Understanding Covariance and Correlation in Multivariate Statistics, Exams of Introduction to Econometrics

Download Understanding Covariance and Correlation in Multivariate Statistics and more Exams Introduction to Econometrics in PDF only on Docsity! Lecture 4: A Review of Probability and Statistics, Beyond the Univariate Case Wooldridge Appendix B&C Review from Last Time  Con…dence Interval Example –Needs to be a 2 tailed test to make hypothesis testing mirror con…dence intervals – In order to reject the null hypothesis  Sample size n; pop var 2; sample mean X  Hypothesis testing: checking against a critical value: X0p n  z 2 or X0p n  z 2  CI: use critical value to construct CI: X  z 2  p n and see if 0 falls outside the CI
For 95% of the random samples we pull, the constructed con…dence interval will contain .  Type I and Type II Errors – In choosing the Type I error to target, its often a matter of consequence/severity –Example from class (people die vs people get too much treatment)  H0 : cancer; HA : no cancer  Type I Error: probability that you say someone doesn’t have cancer when they in fact do, so they don’t get treated  That’s more important than treating people that don’t have cancer –By eliminating the prob of type I error (making it unlikely that the hypothesis is rejected) we must decrease the power of our testing (increase type II error) –Usually we can’t estimate Type II errors because they depend on HA which is usually not well de…ned –Ways to increase power (deal w Type II errors):  Increase sample size  More e¢ cient measure to get smaller  Joint Distribution  We are now interested in looking at two (or more) random variables . Often ask important questions that rely on more than 1 factor.  If we have two discrete random variables X and Y , then the joint distribution is described by the joint probability density function (pdf) of X;Y : fX;Y (x; y) = P (X = x; Y = y). This can easily translate to more than 2 variables.  Show what a marginal pdf is....in discrete case, marginal pdf of x : fX(x) = P y f(x; y)::: sum over all value of Y. Similar for marginal pdf of y.  Example 1.4 –E(X) = P xif(xi) –E(X) = 2( 19 ) + 3( 2 9 ) + 4( 3 9 ) + 5( 2 9 ) + 6( 1 9 ) = 2+6+12+10+6 9 = 36 9 = 4 –E(Y ) = 0( 39 ) + 1( 4 9 ) + 2( 2 9 ) = 8 9 1 Independence  Why is independence important? Knowing the outcomes of X does not change the probabilities of possible outcomes of Y  Random variables (discrete and continuous) X and Y are independent i¤ the joint pdf is the product of the marginal pdfs: fX;Y (x; y) = fX(x)fY (y) – In the discrete case (to match what we wrote as the joint distribution): fX;Y (x; y) = P (X = x)P (Y = y).  Once we establish independence, we can prove this useful conclusion – If X and Y are independent, then E(XY ) = E(X)E(Y ) – Show by using discrete random variables: E(XY ) = P x P y xyfX;Y (x; y) = P x P y xyfX(x)fY (y) =P x xfX(x) P y yfY (y) = E(X)E(Y ) Covariance  Need summary measures of how variables vary with one another...we are looking to use a single number (as we did with expected value and variance) to summarize something about an entire distribution  Population covariance: describes relationship between two variables involving a population – Set up as follows: E(X) = X and E(Y ) = Y –De…ned as the expected value of the product (X X)(Y Y ). [Why does this make sense? If X is above mean, Y above mean yield positive]  Cov(X;Y ) = XY = E[(XX)(Y Y )] = E[(XX)Y (XX)Y ] = E[(XX)Y ] E[(XX)Y ] = E[(XX)Y ]Y [E(X X)| {z } =0 ] = z }| { E[(X X)Y ] = E[XY ] XE(Y ) = E(XY ) XY  Go through PS 1.1b...show similarities to the above
NP t=1 (XtX)(YtY ) = P (XtX)Yt P (XtX)Y = P (XtX)YtY X (Xt X)| {z } =0 from part A = P (Xt X)Yt
Part A: P (Xt X) = P Xt nX = P Xt n( 1n P Xt) = 0 – Important thing to note: be aware of measurements! Covariance can be altered simply by multiplying a rv by a constant. Property: Cov(a1X + b1; a2Y + b2) = a1a2Cov(X;Y ) –Use PS 1.4 as an example of how to calculate covariance  Cov(X;Y ) = E(XY ) E(X)E(Y )  From before we know: (X) = 4; E(Y ) = 89  E(XY ) = (3  1)( 29 ) + (4  2) 2 9 + (5  1)( 2 9 ) = 6+16+10 9 = 32 9  Cov(X;Y ) = 329 4( 8 9 ) = 0  How does covariance relate to independence? –PS 1.3: If X;Y are independent rv, then Cov(X;Y ) = 0  Assume discrete random variables for the …rst case 2