Mathematical Expectation Spiegel et al (2000) - Chapter 3

# Mathematical Expectation Spiegel et al (2000) - Chapter 3

## Mathematical Expectation Spiegel et al (2000) - Chapter 3

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1. Mathematical ExpectationSpiegel et al (2000) - Chapter 3 Examples by Mansoor Al-Harthy Maria Sanchez Sara Russell DP Kar Alex Lombardia Presented by Professor Carol Dahl

2. Introduction • Green Power Co. investment • solar insolation (X) • wind speed (W) Electricity • hybrid (X,W) • uncertainty • characterize • discrete random variable (X) • probability function p(x) • continuous random variable (W) • probability density function f(w)

3. Mathematical Expectation • Expected Value • Functions of Random Variables • Some Theorems on Expectation • The Variance and Standard Deviation • Some Theorems on Variance • Standardized Random Variables • Moments • Variances and Covariance for Joint Distributions • Correlation Coefficient

4. Mathematical Expectation Conditional Expectation & Variance Chebyshev's Inequality Law of Large Numbers Other Measures of Central Tendency Percentiles Other Measures of Dispersion Skewness and Kurtosis

5. Random Variables • value with a probability attached • value never predicted with certainty • not deterministic • probabilistic

6. Solar Insolation (m/s)

7. Mathematical ExpectationDiscrete Case • X is solar insolation in W/ft2 • Want to know averages

8. Mathematical ExpectationDiscrete Case • 3,4,5,6,7 with equal probability • from expectation theory: • n events with equal probability P(X)= 1/n, • Discrete Random Variable X • x1, x2, . . ., xn • x = E(X) = xj*(1/n) = xj/n • = 3*(1/5)+4*(1/5)+5*(1/5)+?*1/5 + ?*? = 5

9. Mathematical ExpectationDiscrete Case • Don't have to be equal probability X P(X) • 3 1/6 • 4 1/6 • 5 1/3 • 6 1/6 • 7 1/6 = 3*(1/6) + 4*1/6+5*(1/3) + ?*1/6 + ?*? = 5

10. Mathematical ExpectationDiscrete Case • don't have to be symmetric • P(X) may be a function • P(Xi) = i/15 = 3*(1/15) + 4*(2/15) + 5*3/15 + 6*? + ?*? = 85/15 = 5.67

11. Mathematical ExpectationContinuous Case • W represents wind a continuous variable • want to know average speed • from expectation theory: • continuous random variable W ~ f(w)

12. Mathematical Expectation Continuous Case • Meteorologist has given us pdf • f(w)=w/50 () <w <10 m/s  = w = 20/3 m/s = 6.67

13. Functions of Random Variables • Electricity from solar X ~ P(X) photovoltaics - 15% efficient Y = 0.15X E(Y)=?

14. Functions of Random Variables • X= {3, 4, 5, 6, 7} • P(xi) = i/15 =0.15* 3*(1/15) + 0.15*4*(2/15) + 0.15*5*3/15 +0.15*6*? + ?*?*? = 0.85 units W/ft2

15. Linear Functions of Random Variables E(g(X)) = 0.15* 3*(1/15) + 0.15*4*(2/15) + 0.15*5*(3/15) + 0.15*6*(4/15) + 0.15*7*(5/15) = 0.15*(3*1/15 + 4*2/15 + 5*3/15 + 6*4/15 + 7*5/15)

16. Mean of Functions of Random Variables Continuous Case - Electricity from wind y = -800 + 200w w>2 with y measured in Watts fix diagram

17. Mean of Functions of Random Variables Continuous Case • w continuous random variable ~ f(w) • f(w)=w/50 0<w <10 m/s  • g(w) = -800 + 200w w>2

18. Mean of Functions of Random Variables Continuous Case • w continuous random variable ~ f(w) • f(w)=w/50 () <w <10 m/s  • y = g(w) = -800 + 200w w>2

19. Mean of Functions of Random Variables Continuous Case = -800*10 + 200*102/2 - (-800*2+200*22) = 533.33

20. Mean of Linear Functions of Random Variables - Continuous Case g(w) = a + bw = ?

21. Functions of Random Variables Hybrid Electricity generation from wind and solar W, S ~ ws/(1600) 0<W<10, 0<S<8 Check it’s a valid pdf

22. Functions of Random Variables Hybrid Electricity generation from wind and solar W, S ~ ws/(1600) 0<W<10, 0<S<8 Electricity generated E = g(w,s) = w2/2 + s2/4 E(E) = E(g(w,s)) = 08 010 g(w,s)f(w,s)dwds

23. Functions of Random Variables work out this integral

24. SUMMARY OF DEFINITIONSExpected Value

25. Some Theorems on Expectation • 1. If c is any constant, then • E(cX) = c*E(X) • Example: • f (x,y) = { xy/96 0<x<4 , 1<y<5 0 otherwise 4 5 4 5 E(x) = ∫ ∫ x f(x,y) dx dy = ∫ ∫ x (xy/96) dx dy = 8/3 x=0 y=1 x=0 y=1 4 5 E(2x) = ∫ ∫ 2x f(x,y) dx dy = 16/3 = 2 E(x) x=0 y=1

