Find the PDF of a random vector

Question

So I have a question in probability theory that's driving me insane. I know it is easy but I can't seem to wrap my head around it.

Assume we have $U_1\sim U[-1,1]$ and $U_2 \sim U[0,2]$ which are two independent random variables. We define $X=\min \{U_1, U_2 \}$ and $U=(U_1,U_2)^T$

What I have to do:

1. I need to show that $U|\{X=U_2\}$ is a continuous random vector and to find its PDF
2. Calculate $\mathbb E U_1|\{X=U_2\}$

I would appreciate any help :)

Thanks

Edit 1

In the first part I meant that I need to show that $U|\{X=U_2\}$ is a continuous random vector (not a continuous random variable as I wrote before) and to find its PDF.

Let $X=(X_1,...X_n)^T$ and $f:\mathbb R^n\rightarrow \mathbb R$ is an integrable function s.t.

$F_X(x)=\int_{\times(-\infty,x_i)}f(t)dt,$ $\forall x\in\mathbb R^n$

In this case we say that $X$ is a continuous random vector with a probability density function $f(.)$

Answer 1

Here's a hint: we have $$P(U_1\leq x_1, U_2\leq x_2 | X=U_2) = P(U_1\leq x_1, U_2\leq x_2, X=U_2)/P(X=U_2) = P(U_1\leq x_1, U_2\leq x_2, U_2\leq U_1)/P(U_2\leq U_1).$$ Can you work out the probabilities on the RHS by integrating the pdfs of $U_1, U_2$?

Answer 2

Without doing any calculation, complicate integration a.s.o., it is enough to observe that, given that $U_1=max[U_1;U_2]$ the random vector $U$ is uniform and continuous on the following purlple triangle

Then the joint density is trivially $2\mathbb{1}_{[0;1]}(u_1)\mathbb{1}_{[0;u_1]}(u_2)$

Concluding:

The marginal density of $U_1$ is $f_{U_1}=2\int_0^{u_1}d u_2=2u_1$ with expectation $\frac{2}{3}$

that's all