test1_sol

test1_sol - 18.05. Test 1. (1) Consider events A = {HHH at...

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Unformatted text preview: 18.05. Test 1. (1) Consider events A = {HHH at least once} and B = {TTT at least once}. We want to find the probability P (A � B ). The complement of A � B will be Ac � B c , i.e. no TTT or no HHH, and P (A � B ) = 1 − P (Ac � B c ). To find the last one we can use the probability of a union formula P (Ac � B c ) = P (Ac ) + P (B c ) − P (Ac � B c ). Probability of Ac , i.e. no HHH, means that on each toss we don’t get HHH. The probability not to get HHH on one toss is 7/8 and therefore, P (Ac ) = 7 �10 8 . The same for P (B c ). Probability of Ac � B c , i.e. no HHH and no TTT, means that on each toss we don’t get HHH and TTT. The probability not to get HHH and TTT on one toss is 6/8 and, therefore, c c P (A � B ) = 6 �10 8 . Finally, we get, P (A � B ) = 1 − 7 �10 8 + 7 �10 8 − 6 �10 � 8 . (2) We have P (F ) = P (M ) = 0.5, P (CB |M ) = 0.05 and P (CB |F ) = 0.0025. Using Bayes’ formula, P (M |CB ) = P (CB |M )P (M ) 0.05 × 0.5 = P (CB |M )P (M ) + P (CB |F )P (F ) 0.05 × 0.5 + 0.0025 × 0.5 (3) We want to find f (y |x) = f (x, y ) f1 (x) which is defined only when f (x) > 0. To find f1 (x) we have to integrate out y , i.e. � f1 (x) = f (x, y )dy. 2 To find the limits we notice that for a given x,≤ < y 2 < 1 − x≤which is not 0 empty only if x2 < 1, i.e. −1 < x < 1. Then − 1 − x2 < y < 1 − x2 . So if −1 < x < 1 we get, � 1−x2 � ≤ y 3 � 1−x2 1 � f1 (x) = � c(x +y )dy = c(x y + )� � = 2c(x2 1 − x2 + (1−x2 )3/2 ). 2 3 − 1−x 3 − 1−x2 � 2 2 2 Finally, for −1 < x < 1, f (y |x) = 2c(x2 ≤ x2 + y 2 c(x2 + y 2 ) ≤ = 1 2 2x2 1 − x2 + 3 (1 − x2 )3/2 1 − x2 + 3 (1 − x2 )3/2 ) ≤ ≤ if − 1 − x2 < y < 1 − x2 , and 0 otherwise. (4) Let us find the c.d.f first. P (Y � y ) = P (max(X1 , X2 ) � y ) = P (X1 � y , X2 � y ) = P (X1 � y )P (X2 � y ). The c.d.f. of X1 and X2 is P (X1 � y ) = P (X2 � y ) = � y f (x)dx. −� If y � 0, this is y P (X1 � y ) = � P (X1 � y ) = � −� and if y > 0 this is 0 �y � ex dx = ex � −� �0 e dx = e � x −� Finally, the c.d.f. of Y , P (Y � y ) = � x� −� e2y , y � 0 1, y > 0. Taking the derivative, the p.d.f. of Y , � 2y 2e , y � 0 f (y ) = 0, y > 0. = ey = 1. and zero otherwise, i.e. for z � 0. f (z ) = � The p.d.f. is P (Z � z ) = � z 3 1 3z 3 + 1, 0<z�1 3 1 + 3z 2 , z>1 + z, 0<z�1 3 1 1− − 3z , z>1 1 6z 2 z2 6 So the c.d.f. of Z is When z ∩ 1, the limits are different �1 2 � 1� 1 �1 y 1 1 � (x + y )dydx = ( + xy )� dx = 1 − 2 − . 2 6z 3z x/z 0 x/z 0 To find the limits, we have to consider the intersection of this set {x � z y } with the square 0 < x < 1, 0 < y < 1. When z � 1, the limits are � 1 � zy �1 2 �1 2 �zy x z z2 z � 2 (x + y )dxdy = ( + xy )� dy = ( + z )y dy = +. 2 2 6 3 0 0 0 0 0 {x�zy } (5) Let us find the c.d.f. of Z = X/Y first. Note that for X, Y ∪ (0, 1), Z can take values only > 0, so let z > 0. Then � P (Z � z ) = P (X/Y � z ) = P (X � z Y ) = f (x, y )dxdy. Figure 1: Region {x � z y } for z � 1 and z > 1.  ��    ��    �         ����������       �   ����� �����  �  �� �� ����� ������ � �� �� ����� ��       ���    ���      ����� �����   �    �����  �   �   �� �� ����� �� � ����� �� � � � �� � ��       ��    ���   ����������   �      �����      �   �� � ����� ����� ������  � �� �� � �����   ���    ���  ��         ����������      �   �����   ������  �� �� �����  ������ � �� �� ����� ���  ���� ������                 ������� PSfrag x z > 1. z�1 y ...
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