Calculate how $\mathbb{E}[V|X]$ distributed. Given $X,Y$ i.i.d where $\mathbb{P}(X>x)=e^{-x}$ for $x\geq0$ and $\mathbb{P}(X>x)=1$ for all $x<0$
and $V=\min(X,Y)$
Calculate how $\mathbb{E}[V|X]$ distributed.
I've found that $F_{V|X=x}(v)=\left\{\begin{array}{rcl} 0&t\leq0\\1-e^{-t}&0\leq t\leq x\\1&else\end{array}\right.$
And I've tried using the formula $\mathbb{E}[V|X]=\int_{\infty}^{\infty}vf_{V|X=x}dv$ and I got that $\mathbb{E}[V|X]=-xe^{-x}-e^{-x}+1$ and in the answer I had to compare with they got $\mathbb{E}[V|X]\sim U(0,1)$
Not sure how to get to this distribution any help?
 A: The conditional pdf $f_{V\mid X}$ that you write is not defined since the joint density $f_{V,X}$ does not exist wrt Lebesgue measure. This is because $V=X$ has a positive probability.
You can write  $$V=\min(X,Y)=X\mathbf1_{X<Y}+Y\mathbf1_{X>Y}\,,$$
where $\mathbf1_A$ is an indicator variable.
Now
\begin{align}
\mathbb E\left[V\mid X\right]&=\mathbb E\left[X\mathbf1_{X<Y}\mid X\right]+\mathbb E\left[Y\mathbf1_{X>Y}\mid X\right]
\\&=X\mathbb E\left[\mathbf1_{X<Y}\mid X\right]+\mathbb E\left[Y\mathbf1_{X>Y}\mid X\right]
\end{align}
For fixed $x>0$, we have $$\mathbb E\left[\mathbf1_{x<Y}\right]=\mathbb P\left(Y>x\right)=e^{-x}$$ and $$\mathbb E\left[Y\mathbf1_{x>Y}\right]=\int_0^xye^{-y}\,\mathrm{d}y=1-e^{-x}-xe^{-x}$$
This suggests that $$\mathbb E\left[V\mid X\right]=Xe^{-X}+1-e^{-X}-Xe^{-X}=1-e^{-X}$$
You can verify this has a uniform distribution on $(0,1)$.
A: $$E[V|X] = E[X|X, Y\ge X]P(Y\ge X|X) + E[Y|X, Y<X]P(Y<X|X) = X\int_{X}^{\infty}e^{-y}dy + \int_{0}^Xye^{-y}dy = Xe^{-X} + 1 - e^{-X}(1+X) = 1 - e^{-X}$$
Which means that $E[V|X]\sim U(0, 1)$ since $F_{X}(X)$ is always distributed $U(0, 1)$ if $F_X(x)$ is a CDF of r.v. $X$ which obviously is the case here.
