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B. First find the cdf of U . FU (u) = P (U ≤ u) = P (min{X, Y } ≤ u) = 1 − P (min{X, Y } > u) = 1 − P ({X > u} ∩ {Y > u}) = 1 − P (X > u)P (Y > u) = 1 − (1 − P (X ≤ u))(1 − P (Y ≤ u)) = 1 − (1 − F (u))2 . d du FU (u) Differentiate the cdf to get the pdf by using the Chain Rule fU (u) = = 2f (u)(1 − F (u)). c. X and Y be independent random variables each having the uniform distribution on [0,1]. Hence, fU (u) = 2(1 − u) for u ∈ [0, 1]. 1 1 E[U ] = ufU (u)du = 0 1 2u(1 − u)du = 2 0 (u − u2 )du 0 1 u2 u3 = 2( − )]10 = .

B. X = 3 − Z where Z is as in part a. c. 3 of missing the target altogether. If he does not miss, he hits an area A with a probability proportional to |A|. The score X is now 0 if the dart misses the target. Otherwise, it is 3 − Z, where Z is as before. i. We see that P (X = 3) = P (Z ≤ 1) = π/9π = 1/9. Similarly, P (X = 2) = P (1 < Z ≤ 2) = (4π − π)/9π = 1/3. Finally, P (X = 1) = P (2 < Z ≤ 3) = (9π − 4π)/9π = 5/9. Hence,   0, if x < 1;      5/9, if 1 ≤ x < 2; FX (x) =  8/9, if 2 ≤ x < 3;      1, if x ≥ 3.

The σ-field generated by A is F. ) (iii). P1 and P2 are not the same. Let A= {{1, 2}, {2, 4}}. Assign probabilities P1 ({1}) = 18 , P1 ({2}) = 18 , P1 ({3}) = 38 , P1 ({4}) = 38 ; and P2 ({1}) = 1 12 , P2 ({2}) = 2 12 , P2 ({3}) = 5 12 , P2 ({4}) 4 12 . = Note that P1 and P2 are not the same, thus satisfying (iii). 1 8 + 1 8 + 2 12 = P1 ({1, 2}) = P1 ({1}) + P1 ({2}) = P2 ({1, 2}) = P2 ({1}) + P2 ({2}) = 1 12 = 1 4 1 4 Hence P1 ({1, 2}) = P2 ({1, 2}). P1 ({2, 4}) = P1 ({2}) + P1 ({4}) = 1 8 P2 ({2, 4}) = P2 ({2}) + P2 ({4}) = 2 12 + 3 8 + = 4 12 1 2 = 1 2 Hence P1 ({2, 4}) = P2 ({2, 4}).