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Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 13: Solutions Functions of Random Variables 1. [A piece of cake? or a sheet cake with a piece missing?] (a) The pdf is nonzero over the shaded region in the lefthand figure shown below. β 1 /α α u v 1 1 1 / 2 1 / 2 u v 1 1 1 / 2 1 / 2 u v 1 1 1 / 2 1 / 2 α (b) f Y ( v ) = Z ∞∞ f X , Y ( u,v ) du = Z 1 1 / 2 4 3 du = 2 3 , < v ≤ 1 2 , Z 1 4 3 du = 4 3 , 1 2 < v < 1 , and f Y ( v ) = 0 otherwise. Use the horizontal lines in the middle figure above if you have difficulty figuring out where the limits came from. By symmetry, X has the same marginal pdf as Y . (c) Y ≤ α X if the random point lies in the deepshaded region in the lower right corner of the rightmost figure above. The area is α 2 α 8 = 3 α 8 and hence P { Y ≤ α X } = 4 3 × 3 α 8 = α 2 . (d) Y > α X if the random point lies in the deepshaded region in the upper left corner of the rightmost figure above. The area is 1 2 α 1 8 α = 3 8 α and hence P { Y ≤ α X } = 1 4 3 × 3 8 α = 1 1 2 α . (e) P { Z ≤ α } = P { Y ≤ α X } = α 2 , < α ≤ 1 , 1 1 2 α , 1 < α < ∞ . (f) f Z ( α ) = d dα F Z ( α ) = d dα P { Z ≤ α } = 1 2 , < α ≤ 1 , 1 2 α 2 , 1 < α < ∞ , , elsewhere. Note that Z has the same pdf as I + 1 (with I = 1) that you found in Problem 3 of Problem Set 11, that is, with I = 1, f Z ( α ) = f I ( α 1)....
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 Spring '10
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 Variance, Probability theory, Exponential distribution

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