ECE 302: Lecture 4.4 Median, Mode, and Mean Prof Stanley Chan School of Electrical and Computer Engineering Purdue University c Stanley Chan 2020. All Rights Reserved. 1 / 16 Median Given a sequence of numbers n 1 2 3 4 5 6 7 8 9 . 100 xn 1.5 2.5 3.1 1.1 -0.4 -4.1 0.5 2.2 -3.4 . -1.4 Find the median. Step 1: You sort the sequence Step 2: You pick the one in the middle If we have a random variable, what is the ideal median? c Stanley Chan 2020. All Rights Reserved. 2 / 16 Median from PMF Definition Let X be a continuous random variable with PDF fX . The median of X is a point c 2 R such that Z c Z 1 fX (x)dx = fX (x)dx: (1) −∞ c c Stanley Chan 2020. All Rights Reserved. 3 / 16 Median from CDF Theorem The median of a random variable X is the point c such that 1 F (c) = : (2) X 2 Proof. R x 0 0 Since FX (x) = −∞ fX (x )dx , we have Z c Z 1 FX (c) = fX (x)dx = fX (x)dx = 1 − FX (c): −∞ c 1 Rearranging the terms shows that FX (c) = 2 . c Stanley Chan 2020. All Rights Reserved. 4 / 16 Example Example 1. (Uniform random variable) Let X be a continuous random variable with 1 PDF: fX (x) = b−a for a ≤ x ≤ b, and is 0 otherwise. x−a CDF: FX (x) = b−a for a ≤ x ≤ b. Find median. Solution: Want FX (c) = 1=2. a + b c = 2 c Stanley Chan 2020. All Rights Reserved. 5 / 16 Example Example 2. (Exponential random variable) Let X be a continuous random variable with −λx PDF: fX (x) = λe for x ≥ 0 −λx CDF: FX (x) = 1 − e for x ≥ 0 Find median. Solution: Want FX (c) = 1=2. c = (log 2)/λ c Stanley Chan 2020. All Rights Reserved. 6 / 16 Mode Given a sequence of numbers n 1 2 3 4 5 6 7 8 9 . 100 xn 50 50 30 10 -40 -10 50 20 -30 . -1 How to find the mode? Step 1: Sort the sequence Step 2: Pick the number that occurs most often? What is the ideal mode? c Stanley Chan 2020. All Rights Reserved. 7 / 16 Mode from PDF and CDF Definition Let X be a continuous random variable. The mode is the point c such that fX (x) attains the maximum: d c = argmax fX (x) = argmax FX (x): (3) x2Ω x2Ω dx c Stanley Chan 2020. All Rights Reserved. 8 / 16 Example Example 1. Let X be a continuous random variable with PDF fX (x) = 6x(1 − x) for 0 ≤ x ≤ 1. Find mode. Solution. The mode of X happens at argmax fX (x). x 1 x = 2 c Stanley Chan 2020. All Rights Reserved. 9 / 16 Mean Given a sequence of numbers n 1 2 3 4 5 6 7 8 9 . 100 xn 1.5 2.5 3.1 1.1 -0.4 -4.1 0.5 2.2 -3.4 . -1.4 Find the mean. N 1 X X = X : N n n=1 How to find mean from CDF? c Stanley Chan 2020. All Rights Reserved. 10 / 16 Mean from CDF: Positive Case Lemma Let X > 0. Then, E[X ] can be computed from FX as Z 1 E[X ] = (1 − FX (t)) dt: (4) 0 Proof: Z 1 Z 1 Z 1 (1 − FX (t)) dt = [1 − P[X ≤ t]] dt = P[X > t]dt 0 0 0 1 1 1 x Z Z (a) Z Z = fX (x)dxdt = fX (x)dtdx 0 t 0 0 Z 1 Z x Z 1 = dtfX (x)dx = xfX (x)dx = E[X ]: 0 0 0 c Stanley Chan 2020. All Rights Reserved. 11 / 16 Interchange the integrations Figure: A double integration can be evaluated in two ways: x then t, or t then x. c Stanley Chan 2020. All Rights Reserved. 12 / 16 Negative Case Lemma Let X < 0. Then, E[X ] can be computed from FX as Z 0 E[X ] = FX (t)dt: (5) −∞ Proof. Z 0 Z 0 FX (t)dt = P[X ≤ t]dt −∞ −∞ Z 0 Z t = fX (x)dxdt −∞ −∞ Z 0 Z 0 Z 0 = fX (x)dtdx = xfX (x)dx = E[X ]: −∞ x −∞ c Stanley Chan 2020. All Rights Reserved. 13 / 16 The overall result Theorem The mean of a random variable X can be computed from the CDF as Z 1 Z 0 E[X ] = (1 − FX (t)) dt − FX (t)dt: (6) 0 −∞ Proof. Let X = X + − X −. Then, + − E[X ] = E[X − X ] + − = E[X ] − E[X ] Z 1 Z 0 = (1 − FX (t)) dt − FX (t)dt: 0 −∞ c Stanley Chan 2020. All Rights Reserved. 14 / 16 Summary We have learned three things: Median 50% of the area, from left and from right. Mode Peak of PDF, steepest slope of CDF Mean R 1 PDF: 0 tfX (t)dt, for X > 0 R 1 CDF: 0 (1 − FX (t)) dt, for X > 0 c Stanley Chan 2020. All Rights Reserved. 15 / 16 Questions? c Stanley Chan 2020. All Rights Reserved. 16 / 16.