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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 3.7.1 3.7.2 3.7.5 and 3.7.7 Problem 3.7.1 Solution Since 0 X 1, Y = X 2 satisfies 0 Y 1. We can conclude that F Y ( y ) = 0 for y < and that F Y ( y ) = 1 for y 1. For 0 y < 1, F Y ( y ) = P bracketleftbig X 2 y bracketrightbig = P [ X y ] (1) Since f X ( x ) = 1 for 0 x 1, we see that for 0 y < 1, P [ X y ] = integraldisplay y dx = y (2) Hence, the CDF of Y is F Y ( y ) = y < y y < 1 1 y 1 (3) By taking the derivative of the CDF, we obtain the PDF f Y ( y ) = braceleftbigg 1 / (2 y ) 0 y < 1 otherwise (4) Problem 3.7.2 Solution Since Y = X , the fact that X is nonegative and that we asume the squre root is always positive implies F Y ( y ) = 0 for y < 0. In addition, for y 0, we can find the CDF of Y by...
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- Spring '08