P( X < 3 | X +Y) and E(X | min(X,t) ) X,Y iid exponetia(1) t > 0 Let $X, Y$ be i.i.d exponential random variables with $\lambda = 1 $and $t$ is some positive constant??
How to find the random variables:
a) $P(X < 3\mid X+Y) $
b) $E(X \mid \min(X,t))$
So far my attempts are:
a) let $K > 3$
$P(X < 3 \mid X + Y) =  P(X < 3\mid X + Y \leq k) $
$= P( [X < 3] \cap [X + Y \leq k ] )/P(X + Y \leq k)$
$= P( [X < 3] \cap [Y \leq k - 3] )/P(X + Y \leq k)$
$= P(X < 3) \cdot P(Y \leq k-3)/P(X + Y \leq k)$ by independence
b)
since the RV $\min(X,t) = X$ if $X < t$ and $t$ if $X \geq t$
so if $X < t$
$E(X \mid \min(X,t)) = E(X \mid X) = X$
if $X \geq t$
$E(X \mid \min(X,t)) = E(X \mid t) = E(X \mid X \geq t) = t+1$
and the problem I have with this is that I can't verify the this expression of 
$E(X | \min(X,t))$ satisfies the definition of conditional expectation 
i.e $E(X I_A) = E(E(X \mid \min(X,t)) I_A)$ for all A in $σ(\min(X,t))$
 A: For the first part I don't see why $P(X<3\mid X+Y) = P(X<3\mid X+Y\leq k)$?
Clearly $P(X<3\mid X+Y)=0$ if $X+Y<3$. 
For first part, I consider change of variable. Let $Z=X+Y$. 
$f(x,y) = e^{-(x+y)}\mathbb{1}_{\{x>0,y>0\}}$ is the joint density of $(X,Y)$. Then doing a change of variable. $z = x+y, x = x$, we have
$g(x,z) = e^{-z}\mathbb{1}_{\{z>0,x<z\}}$
Conditional on $z$, $g(x\mid z) = g(x,z)/g(z)$ where $g(z)$ is the density of $g$.
$g(z) = \int^\infty_0 e^{-z}\mathbb{1}_{\{z>0,x<z\}}\text{d}x = \int^z_0 e^{-z}\mathbb{1}_{\{z>0\}} \text{d}x= ze^{-z}\mathbb{1}_{\{z>0\}}$ i.e. $Z$ is a $\Gamma(2,1)$ distribution.
Now $g(x\mid z)= 1/z \mathbb{1}_{\{x<z\}}$, i.e. $X\mid Z$ is a uniform distribution $U(0,Z)$, so the answer to your question is
$3/X+Y$ if $X+Y>3$, $1$ otherwise.
For part 2
$E(X\mid\min(X,t)) = E(X\mathbb{1}_{\{X<t\}}\mid\min(X,t))+E(X\mathbb{1}_{\{X\geq t\}}\mid\min(X,t))= X{1}_{\{X<t\}}+E(X\mathbb{1}_{\{X\geq t\}}\mid X>t) $
The second bit calculate to $\dfrac{\int^\infty_t e^{-s}s \,\text{d}s}{\int^\infty_te^{-s}\,\text{d}s}$
PS I am not sure if this is just $1+t$, I think you are right. For any event $A$ measurable with respect to $\sigma(\min(X,t))$, write $A$ as $(A\cap\{X\geq t\}) \cup (A\cap\{X<t\})$. Take it from there.
