6.041/6.43112.Iterated Expectations;Sum of a random number of r.v

LECTURE 12 Conditional expectations • Readings: Section 4.3; • Given the value y of a r.v. Y : parts of Section 4.5 E[X | Y = y]= ! xpXY (x | y)(mean and variance only; no transforms) x |(integral in continuous case) Lecture outline • Stick example: stick of length ! break at uniformly chosen point Y • Conditional expectation break again at uniformly chosen point X – Law of iterated expectations E[X | Y = y]= y (number) • 2 – Law of total variance Sum of a random number • Y of independent r.v.’s E[X | Y ]= 2 (r.v.) – mean, variance • Law of iterated expectations: E[E[X | Y ]] = ! E[X | Y = y]pY (y)= E[X] y • In stick example: E[X]= E[E[X | Y ]] = E[Y/2] = !/4 var(X | Y ) and its expectation Section means and variances var(X | Y = y)= E "(X − E[X | Y = y])2 | Y = y # Two sections: • y = 1 (10 students); y = 2 (20 students) var(X | Y ): a r.v. • 1030with value var(X | Y = y) when Y = yy =1: 1 ! xi = 90 y =2: 1 ! xi = 60 10 i=1 20 i=11 Law of total variance: • var(X)= E[var(X | Y )] + var(E[X | Y ]) 1 3090 10+60 20 E[X]= ! xi = ·· = 70 30 i=1 30 Proof: E[X | Y = 1] = 90, E[X | Y = 2] = 60 (a) Recall: var(X)= E[X2] − (E[X])2 $%90, w.p. 1/3 (b) var(X | Y )= E[X2 | Y ] − (E[X | Y ])2 E[X | Y ]= &60, w.p. 2/3 E[E[X | Y ]]= 1 90+ 2 60=70= E[X]3 · 3 · (c) E[var(X | Y )] = E[X2] − E[(E[X | Y ])2 ] (d) var(E[X | Y ]) = E[(E[X | Y ])2 ]−(E[X])2 var(E[X | Y ]) = 1(90 − 70)2 + 2(60 − 70)2 33600 Sum of right-hand sides of (c), (d): = 3 = 200 E[X2] − (E[X])2 = var(X) Section means and variances (ctd.) 1 10 10! i=1 (xi−90)2 = 10 1 20 30! i=11 (xi−60)2 = 20 var(X | Y = 1) = 10 var(X | Y = 2) = 20 var(X | Y )= $% & 10, w.p. 1/3 20, w.p. 2/3 E[var(X | Y )] = 1 3 ·10 + 2 3 ·20 = 50 3 var(X)= E[var(X | Y )] + var(E[X | Y ]) = 50 3 + 200 = (average variability within sections) + (variability between sections) Example var(X)= E[var(X | Y )] + var(E[X | Y ]) x1 fX(x) 2 1/3 2/3 Y=1 Y=2 E[X | Y = 1] = E[X | Y = 2] = var(X | Y = 1) = var(X | Y = 2) = E[X]= var(E[X | Y ]) = Sum of a random number of independent r.v.’s • N: number of stores visited (N is a nonnegative integer r.v.) • Xi: money spent in store i – Xi assumed i.i.d. – independent of N • Let Y = X1 + ···+ XN E[Y |N = n]= E[X1 + X2 + ···+ Xn |N = n] = E[X1 + X2 + ···+ Xn] = E[X1]+ E[X2]+ ···+ E[Xn] = n E[X] • E[Y | N]= N E[X] E[Y ] = E[E[Y | N ]] = E[N E[X]] = E[N] E[X] Variance of sum of a random number of independent r.v.’s • var(Y )= E[var(Y |N)] + var(E[Y |N]) • E[Y | N]= N E[X] var(E[Y |N]) = (E[X])2 var(N) • var(Y |N = n)= n var(X) var(Y |N)= N var(X) E[var(Y |N)] = E[N] var(X) var(Y ) = E[var(Y |N)] + var(E[Y |N]) = E[N] var(X)+(E[X])2 var(N) 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.