Choose independently two numbers B and C at random from the interval [0, 1] with uniform density. Prove that B and C are proper probability distributions. Note that the point (B,C) is then chosen at random in the unit square. Find the probability that
Note that the point (B,C) is then chosen at random in the unit square <=> \(B(x, y) + C(x, y) = 1\) => \(B(0, 1/2) + C(1/2, 0) = 1\)
if $B + C < 1/2: x < 1/2, y < 1/2 - x $, then:
\[ P\Bigg(B + C < 1/2\Bigg) = P\Bigg( 0 < x < 1/2, \space 0 < y < 1/2 - x \Bigg) \\ =\int_{0}^{1/2}\int_{0}^{1/2-x} dydx =\int_{0}^{1/2}[1/2-x] dx \\ =\Bigg[ x/2 - x^2/2 \Bigg]_{0}^{1/2} =1/4 - 1/8 = 1/8 =0.125 \]
\(\begin{cases} B=1 * C=1/2 < 1/2 \\ or\\ B=1/2 * C=1 < 1/2 \end{cases}\)
\[ P\Bigg( B*C < 1/2\Bigg) = P\Bigg( 0 < x < 1/2, \space 0 < y < 1 - x \Bigg) \\ =\int_{0}^{1/2}\int_{0}^{1-x} dydx =\int_{0}^{1/2}[1-x] dx \\ =\Bigg[ x - x^2/2 \Bigg]_{0}^{1/2} =1/2 - 1/8 = 3/8 =0.375 \]
\(\begin{cases} B + C = 1 (given) \\ |B - C| < 1/2 => 1-C-C < 1/2 => \begin{cases} C > 1/4 \\ B = 1-C < 3/4\end{cases} \end{cases}\)
\[ P \Bigg(|B - C| < 1/2\Bigg) = P\Bigg( C > 1/4, \space B < 3/4 \Bigg) = P\Bigg( 1/4 < x < 1, \space 1/4 + x < y < 3/4 \Bigg) \\ =\int_{1/4}^{1}\int_{0}^{3/4-x} dydx =\int_{0}^{1/4}[y]_{0}^{3/4-x} dx =\int_{1/4}^{1}(3/4-x) dx\\ =[3x/4 - x^2/2]_{1/4}^{1} = 3/4 - 1/32 - 3/4 + 1/2 =15/32 =0.46875 \]
\(max(B, C) = 1/2(B+C+|B-C|)\)
\[ P\Bigg( max\left\{\begin{array}{lr} B, C\end{array}\right\} < 1/2\Bigg) = P\Bigg( 1/2*(B+C+|B-C|)\Bigg) = 1/2 * P\Bigg( P(B+C)+ P(|B-C|) \Bigg) \\ = 1/2 * (1/8 + 15/32) = 19/64 = 0.296875 \]
\[ P\Bigg( min\left\{\begin{array}{lr}B, C\end{array}\right\} < 1/2\Bigg) = P\Bigg( 1/2*(B+C-|B-C|)\Bigg) = 1/2 * P\Bigg( P(B+C)- P(|B-C|) \Bigg) \\ = 1/2 * (1/8 - 15/32) = 11/64 = 0.171875 \]