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Unformatted text preview: Mathematics Department, UCLA T. Richthammer winter 09, sheet 6 Feb 06, 2009 Homework assignments: Math 170A Probability, Sec. 1 064. Calculate and sketch the CDF of the RV X , where the distribution P X (a) is a discrete probability law with weights p 1 = 1 2 , p 3 = 1 6 , p 4 = 1 3 . (b) is the equidistribution on [0 , 1]. Answer: (a) F X ( x ) = 0 for x &lt; 1, F X ( x ) = 1 2 for 1 x &lt; 3, F X ( x ) = 2 3 for 3 x &lt; 4, F X ( x ) = 1 for 4 x . The graph of F X is a jump function . . . . (b) For 0 x 1: F X ( x ) = P ( X x ) = P X ([0 , x ]) = ([0 ,x ]) ([0 , 1]) = x 1 = x . Furthermore F X ( x ) = 0 for x 0 and F X ( x ) = 1 for x 1. Picture . . . 065. Consider the CDFs F X ( x ), F Y ( x ), F Z ( x ) defined by: for x &lt; . 5 x 2 for 0 x &lt; 1 . 5 x for 1 x &lt; 2 1 for 2 x , for x &lt; . 5 for 0 x &lt; 1 . 75 for 1 x &lt; 2 1 for 2 x , for x &lt; x for 0 x &lt; 1 . 5 for 1 x &lt; 2 1 for 2 x . (a) In fact, one of these functions is not a CDF of a RV, and one of them is the CDF of a discrete RV. Which ones and why? (b) For the two functions that are CDFs calculate the probabilities that the corresponding RV takes a value in the set (i) { 1 , 2 } , (ii) [1 , 2 . 5], (iii) [0 . 5 , 2), (iv) (0 . 5 , 1 . 5). Answer: (a) F Z is not increasing: F Z (0 . 75) &gt; F Z (1), but every CDF has to be increasing, so F Z is not the CDF of a RV. F Y is a jump function, and thus the CDF of a discrete RV. (b) P ( X { 1 , 2 } ) = 0, P ( Y { 1 , 2 } ) = 0 . 5, P (1 X 2 . 5) = 1 . 5 = 0 . 5, P (1 Y 2 . 5) = 0 . 5, P (0 . 5 X &lt; 2) = 1 . 125 = 0 . 875, P (0 . 5 Y &lt; 2) = 0 . 75 . 5 = 0 . 25, P (0 . 5 &lt; X &lt; 1 . 5) = 0 . 75 . 125 = 0 . 625, P (0 . 5 &lt; X &lt; 1 . 5) = 0 . 75 . 5 = 0 . 25. 066. Let X be a RV, Y := 2 X + 3 and Z := e X . Express the CDFs F Y and F Z in terms of F X . Answer: F Y ( c ) = P ( Y c ) = P (2 X + 3 c ) = P ( X c 3 2 ) = F X ( c 3 2 ). Similarly for c &gt; 0 we have F Z ( c ) = P ( Z c ) = P ( e X c ) = P ( X ln c ) = F X (ln c ) and F Z ( c ) = 0 for c 0. 067. Let X be a RV with PMF p ( n ), n { 1 , . . . , 5 } , where (a) p ( n ) = c (b) p ( n ) = cn . In both cases determine the value of the constant c, and sketch a picture of the PMF. Explain from this picture that in (b) the expectation should be larger, but the standard deviation should be smaller than in (a). Confirm this by calculating E ( X ) and ( X ). Answer: (a) 1 = n p ( n ) = 5 c , so c = 1 5 , so p ( n ) = 1 5 . E ( X ) = 3, E ( X 2 ) = 11, so V ( X ) = 2 and ( X ) = 2....
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 Winter '10
 RICHTHAMMER

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