HW9.1. Function of a random variable (a) Let X have pdf fx(u) = = {au Jau u0.25 |u| 0.25 with parameter a = 96. Let Y = |X|. Using LOTUS, find E[Y]. (Hint: the answer is simplified as a fraction.) E[Y] = fraction (b) The pdf of Y (in part a) has the form: fy(y) = {hly d Sh(y) 0 y c otherwise Find h(0.5), d, and c. Note that h(0.5), d, and c are all scalars. h(0.5) fraction d = number (4 significant figures) c = number (4 significant figures)