Question

4. A random point (X, Y ) is chosen uniformly from within the unit disk in R2, {(x, y)|x2+y2< 1} (a) Let (R, O) denote the polar coordinates of the point (X,Y). Find the joint p.d.f. of R and . Compute the covariance between R and 0. Are R and e are independent? (b) Find E(XI{Y > 0}) and E(Y|{Y > 0}) (c) Compute the covariance between X and Y, Cov(X,Y). Are X and Y are independent?

Answer 1

Given the joint PDF

xY (x, y);x2 y2<1

a)The joint CDF of is

X R cos, Y = Rsin O

FRe (r,0)= fxy (r cose, rsine) rdedr 0 rdedr FRe (r,e) = TT e rdr 0 7T or2 0r<1,0 0< 2T 2T FRe (r,0) FRe (r, 0)

The joint PDF is

e e orae Re (r, 0) fR,e (r,0) R,e (r,);0 <r< 1,0 0 2

The marginal PDFs are

2T fRe (r,0) d fR (r)= 2T rde fR (r) 2r; 0<r<1 fR (r)

1 fe (0) fRe (r, 0) dr -dr fe () T 1 fe (0)= ;0 0 2 2T

WE see
R, (r,0)= fr(r)fe (0)
. Hence
R, 6
are independent.

b)The marginal PDF of . Dependent on , varies from $- \sqrt{1-X^2}$ to $\sqrt{1-X^2}$

xY (x, y);x2 y2<1

V1-x2 (x) -V1-x2 V1-x2 fxx (x, y) dy -(x) X J-V1-x2 dy T fx (x) 2/1- x2 ;-1 <x 1 =

Similarly,

$f_Y\left ( y \right )=\frac{2\sqrt{1-y^2}}{\pi } ;-1\leqslant y\leqslant 1$

. fy (y) dy P(Y 0) 0 2 /1-y -dy P(Y> 0)= TT P(Y 0) 1/2

Now the expected value,

$E(X|Y>0)=\frac{\int_{-1}^{1}xf_X\left ( x \right )dx}{P(Y>0)}\\ E(X|Y>0)=\frac{\int_{-1}^{1}x\frac{2\sqrt{1-x^2}}{\pi } dx}{P(Y>0)}\\ E(X|Y>0)=0$

Since $x\frac{2\sqrt{1-x^2}}{\pi }$ is an odd function.

$E(Y|Y>0)=\frac{\int_{0}^{1}yf_Y\left ( x \right )dy}{P(Y>0)}\\ E(Y|Y>0)=\frac{\int_{0}^{1}y\frac{2\sqrt{1-y^2}}{\pi } dx}{0.5}\\ E(X|Y>0)=\frac{\left ( 2/3\pi \right )}{0.5}\\ {\color{Blue} E(X|Y>0)=\frac{4}{3\pi }}$

c)

$E(XY)=\int \int _Rxyf_{X,Y}\left ( x,y \right )dydx\\ E(XY)=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac{xy}{\pi }dydx\\ E(XY)=0\\$

Similarly,

The covariance is

Cov(X, Y) E(XY) - E(X)E(Y) = 0

Since $f_{X,Y }\left (x,y\right )\neq f_{X}\left ( x \right )f_{Y }\left ( y \right )$ , X, Y are not independent.