6.041/6.431 5.Discrete random variables; probability mass functions

LECTURE 5 Random variables • Readings: Sections 2.1-2.3, start 2.4 • An assignment of a value (number) to every possible outcome Lecture outline • Mathematically: A function • Random variables from the sample space ! to the real numbers • Probability mass function (PMF) – discrete or continuous values • Expectation Can have several random variables Variance • • defined on the same sample space Notation:• – random variable X – numerical value x Probability mass function (PMF) How to compute a PMF pX(x) – collect all possible outcomes for which • (“probability law”, X is equal to x “probability distribution” of X) – add their probabilities – repeat for all x Notation:• • Example: Two independent rools of a pX(x)= P(X = x) fair tetrahedral die= P({! ! ! s.t.X(!)= x})F: outcome of first throwpX(x) " 0 !xpX(x)=1 S: outcome of second throw• X = min(F,S) • Example: X=number of coin tosses until first head 4 – assume independent tosses, 3 P(H)= p> 0 S = Second roll2pX(k)= P(X = k) = P(TT TH) 1 ···= (1 − p)k−1 p, k =1,2,... 1 2 3 4 F = First roll – geometric PMF pX(2) = Binomial PMF • X: number of heads in n independent coin tosses • P(H)= p • Let n =4 pX(2) = P(HHT T )+ P(HT HT )+ P(HT T H) +P(T HHT )+ P(T HT H)+ P(T T HH) =6p 2(1 − p)2 = "4 2 #p 2(1 − p)2 In general: pX(k)= "n k # p k(1−p)n−k , k =0, 1, . . . , n Expectation • Definition: E[X]= $ x xpX(x) • Interpretations: – Center of gravity of PMF – Average in large number of repetitions of the experiment (to be substantiated later in this course) • Example: Uniform on 0, 1, . . . , n 1 x pX(x ) . . . 0 n-1 n 1/(n+1) E[X]=0× 1 n +1 +1× 1 n +1 +···+n× 1 n +1 = Properties of expectations • Let X be a r.v. and let Y = g(X) – Hard: E[Y ]= $ y ypY (y) – Easy: E[Y ]= $ x g(x)pX(x) • Caution: In general, E[g(X)] %= g(E[X]) Properties: If ", # are constants, then: • E["]= • E["X]= • E["X + #]= Variance Recall: E[g(X)] = $ x g(x)pX(x) • Second moment: E[X2]= !x x2pX(x) • Variance var(X)= E %(X − E[X])2& = $ x (x − E[X])2 pX(x) = E[X2] − (E[X])2 Properties: • var(X) " 0 • var("X + #)= "2var(X) MIT OpenCourseWare http://ocw.mit.edu 6.041 /6.431 Probabilistic Systems Analysis and Applied Probability Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.