Conditional Expectation of multiple random variables (i.i.d) with inequalities on different variables Suppose $X$ ~ exp($\theta_1$), and $Y_i$ ~ exp($\theta_2$) for $i=1,2, i.i.d$
Calculate $\mathbb{E}[\min\{X, Y_1 + Y_2\}|Y_1 \leq X, a \leq X]$ for some $a \geq 0$.
Can someone guide me here? Thanks!
My attempt: Let $Y =Y_1 + Y_2$, then $Y$ ~ Erlang$(2, \theta_2)$, and the problem can be re-written as:
$\mathbb{E}[\min\{X, Y\}|Y_1 \leq X, a \leq X]$

HINT: Condition on $Y_2$, $X$, and $Y_1$ respectively (and use independence):
$$E
[\min\{X,Y_1+Y_2\}|Y_1\le X,a\le X]\\
=\int_0^\infty E[\min\{X,Y_1+Y_2\}|Y_1\le X,a\le X,Y=y_2]\,f_{Y_2}(y_2)\,dy_2\\
=\int_0^\infty E[\min\{X,Y_1+y_2\}|Y_1\le X,a\le X]\,f_{Y_2}(y_2)\,dy_2\\
=\int_0^\infty\int_a^\infty E[\min\{X,Y_1+y_2\}|Y_1\le X,a\le X, X=x]\,f_{X|X\ge a}(x)f_Y(y)\,dxdy_2\\
=\int_0^\infty\int_a^\infty E[\min\{x,Y_1+y_2\}|Y_1\le x]\,f_{X|X\ge a}(x)f_{Y_2}(y_2)\,dxdy_2\\
=\int_0^\infty\int_a^\infty\int_0^x \min\{x,y_1+y_2\}\,f_{Y_1|Y_1\le x}(y_1)\,dy_1f_{X|X\ge a}(x)\,dxf_{Y_2}(y_2),dy_2
$$
The last integral is just a triple definite integral (albeit messy and might not have a closed form).
Another idea would be to let $Z=\min\{X,Y_1+Y_2\}|Y_1\le X,a\le X$
Then:
$$1-F_Z(z)=P(Z>z)=P(X>z,Y_1+Y_2>z|Y_1\le X,a\le X)\\
=\int_0^\infty P(X>z,Y_1>z-y_2|Y_1\le X,a\le X)f_{Y_2}(y_2)\,dy_2
$$
and:
$$P(X>z,Y_1>z-y_2|Y_1\le X,a\le X)=\frac{P(X>z,Y_1>z-y_2,Y_1\le X,a\le X)}{P(Y_1\le X,a\le X)}\\
=\frac{P(X>\max\{z,a\},z-y_2\le Y_1\le X)}{P(Y_1\le X,a\le X)}
$$
Which again seems unlikely to be computed in closed form.
