Conditional expectation of random variables

Question

I have a problem to solve. Variables $X_1,X_2,...,X_n$ are i.i.d. and have exponential distribution with parameter $2$.

a) Find the distribution of variable $Y$=min{$X_1,X_2,...,X_{n-1}$}

b) Compute $E(\min${$X_1,X_2,...,X_n$}|$X_n$)

In a) we know that distribution of $X_i$= $1-e^{-2t}$ so

$F_Y(t)\\=P(Y<t)\\=P(\min[ X_1,X_2,...,X_{n-1}]<t)\\=1-P(\min[ X_1,X_2,...,X_{n-1}]>t)\\=1-(P(X_1>t))^{n-1}\\=1-(1-(X_1<t))^{n-1}\\=1-e^{-2(n-1)t}$

What about $b$? I don't know the joint distribution of $X_n$ and $ \min${$X_1 , X_2 , \dots ,X_n$} and I don't know how to solve it in different way :(

Answer 1

For random variables $X,Y$, where $X$ is measurable respectivly to a $\sigma$-Algebra $\mathcal{X}$ and $Y$ is independent from $\mathcal{X}$, and a measurable function $\phi$ holds:

$$\Bbb{E}[\phi (X,Y) | \mathcal{X}](\omega) = \big(\Bbb{E}[\phi(x,Y)]\big)_{x=X(\omega)}$$

This indexnotation means that we don't integrate over $X$ anymore.