# Econ7630_Exam1a

 .remove_background_ad { border: 1px solid #555555; padding: .75em; margin: .75em; background-color: #e7e7e7; } .rmbg_image { max-height: 80px; } PDF Facts ﻿﻿P[X = x] >= 0 Sum[f(x)] = 1 Random Variable A function that assigns a unique numerical value to each sample space outcome. CDF ﻿﻿ Given this discrete CDF, what is P[X<=1.5]? x f(x) 1 0.1 2 0.2 3 0.3 4 0.4 0.1 Joint PDF: f (X, Y) for discrete r.v. P[X = x, Y = y] Joint PDF: f (X, Y) for continous r.v. over a certain range. Marginal PDF: f (x) for discrete r.v. f(x) = sumy[f(x,y)] f(y) = sumx[f(x,y)] ﻿﻿Marginal PDF: f (X) for continous r.v. Conditional Probability: P[X = x | Y = y] (i.e. f(X|Y)) = P[X = x, Y = y]/P[Y = y] = f(x, y)/f(y) The conditional probabilty that X equals x given that Y equals y is equal to the joint PDF of X and Y divided by the marginal PDF of y. X, Y are independent IFF (1) f(x, y) = f(x)f(y) for all x, y. Or (derived from (1))(2) f(x|y) = f(x),(3) f(y|x) = f(y) for all x, y. Given that X, Y are independent show that f(x|y) = f(x) for all x, y. We know that (i) X,Y are independent if f(x,y) = f(x)f(y) for all x,y.(ii) f(x|y) = f(x,y)/f(y) So by (i) f(x|y) = f(x)f(y)/f(y) -> f(x|y) = f(x). Expected Value: E[X] for (1) Discrete (2) Continous What does Expected Value mean for a discrete random variable? The average value of X on an infinite number of experimental trials. For example, E[X] for a six sided fair die is 3.5. Clearly saying something like the likely value of a die role is 3.5 does not make sense. Average makes sense. Let g(X) be a function of X. What is E[g(X)]? ﻿ Remember: even though X is being transformed by g(X), you still use the original PDF f(x). If g(X) equals c, a constant, then E[g(X)] = E[c] = c Remember: The Expected Value of a constant is that constant. This allows you to pull constants out of the E[ ] operator. Show that E[c] = c, where c is a constant. Let g(X) = c -> E[g(X)] = integral[g(x)f(x)dx] = integral[cf(x)dx] = c*integral[f(x)dx] = c(1) = c If g(X) = aX + b, then E[g(X)] = E[aX + b] = aE[X] + b Show that E[aX + b] = aE[X] + b Let g(X) = aX + b -> E[g(X)] = E[aX + b] = E[aX] + E[b] -> aE[X] + b Let g(X) = g1(X) + g2(X) + . . . + gn(X), what is E[g(X)]? Remember: the expected value of the sum is the sum of the expected values. Var[X] = E[(X - mu)2] = E[X2] - mu2, where mu = E[X]. Proof:E[(X - mu)2] = E[(X - mu)(X - mu)] = E[X2 - 2muX + mu2] = E[X2] - 2muE[X] + mu2 = E[X2] - 2mu*mu + mu2 = E[X2] - 2mu2 + mu2 = E[X2] - mu2 Let Y = a + bX, what is Var[Y]? = b2Var[X] Proof:Let Y = a + bX -> E[Y] = a + bE[Y]. Then Var(Y) = Standardized Variable: z = (x - mu)/sigwhere sig = sqrt[Var(X)] (i.e. the standard deviation) Show that E[z] = 0 ﻿E[z] = E[(1/sig)X - mu/sig] = (1/sig)E[X] - E[mu/sig] = mu/sig - mu/sig = 0 Show that Var(z) = 1 ﻿Var(z) = Var[(1/sig)X - mu/sig] = (1/sig)2Var(X) = sig2/sig2 = 1 Remember: to get to line two, Var(aX +- b) = a2Var(X). Let g(X1, X2) have joint PDF f(X1, X2). E[g(X1, X2)] = ﻿ cov(X1, X2) ﻿= E[X1, X2] - mu1mu2Derivation:﻿ If cov(X,Y) 1. > 0 2. < 0 3. = 0 1. (X, Y) pairs tend to be both greater than their means or both less than their means.2. (X, Y) pairs tend to be mixed about their means (one greater and one less)3. (X, Y) pairs "evenly" spread about their means. Correlation ﻿ If p (for correleation) 1. = 1 2. = -1 3. = 0 1. perfect positive relation2. perfect negative relation3. no linear relationship Remember: absolute_value(p) measure the strength of the linear relationship. When X1 and X2 are independent, E[X1, X2] = mu1mu2 Proof: If X, Y are independent then cov(X, Y) = 0 Remember: the converse is not true. Proof:We know E[X, Y] = muXmuY, when X, Y ind. cov[X, Y] = E[X, Y]-muXmuY = muXmuY-muXmuY = 0 Let c, d be constants E[cX + dY] = cE[X] + dE[Y] = c*muX + d*muY Var(c1X1 + c2X2) c12Var(X1) + c22Var(X2) + 2c1c2Cov(X1, X2)Proof:Var(c1X1 + C2X2) E[X|Y=y] for continous = integral(x*f(x|y)dx)Note: f(x|y) = f(x,y)/f(y) E[a + bX | X] = a + b*mu E[g(X)|X] = E[g(X)] Law of Iterated Expectations E[Y] = Ex[Ey(Y|X)]Proof: If E[Y|X] = E[Y] then cov(X,Y) = 0Proof:﻿ Var(X|Y) = E[X2|Y] - E[X|Y]2 Normal Distribution Standard Normal Distribution Z~(0,1) ﻿ If X~N(mu,sig2) and Y=aX + b, then Y~ Y~N((a*mu +b), a2sig2)﻿ If X,Y are normal then cov(X,Y) cov(X,Y) = 0 <=> X,Y are independent. .remove_background_ad { border: 1px solid #555555; padding: .75em; margin: .75em; background-color: #e7e7e7; } .rmbg_image { max-height: 80px; } Authormattstam ID65858 Card SetEcon7630_Exam1a DescriptionExam 1 (cards part a) for graduate level Econometrics 1 at LSU for Spring 2011. Includes many not too terribly rigorous "proofs." Updated2011-02-13T18:32:12Z Show Answers