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X; Y; U are random independent variables. X; Y are exponential distributed
with $lambda$ and U $sim$ Uni[0; 1].
How i find the probability:
P(X $le$ U $le$ Y )

Thanks

For given positive real numbers $xleq yleq 1$, we have that
$
mathbb P(xleq Uleq y)=y-x,
$
whereas if $xleq 1leq y$ we have $mathbb P(xleq Uleq y)=1-x$, and otherwise this quantity is zero. Therefore

$$mathbb P(Xleq Uleq Y)=mathbb E[Y-X; Xleq Yleq 1] + mathbb E[1-X; Xleq 1leq Y].$$
Since $X$ and $Y$ are independent, this simplifies to
$$
mathbb P(Xleq Uleq Y)=mathbb E[Y-X; Xleq Yleq 1] + mathbb E[1-X; Xleq 1] cdot mathbb P(Ygeq 1).
$$
The second term is easy to calculate, since $mathbb P(Ygeq 1)=e^{-lambda}$ by definition and
$$
mathbb E[1-X; Xleq 1] =int_{0}^1(1-x) lambda e^{-lambda x} dx=1+frac{e^{-lambda}-1}{lambda}.
$$
For the first term, we have
$$begin{align*}
mathbb E[Y-X; Xleq Yleq 1]&=int_0^1int_0^y(y-x) lambda^2 e^{-lambda(x+y)} dx dy\
&=int_0^1 e^{-lambda y}(lambda y-1)+e^{-2lambda y} dy\
&=frac{1-e^{-2lambda}}{2lambda}-e^{-lambda}.
end{align*}$$
Therefore
$$
mathbb P(Xleq Uleq Y)=frac{1-e^{-2lambda}}{2lambda}+e^{-lambda}cdot frac{e^{-lambda}-1}{lambda}=frac{(1-e^{-lambda})^2}{2lambda}.
$$