Probability of independent random variables that are uniformly distributed

Question

$$ \underline{ \bf Attempt } $$

Let $X_i$ be the insurer bids $i=1,2$. Now, the company decides two bids if the difference betweem them is $\bf more$ than $20$. In other words, we want

$$ P( |X_1-X_2| > 20 ) $$

Let $A = \{ (x_1,x_2) : |x_1-x_2| > 20 \} $. Since $X_i$ are uniform and indepedent, we have

$$ f_{X_1 X_2} (x_1,x_2) = f_{X_1}(x_1) f_{X_2} (x_2) = \frac{1}{200} \times \frac{1}{200} = \frac{1}{4 0000} $$

Notice the regions $[2000,2200] \times [2000,2200]$ is contained in $A$, thus

$$ P( |X_1-X_2| > 20 ) = \int\limits_{2000}^{2200} \int\limits_{2000}^{2200} \frac{1}{4 0000} = 1 $$

Now, this is obviously not the correct answer. My question is, when solving this do we to integrate over the intersection of the regions? That is we have to integrate over $A \cap [2000,2200]^2$?

Answer 1

The square is not contained in $A$; rather, the intersection $A\cap[2000;2200]^2$ is a pair of triangles contained within the square.

Thus you want:$$\begin{split}\mathsf P(\lvert X_1-X_2\rvert\geqslant 20) &= \mathsf P(X_1\geqslant X_2+20)+\mathsf P(X_2\geqslant X_1+20) \\ &= 2 \iint_{2000+20\leqslant x+20\leqslant y\leqslant 2200} \tfrac1{200^2}~\mathsf d(x,y)\end{split}$$

Answer 2

Tip (your question has already been answered):

Observe that $X_i=2000+200U_i$ where $U_i$ is uniformly distributed over $[0,1]$.

We find easily that: $$|X_1-X_2|>20\iff|U_1-U_2|>0.1$$ and calculating $\mathsf P(|U_1-U_2|>0.1)$ instead of $\mathsf P(|X_1-X_2|>20)$ gives less opportunities to make mistakes.