26. Some Theorems on Expectation • 2. X and Y any random variables, then  • E(X+Y) = E(X) + E(Y) • Example: • E(y) = ∫ ∫ y f(x,y) dx dy = 31/9 • E(2x+3y) = ∫ ∫ (2x+3y) f(x,y) dx dy • = ∫ ∫ (2x+3y) (xy/96) dx dy = 47/3 • E(2x+3y) = 2 E(x) + 3 E(y) = 2*(8/3)+ 3*(31/9) = 47/3 4 5 x=0 y=1 equivalent

27. Some Theorems on Expectation • Generalize • E(c1*X+ c2*Y) = c1*E(X) + c2* E(Y) • added after check • add simple numerical mineral economic example here • 3. If X & Y are independent variables, then • E(X*Y) = E(X) * E(Y) • add simple numerical mineral economic example here

28. Some Theorems on Expectation slide 3-29

29. Some Theorems on Expectation slide 3-29

30. Some Theorems on Expectation slide 3-29 • 2E(X) + 3E(Y) = 2*18 + 3*12 = 36 + 36 = 72 • So, • E(2X+3Y) = 2E(X) + 3E(Y) = 72

31. Some Theorems on Expectation slide 3-29 • If X & Y are independent variables, then: • E(X*Y) = E(X) * E(Y)

32. Some Theorems on Expectation slide 3-29 • E(X) * E(Y) = 18 * 12 = 216 • so, E(X*Y) = E(X) * E(Y) = 216

33. Variance and Standard Deviation • variance measures dispersion or risk of X distributed f(x) • Defn: Var(X)= 2 = E[(X-)2] • Where  is the mean of the random variable X •  X discrete • X continuous

34. The Variance and Standard Deviation • Standard Deviation

35. Discrete Example Expected Value • An example: Net pay thickness of a reservoir • X1 = 120 ft with Probability of 5% • X2 = 200 ft with Probability of 92% • X3 = 100 ft with Probability of 3% • Expected Value = E(X)=xP(x) • = 120*0.05 + 200*0.92 + 100*0.03 = 193.

36. Discrete Example of Variance and Standard Deviation • An example: Net pay thickness of a reservoir • X1 = 120 ft with Probability of 5% • X2 = 200 ft with Probability of 92% • X3 = 100 ft with Probability of 3% • Variance = E[(X-)2] =

37. Variance and Standard Deviation •  Standard deviation • = (571)0.5 = 23.896

38. Continuous Example Variance and Standard Deviation • add continuous mineral econ example of mean and variance

39. Variance and Standard Deviation Theorems • \$1,000,000 investment fund available, • Risk of Solar Plant Var(X)=69, • Risk of Wind Plant Var(Y)=61, • X and Y independent • Cov(X, Y)= XY = E[(X-X)(Y-Y)]=0 • Expected Risk, if 50% fund in plant 1? • 1. Var(cX) = c2Var(X) • So, as c= 0.5 • Var(cX) = 0. 52Var(x) = 0.52*69 = 17.25

40. Variance and Standard Deviation Theorems 2. Var(X+Y) = Var(X) + Var(Y) An example: Var(X+Y)= 61+ 69= 130 3. Var(0.5X+0.5*Y)= 0.52*Var(X) +0.52*Var(Y) = 0.25* 69 + 0.25*61 = 32.5 Later we will generalize to non-independent4. Var(c1X+ c2Y) = c12Var(X) + c22Var(Y) +2 c1c2Cov(X,Y)

41. Summary Theory Expectations & Variances Expectations Variances E(cX) =c E(X) Var(cX) = c2Var(X) Independent Var(X+Y) = Var(X) + Var(Y) E(X+Y)= E(X)+ E(Y) Independent Var(X-Y) = Var(X) +Var(Y) E(X-Y)= E(X)- E(Y) Var(c1X+ c2Y) = c12Var(X) + c22Var(Y)+2c1c2Cov(X,Y) E(c1*X+ c2*Y) = c1*E(X) + c2*E(Y)

42. Standardized Random Variables • X random variable • mean  standard deviation . • Transform X to Z • standardized random variable

43. Standardized Random Variables • Z is dimensionless random variable with

44. Standardized Random Variables

45. Standardized Random Variables • X and Z often same distribution • shifted by  • scaled down by  • mean 0, variance 1

46. Standardized Random Variables • Example: • Suppose wind speed W ~ (5, 22) • Normalized wind speed

47. Measures of Variation WINDSOLAR Discrete r.v. Continuous r.v. X1 ~ p(x1) X2 ~ f(x2) Variation is a function of Xi 1st measure of variation • (Xi – µ)2 • g(x1) = (x1 - µ)2 g(x2) = (x2 - µ)2 E(g(x1)) = Σ g(x1)p(x1)