This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mathematics Department, UCLA T. Richthammer winter 09, sheet 10 Mar 06, 2009 Homework assignments: Math 170A Probability, Sec. 1 120. Express the following probabilities in terms of the joint CDF F of X = ( X 1 , X 2 ): (a) P ( X 1 ≤ a, X 2 > b ) (b) P ( X 1 > a, X 2 > b ) (c) P ( X 1 ≤ a, X 2 < b ) Answer: (a) P ( X 1 ≤ a, X 2 > b ) = P ( X 1 ≤ a ) P ( X 1 ≤ a, X 2 ≤ b ) = F ( a, ∞ ) F ( a, b ) (b) P ( X 1 > a, X 2 > b ) = 1 P ( X 1 ≤ a or X 2 ≤ b ) = 1 P ( X 1 ≤ a ) P ( X 2 ≤ b )+ P ( X 1 ≤ a, X 2 ≤ b ) = 1 F ( a, ∞ ) F ( ∞ , b ) + F ( a, b ) (c) P ( X 1 ≤ a, X 2 < b ) = F ( a, b ) 121. Show that 3 RVs X, Y, Z are independent if and only if for all sets A, B, C ⊂ R we have P ( X ∈ A, Y ∈ B, Z ∈ C ) = P ( X ∈ A ) P ( Y ∈ B ) P ( Z ∈ C ). Answer: If X, Y, Z are independent, then by definition we have the product property. Now suppose we have the product property. To show that X, Y, Z are independent we need to show that P ( X ∈ A, Y ∈ B ) = P ( X ∈ A ) P ( Y ∈ B ), P ( X ∈ A, Z ∈ C ) = P ( X ∈ A ) P ( Z ∈ C ), P ( Y ∈ B, Z ∈ C ) = P ( Y ∈ B ) P ( Z ∈ C ) for all subsets A, B, C ⊂ R . This can be obtained from the product property by plugging in C = R , B = R and A = R respectively. 122. Suppose that 3 balls are chosen without replacement from an urn consisting of 13 balls; 5 of them are labelled with ”0”, 8 of them with ”1”. Let X i denote the number of the i th ball chosen. (a) Find the joint PMF of X = ( X 1 , X 2 , X 3 ). (b) Calculate the PMF of X 3 . Answer: (a) p (0 , , 0) = 5 13 4 12 3 11 , p (0 , , 1) = 5 13 4 12 8 11 , p (0 , 1 , 0) = 5 13 8 12 4 11 , p (1 , , 0) = 8 13 5 12 4 11 , p (0 , 1 , 1) = 5 13 8 12 7 11 , p (1 , , 1) = 8 13 5 12 7 11 , p (1 , 1 , 0) = 8 13 7 12 5 11 , p (1 , 1 , 1) = 8 13 7 12 6 11 . (b) p (0) = p (0 , , 0) + p (0 , 1 , 0) + p (1 , , 0) + p (1 , 1 , 0) = 60+160+160+280 13 · 12 · 11 = 5 13 , p (1) = 8 13 . 123. An experiment with success rate p is repeated infinitely often. Let X 1 be the number of failures preceding the first success, and let X 2 be the number of failures between the first two successes. (a) Find the joint PMF of X = ( X 1 , X 2 ). (b) Calculate the PMFs of X 1 and X 2 , and show that X 1 and X 2 are independent. Answer: (a) { X 1 = x 1 , X 2 = x 2 } means that we have x 1 failures, then one success, then x 2 fail ures, then one success, and after that an arbitrary sequence of failures and successes. So p ( x 1 , x 2 ) = P ( X 1 = x 1 , X 2 = x 2 ) = (1 p ) x 1 p (1 p ) x 2 p = p 2 (1 p ) x 1 + x 2 . (b) p 1 ( x 1 ) = ∑ ∞ x 2 =0 p ( x 1 , x 2 ) = p (1 p ) x 1 and similarly p 2 ( x 2 ) = p (1 p ) x 2 . As p ( x 1 , x 2 ) = p 1 ( x 1 ) p 2 ( x 2 ) the RVs are independent. 124. Let X 1 , X 2 be independent Poisson RVs with parameters λ 1 , λ 2 , and let X := X 1 + X 2 ....
View
Full Document
 Winter '10
 RICHTHAMMER
 XY sexdetermination system, CDF, Emoticon, X&Y, T. Richthammer

Click to edit the document details