101}{108} \end{eqnarray}
The third integrand should be $1\cdot0$ rather than $0f_X(x)$ (for the same net result, of course). – Did Mar 19 '13 at 6:07