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$U_1$ and $U_2$ are identically independently distributed ~uni[0,1]. I'm trying to find the joint PDF between the $P_1$=min($U_1$, $U_2$) and $P_2$=max($U_1$, $U_2$). I have found the marginal pdfs:

$f_{p1}(y)=2y^2-4y+2$

$f_{p2}(x)=2x$

Since the min and max are not independent, I think the best course of action would be to find the conditional probability by conditioning on $U_1$ and then multiplying by the pdf of $U_1$. $$\int_0^1 P(min(U_1, U_2)<p_{1} ,P(max(U_1, U_2)<p_{2}\ \ |\ \ U_1=u)\ f_u(u) \;\mathrm{dx}$$

However, I am stuck now.

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1 Answer 1

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Going back to the cdf is always an option (and I don't believe your marginal pdf for $P_1$ as it doesn't integrate to 1).

We have, for $0\leq p_1\leq p_2\leq 1$ $$ \begin{align*} F_{P_1,P_2}(p_1,p_2)&=\mathbb{P}(P_1\leq p_1, P_2\leq P_2)\\ &=\mathbb{P}(\min(U_1,U_2)\leq p_1,\max(U_1,U_2)\leq p_2)\\ &=\mathbb{P}((U_1,U_2)\in[0,p_1]\times[0,p_2]\cup[0,p_2]\times[0,p_1])\\ &=\mathbb{P}((U_1,U_2)\in[0,p_1]\times[0,p_2]\amalg(p_1,p_2]\times[0,p_1])\\ &=2p_1p_2-p_1^2 \end{align*} $$ so differentiating gives the pdf $$ f_{P_1,P_2}(p_1,p_2)=\frac{\partial^2 F_{P_1,P_2}(p_1,p_2)}{\partial p_1\partial p_2}= \begin{cases} 2&\text{if }0<p_1<p_2<1\\ 0&\text{otherwise} \end{cases}. $$

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