= Expected value of a capped nonnegative random variable
{title2=$\mathbb E[\min(Y,a)]=\int_0^a\mathbb P(Y>t)dt$}
For a nonnegative <random variable> $Y$ and a fixed cap $a\geq0$, the <tail-sum formula> gives
$$
\mathbb E[\min(Y,a)]=\int_0^a\mathbb P(Y>t)dt,\qquad
\mathbb E[(Y-a)_+]=\int_a^\infty\mathbb P(Y>t)dt.
$$
Below the cap, the <events> $\{\min(Y,a)>t\}$ and $\{Y>t\}$ coincide; at and above the cap, the former <event> is empty. For the second identity, the <positive part> $(Y-a)_+$ exceeds $t\geq0$ precisely when $Y>a+t$. Thus the two <integrals> divide the <area> under the <survival function>, just as $Y=\min(Y,a)+(Y-a)_+$ divides each outcome. The capped <expected value> is always finite, while the excess <expected value> may be infinite.
For the <uniform distribution> on the <interval> $[0,4]$ and cap $a=2$, the capped <expected value> is $\int_0^2(1-t/4)dt=3/2$ and the excess <expected value> is $\int_2^4(1-t/4)dt=1/2$. Directly averaging the capped values also gives $\int_0^2y(1/4)dy+2\mathbb P(Y>2)=1/2+1=3/2$. Capping therefore measures a different average from the uncapped <expected value> two, even when both are finite.
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