Question

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable-X for the right tire and Y for the left tire, with joint pdf given below. f(x, y) = R(x2 + y2) " 20 SX 530, 20 Sy s 30 O otherwise (a) Compute the covariance between X and Y. (Round your answer to four decimal places.) Cov(x, y) = (b) Compute the correlation coefficient p for this X and Y. (Round your answer to four decimal places.)

Answer 1

For a valid pdf following must be true:

..................(1)

30 30 f(z,y)dyd.r = 1 -J20 J20

Now

30 30 30 730 1 f(,y)dydx = k(1? + y dydx = J20 J20 J20 J20 20

203 303 k 1 30.72 + "+ -20.22 - 90.2 = 1 J 20 k 10.12 + + 19000 L dr = 20

...........(2)

10.30 19000.30 10.20 19000.20] | 380000] = 十 3

Equating equation (1) and (2) gives

k = 3/380000

------------------------------------------

(a)

UG D J - Topp (48+44 I J »-(4x)a

= k | 25023 + 162500.x) dx = k [62.524 + 81250.22) = 81250000k = 641.447420

The marginal pdf of X is

30 fx(x) = 460) [ 15.do = * [*e*o%iks = [2o + *D* -- [un 4 y 2012 f(x, y)dy = k / (x2+y?)dy = k ’y+ = k 303 30.02 + 80,72 3 - 20. - 2002

The mean of X is

30 30 E(X) = xfx(c) dx = k 3208333.3333k = 25.3289 Eix) – L'astek I[van 4 bogat) do a los usuan 1 1023_190007 102 + = 10.04 9500.r21 43 J20 J20 J20 L

By symmetry we have

$E(Y)=3208333.3333k=25.3289$

So covariance of X and Y is

Cou(X,Y)= E(XY)-E(X)E(Y) = 641.4474–25.32892 = -0.10577521 ~ -0.1058

(b)

Now

Therefore

730 19000.cº [10.5 1900023710 dr =k + E(X) = fx () dr =k 3208333.3333k = 649.8246 9 120

var (X) = 649.8246 - 25.328928.2714

By symmetry

var(Y) = 8.2714

So

p=- Cou(X,Y) Var(X)Var(Y) -0.0128