relation in uniform joint distribution function

Question

Define $R(x_1, x_2) := \lim\limits_{p \rightarrow 0} \frac{1}{p} \mathbb{P}(1-F_1(X_1)\leq px_1, 1-F_2(X_2)\leq px_2 )$ for $x_1\geq 0$, $x_2 \geq 0$, where $X_1$ and $X_2$ are random variables, not necessary independent, and $F_1$ and $F_2$ are their marginal distribution functions.

Note that $1-F_i(X_i)$ itself has an uniform distribution on $[0,1]$ .

For $t>0$, the relation $R(tx_1,tx_2) = tR(x_1,x_2)$ holds. Why is this true?

Answer 1

$\frac1p\mathbb P(1-F_1(X_1)\leq ptx_1,1-F_2(X_2)\leq ptx_2)=t\frac1{pt}\mathbb P(1-F_1(X_1)\leq ptx_1,1-F_2(X_2)\leq ptx_2)$

Now on both sides let $p\to0$ (so that also $pt\to0$).

Then you find: $R(tx_1,tx_2)=tR(x_1,x_2)$ as a result.

Answer 2

Note that $$R(x_1,x_2)=\lim\limits_{p\to0}\frac1pg(px_1,px_2)$$ for some given function $g$, hence $$R(tx_1,tx_2)=\lim\limits_{p\to0}\frac1pg(ptx_1,ptx_2)=t\lim\limits_{pt\to0}\frac1{pt}g((pt)x_1,(pt)x_2)=tR(x_1,x_2)$$