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Examples on Transformations of Random Variables 1. Let X ∼ U ([−π, π]). Find the distribution of the random variable Y = cos X. The density of X is given by fX (x) = ½ 1 2π 0 if x ∈ [−π, π] otherwise Method 1: Note that the range of random variable Y is [−1, 1]. There are two solutions to the equation y = cos x for x ∈ [−π, π], one in [−π, 0] and the other in [0, π]. Hence, the density of Y = cos X is given by ¯ ¯ X ¯ dx ¯ −1 fX (cos y) ¯¯ ¯¯ fY (y) = dy cos x=y ¯ ¯ X ¯ ¯ 1 −1 ¯ ¯ fX (cos y) ¯ = ¯ −1 − sin (cos y) cos x=y ½ 2 1 if cos−1 y ∈ [0, π] 2π sin (cos−1 y) = 0 otherwise Thus, ½ 1 π sin (cos−1 y) if y ∈ [−1, 1] (1) otherwise p If we further use the fact that sin(cos−1 (y)) = 1 − y 2 we obtain the following expression for the PDF: ( √1 if y ∈ [−1, 1] π 1−y 2 fY (y) = (2) 0 otherwise fY (y) = 0 Method 2: The CDF of X is FX (x) = x+π x 1 x−a = = + b−a 2π 2π 2 The sets that are equivalent to the event {Y ≤ y} are {X < − cos−1 (y)} and {X > cos−1 (y)}. The CDF of Y is given by FY (y) = P (Y ≤ y) = P (cos X ≤ y) 0 P (X ≤ − cos−1 y) + P (X ≥ cos−1 y) = 1 y ∈ [−∞, −1) 0 −1 cos y = if y ∈ [−1, 1] 1− π 1 otherwise 1 y ∈ [−∞, −1) if y ∈ [−1, 1] otherwise assuming that cos−1 y ≥ 0. The probability density function of Y is obtained as the derivative of this CDF expression. 2. Square law : Let X ∼ U ([−1, 1]). Find the distribution of the random variable Y = X 2 . The PDF of X is given by fX (x) = ½ 1 2 0 if x ∈ [−1, 1] otherwise Method 1: Note that the range of random variable Y is [0, 1]. There are two solutions to the equation y = x2 . Hence, the density of Y = X 2 is given by ¯ ¯ X ¯ dx ¯ fX (x) ¯¯ ¯¯ fY (y) = dy x2 =y ¯ ¯ ¯ ¯ 1 ¯¯ 1 ¯¯ 1 ¯¯ 1 ¯¯ = √ + √ 2 ¯ −2 y ¯ 2 ¯ 2 y ¯ ½ 1 √ if y ∈ [0, 1] 2 y = (3) 0 otherwise Method 2: The CDF of X is FX (x) = x 2 The CDF of Y is given by FY (y) P (Y ≤ y) = P (X 2 ≤ y) y ∈ [−∞, 0) 0 √ √ P (− y ≤ X ≤ y) if y ∈ [0, 1] = 1 otherwise y ∈ [−∞, −1) 0√ √ √ − y y = − 2 = y if y ∈ [−1, 1] 2 1 otherwise = Hence, the density of Y is given by (2). 3. Full-wave rectifier: Let X ∼ U ([−1, 1]). Find the distribution of the Y = Xu(X). The density of X is given by fX (x) = ½ 1 2 0 if x ∈ [−1, 1] otherwise Note that this transformation is not differentiable so the gradient method is not applicable in this case. Instead let us look at the set equivalence method. 2 – For y < 0, the event {Y ≤ y} does not have a solution on the real line and hence reduces to a null event. Consequently the probability of this event is 0. – For y = 0, the event {Xu(X) ≤ 0} has only one solution X = 0 and the probability of this event is FX (0+ ) − FX (0− ). – For y > 0, the event that {Xu(X) ≤ y} reduces to the event {X ≤ y} and the probability of this event is just FX (y). The CDF of the transformed random variable can then be summarized as: y<0 0 FX (0+ ) − FX (0− ) y = 0 FY (y) = FX (y) y>0 Since the input distribution is uniform this reduces to: ½ 0 y≤0 FY (y) = FX (y) y > 0 3