Introduction to Probability & Statistics Expectations

Expectations    Mean:   E X xdF x [ ] ( )   xp x discrete ( ) ,    E X xdF x [ ] ( )    xp x discrete ( ) ,      xf x dx continuous ( ) ,

Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6  = E[X] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 3.5

Expected Life   E [ x]  x  e dx   x e dx    ( ) 2  1  For a producted governed by an exponential life distribution, the expected life of the product is given by  2.0 E [ x]   x  e   x dx 1.8 1.6 1.4 1.2 f (x t )   e   x 1.0      x e dx 2 1 Density 0.8 0.6 0.4 0.2 0.0 X 0.5 1 1.5 2 2.5 3    ( ) 2 1/  1 

Variance      ( ) x dF     E x [( ) ] =     ( ) x p    2    ( ) x dF   2   E x [( ) ] =   2    ( ) x p   2      ( ) x f dx

Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 2 = E[(X-)2] = (1-3.5)2(1/6) + (2-3.5)2(1/6) + (3-3.5)2(1/6) + (4-3.5)2(1/6) + (5-3.5)2(1/6) + (6-3.5)2(1/6) = 2.92

Property        ( ) x f dx    ( ) x f dx     x f dx xf ( 2      ( ) x f dx     ( ) x f dx 2      x f dx xf 2 ( ) 

Property        ( ) x f dx    ( ) x f dx     x f dx xf ( 2      ( ) x f dx     ( ) x f dx 2      x f dx xf 2 ( )     E X [ ] 2    E X [ ] 2 

Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 2 = E[X2] - 2 = 12(1/6) + 22(1/6) + 32(1/6) + 42(1/6) + 52(1/6) + 62(1/6) - 3.52 = 91/6 - 3.52 = 2.92

Exponential Example       E X [ ]   1   x e dx ( )   x e For a producted governed by an exponential life distribution, the expected life of the product is given by   2   E X [ ] 2.0 1.8   2 1    x e dx ( ) 1.6 1.4 1.2 f (x t )   e   x 1.0 Density 0.8      x e dx 3 1 0.6  1 2  0.4 0.2 0.0 X 0.5 1 1.5 2 2.5 3    ( ) 3 1/ 1 2   1 2  =

Properties of Expectations 1. E[c] = c 2. E[aX + b] = aE[X] + b 3. 2(ax + b) = a22 4. E[g(x)] = g(x) E[g(x)] X (x-)2 e-tx g x dF ( ) 

Class Problem Total monthly production costs for a casting foundry are given by TC = $100,000 + $50X where X is the number of castings made during a particular month. Past data indicates that X is a random variable which is governed by the normal distribution with mean 10,000 and variance 500. What is the distribution governing Total Cost?

Class Problem Soln: TC = 100,000 + 50X is a linear transformation on a normal TC ~ Normal(mTC, s2TC)

Class Problem Using property E[ax+b] = aE[x]+b mTC = E[100,000 + 50X] = 100,000 + 50(10,000) = 600,000

Class Problem Using property s2(ax+b) = a2s2(x) s2TC = s2(100,000 + 50X) = 502 s2(X) = 502 (500) = 1,250,000

Class Problem TC = 100,000 + 50 X but, X ~ N(100,000 , 500) TC ~ N(600,000 , 1,250,000) ~ N(600000 , 1118)