How to find $P\left(\sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2} > X_2 + X_1\right)$ where $X_1, X_2, Y_1, Y_2$ are uniformly distributed from [0,5]

Question

How to find $P((X_2 - X_1)^2 + (Y_2 - Y_1)^2 > X_2 + X_1)$ where $X_1, X_2, Y_1, Y_2 \sim U[0,5]$ and iid.

Rearranging the probability, we have:

$$P\left(\sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2} > X_2 + X_1\right) = P((Y_2 - Y_1)^2) > (X_2 + X_1)^2 - (X_2 - X_1)^2 )$$

Then letting $T=Y_2 - Y_1,U=X_2+X_1,V=X_2-X_1$, we have:

$$ P((Y_2 - Y_1)^2 > X_2 + X_1 - (X_2 - X_1)^2 ) = P(T^2 > U^2 - V^2 ) $$

where

$$ f(t)= \begin{cases}t+5 & \text { for } -5<t<5 \\ 5-t & \text { for } 0 \leq t<5 \\ 0 & \text { otherwise. }\end{cases} $$

$$ f(u)= \begin{cases}u & \text { for } 0<u<5 \\ 5-u & \text { for } 5 \leq u<10 \\ 0 & \text { otherwise. }\end{cases} $$

$$ f(v)= \begin{cases}v+5 & \text { for } -5<v<5 \\ 5-v & \text { for } 0 \leq v<5 \\ 0 & \text { otherwise. }\end{cases} $$

But I am stuck here, I am not sure what the joint distribution would be. Further, I am not sure how the marginal would be written out since we have $U^2 - V^2$ on the RHS instead of 1 variable…

Do we have to make a new variable i.e. $Z = U^2 - V^2$?

$$P(T^2 > Z) = \int_0^5 \int_0^z f(t,z)dtdz$$

I am not sure where to go from here

Answer 1

Squaring the inequality and rearranging yields $\left(Y_2-Y_1\right)^2\gt4X_1X_2$. By separately considering $\frac15(X_1,X_2)$ and $\frac15(Y_1,Y_2)$ on the unit square and determining the relevant areas, you can obtain $P(4X_1X_2\gt a)=P(X_1X_2\gt\frac a4)=1+\frac a{100}\left(\log\left(\frac a{100}\right)-1\right)$ (the area above the hyperbola) and $P((Y_2-Y_1)^2\gt a)=P(|Y_2-Y_1|\gt \sqrt a)=\left(1-\frac{\sqrt a}5\right)^2$ (the area of two right iscosceles triangles).

The desired probability is

\begin{eqnarray*} \int_0^{25}P(4X_1X_2\le a)\frac{\partial P((Y_2-Y_1)^2\le a)}{\partial a}\,\mathrm da &=& -\int_0^{25}\frac a{100}\left(1-\log\left(\frac a{100}\right)\right)\frac\partial{\partial a}\left(1-\frac{\sqrt a}5\right)^2\,\mathrm da \\ &=& \int_0^{25}\frac a{100}\left(1-\log\left(\frac a{100}\right)\right)\frac1{5\sqrt a}\left(1-\frac{\sqrt a}5\right)\,\mathrm da \\ &=& \frac1{2500}\int_0^{25}\left(1-\log\left(\frac a{100}\right)\right)\left(5\sqrt a-a\right)\,\mathrm da \\ &=& \frac1{144}\left(13+12\log2\right) \\[4pt] &\approx& 0.148\;. \end{eqnarray*}

Here’s Java code that checks the result by simulation.