Expectations Introduction to Probability & Statistics Expectations

Expectations Mean :   xpxxdiscrete x (),     xfxdxxcontinuous(),     EXxdFx[]()

Example Consider the discrete uniform die example: x 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

  [x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x   0 ft )e x (x   Density X

  [x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x   0      xedx x21 0 ft )e x (x   Density X

  [x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x   0      xedx x21 0   ()2 2  1 ft )e x (x   Density X

  [x] Expected Life For a producted governed by an exponential life distribution, the expected life of the product is given by Exedx x   0      xedx x21 0   ()2 2  1 ft )e x (x   Density X 1/

Variance   2    ()()xdFx  22   ()()xpx x  22     ()()xfxdx  22  Ex[()] =

Example Consider the discrete uniform die example: x 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  22     ()()xfxdx   ()()xxfx 22 2    xfx xfxdxfx 2 2 2()()() 

Property  22     ()()xfxdx   ()()xxfx 22 2    xfx xfxdxfx 2 2 2()()()   EXEX[][] 22 2 

Property  22     ()()xfxdx   ()()xxfx 22 2    xfx xfxdxfx 2 2 2()()()   EXEX[][] 22 2   EX[] 22 

Example Consider the discrete uniform die example: x p(x) 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6 1 / 6  2 = E[X 2 ] -  2 = 1 2 (1/6) (1/6) (1/6) (1/6) (1/6) (1/6) = 91/ = 2.92

Exponential Example For a producted governed by an exponential life distribution, the expected life of the product is given by ft )e x (x   Density X     xedx x ()  222  EX[] /

Exponential Example For a producted governed by an exponential life distribution, the expected life of the product is given by      xedx x31 0 ft )e x (x   Density X 1/     xedx x ()  222  EX[]  1 2

Exponential Example For a producted governed by an exponential life distribution, the expected life of the product is given by      xedx x31 0   ()3 3 ft )e x (x   Density X 1/     xedx x ()  222  EX[]  1 2  1 2 = 1 2

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

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

Property Derviation Prove the property: E[ax+b] = aE[x] + b

Property Derivation

 1

 1 = a  + b1 = aE[x] + b

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, X is a linear transformation on a normal TC ~ Normal(  TC,  2 TC )

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

Class Problem Using property  2 (ax+b) = a 2  2 (x)  2 TC =  2 (100, X) = 50 2  2 (X) = 50 2 (500) = 1,250,000

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