I recently faced a probability problem that is puzzling me. I would like to ask you if you could help me.
I have two random variables X1 and X2 i.i.d with uniform distribution U[64,96] and other two random variables Y1 and Y2 i.i.d. distributed according to the sum of two uniform distributions, i.e., Y1,Y2 ~ U[16,32] + U[64,96].
I need to compute the probability that at least one between X1 and X2 happens before both Y1 and Y2.
I worked quite a bit on it, but cannot derive the exact probability, which should be around 0.98, according to simulations in matlab:
nsims = 1000000;
X1 = 64 + (96-64).*rand(nsims,1);
X2 = 64 + (96-64).*rand(nsims,1);
Y1 = 16 + (32-16).*rand(nsims,1) + 64 + (96-64).*rand(nsims,1);
Y2 = 16 + (32-16).*rand(nsims,1) + 64 + (96-64).*rand(nsims,1);
P = sum((X1<min(Y1,Y2)) |(X2<min(Y1,Y2)))/nsims;
I also derived both analytically and empirically the P(X1<Y1) = 0.958
Also, how would the problem change if we have X1,X2...Xm and Y1,Y2,...Yn?
This is not an homework, just a personal project. Thanks in advance
If you let $X = \min(X_1,X_2)$ then it has density $p(x) = \frac{96-x}{512}$ for $64 \le x \le 96$ and zero elsewhere.
Meanwhile $Y_1$ has density $p(y_1) = \frac{y_1-80}{512}$ for $80 \le y_1 \le 96$ plus a probability of being greater than $96$, and similarly for $Y_2$.
So you want $$\int_{64}^{96} \frac{96-x}{512} P(Y_1\gt x) P(Y_2 \gt x)\, dx$$ $$=\int_{64}^{80} \frac{96-x}{512} \, dx +\int_{80}^{96} \frac{96-x}{512} \left( 1- \frac{(x-80)^2}{2\cdot 512} \right)^2\, dx$$ $$=\frac34+\frac{221}{960} = \frac{941}{960} \approx 0.9802$$