Question

2. Properties of Correlation and Covariance: Two random variables Y and Z are represented by the following relationships Y = 0.5+0.6X Z = 0.2+0.3x where X is another random variable. You can treat the variance, Var(X), as a given constant. It may help to give Var(X) a name, ie. Var(x)ox2 a. Calcuate var(Y) and Var(Z) as a function of Var(X). Which is hrger? b. Calcuate Cov(Y,Z), Cov(X,Z) and Cov(X,Y) as a function of var(X). c. Calcuate Corr(Y,Z), Corr(X,Z) and Corn(X,Y) (as numbers. Explain the results intuitively.

Answer 1

TOPIC:Variance,covariance and correlation coefficients between random variables.

results as - the following 7 At first, consider ; where, 'a' and 14 Id, y = a + bx b are constants i then, var (y) = 6², var (x). Proof - vor (Y). = var (a + bx). = E[a+bx ² - [(a+bx)? = E(2²+ 2ab + b 2 - (a +6E(47) = o(aka + 2ab EGY-EL +62 [ (x²) – E²(x)), = b? Varcx). (enovely 2. It, y = a tbx, 2 = c+dx. Then, Cor (1, 2) a bd. var (x). Proof Here, Con (7,2).

- E (72) - E(y). E (2) . = [(a+bx) (c + dx] - E[a+bx). E CC + 2x). [ac + adx + be * + bd x2 -Cato bec) (c+den = (acht a) + ad (E(XT- ESA) + bc. (E CA) -ECKT) + bd (E (x4 - E*C*). = bd var (x). (proved)

and 2 au 2x Define the mandom variables X, Y S 7 20.5 +0.68 land, 2 = 0.2 + 0.3x: Given that var (x) = 52. @ we need var (Y). = var (0.5 +0,6x). Resultat = (0.6) ². var (x), = 0.36 ?. and var (2). avar (0.2 +0.3 x). = (0,373 var. (x) = (0.09) or 2 [By Result-? .

so clearly, I var (4) y van (2). (Oo, o ad van(y) is larger than var (Z). [onse 6 we need, Cov (7, 2) , - Cov (0.5 +0. 6x 0,2 +0.3x). = (0,2).(0.9. var (X). [ By Resust -2). and Cor (x, 2). = cav (x, 0.2 +0.3 X). - L. (0.2) is a = Cov (2,0 đ) + Cov (x, O 3 ). t 0.3. var (x). Cor (x, ax), =aivar (x). constant

and cow (x, y). = Con (x, 0.5 +0.6x). Con (X, 0,5 at Cor (x, 0.6x). = 0 + 0.6. Var (*) [ Cor (x, 0.6x). I (0.6) war (x)] t. Tanss Como (y, z) = con (y, z) Ivan (Y). var (2) (0.185 Vlo. 36) $43.{10.09.17 0.18 0:18 0 2 Conn (x, 2). Cov (1, 2). Ivar (x). Van (2) (0i 3) oc? (0.30

and coor (x, y). CON (x, y) var (xl. var(y) (0.6) 8 2 J?. (0.36) 70.6704? (0.6 & 2 - Explanation in a observe that all the pairs (1, 2), (x, y) and (x, z) are associated with a linear relationship among each other, So, since the relationships between the variables can be fully explained by the linear relationships, the linear connelation coefficient becomes t. [ons

(Also,the slope of the equations , Y=0.5+0.6X and Z=0.2+0.3X are positive. Hence,there is a perfectly linear and positive correlation between the variables and hence the correlation coefficients become 1).