Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITY◮A biased die gives probabilities 1 p, p, p, p, p, 2p of throwing 1, 2, 3, 4, 5, 6 respectively.2If the random variable X is the number shown on the die and the random variable Y isdefined as X 2 , calculate the covariance and correlation of X and Y .We have already calculated in subsections 30.2.1 and 30.5.4 thatp = 253, E[X] =13Using (30.135), we obtainNow E[X 3 ]isgivenby13 , E[ X 2] = 25313, V[X] =480169 .Cov[X,Y ]=Cov[X,X 2 ]=E[X 3 ] − E[X]E[X 2 ].E[X 3 ]=1 3 × 1 p 2 +(23 +3 3 +4 3 +5 3 )p +6 3 × 2p= 13132 p = 101,and the covariance of X and Y is given byCov[X,Y ] = 101 − 5313 × 25313 = 3660169 .The correlation is defined by Corr[X,Y ]=Cov[X,Y ]/σ X σ Y . The standard deviation ofY may be calculated from the definition of the variance. Letting µ Y = E[X 2 ]= 253 gives13σY 2 = p ( )1 2 2 ( )− µ Y + p 2 2 2 ( )− µ Y + p 3 2 2 ( )− µ Y + p 4 2 2− µ Y2+ p ( )5 2 2 ( )− µ Y +2p 6 2 2− µ Y187 356 28 824= p =169 169 .We deduce thatCorr[X,Y ]= 3660√ √169 169169 28 824 480 ≈ 0.984.Thus the random variables X and Y display a strong degree of positive correlation, as wewould expect. ◭We note that the covariance of X and Y occurs in various expressions. Forexample, if X and Y are not independent thenV [X + Y ]=E [ (X + Y ) 2] − (E[X + Y ]) 2= E [ X 2] +2E[XY ]+E [ Y 2] −{(E[X]) 2 +2E[X]E[Y ]+(E[Y ]) 2 }= V [X]+V [Y ]+2(E[XY ] − E[X]E[Y ])= V [X]+V [Y ]+2Cov[X,Y ].1202