Correlation and cov doubt

[The Villain]

Suppose that the pair (X, Y ) is uniformly distributed on
the interior of a circle of radius 1. Compute ρ(X, Y).

[Dreyfus]

Ron .....I m getting E(x) = E(Y)=0 and Var(X)=Var(Y)=1/3
But I m not able to get the limits of integral for computing cov(X,Y)

[Akshay Jain]

since all the points X,Y lie in a circle with equal probability there is no linear relationship b/w dem...so correlation must be 0

[kangkan]

@vaibhav..what is the density function for this?

[Akshay Jain]

the density function will be f(X,Y)=1/pi for x^2 +y^2<1

[Dreyfus]

Akshay can u plz tell how to proceed for cov(x,y) for more formal proof....

[kangkan]

thanks akshay..i have never seen a bivariate uniform distribution before :)

[Akshay Jain]

let me confirm my ans 1st......@ron do u no the ans???

[The Villain]

Yeah ans is indeed 0.Bt i couldnt understand the explaination.

[Dreyfus]

Ron the location of circle will only change the scale or origin or both but in any of the case r will remain independent of change in origin and scale, that's y circle is assumed to be centered at origin to make calculations easier...

[The Villain]

Acha i missed that pt.Got it now...thanxx Vaibhav :))

[Akshay Jain]

ok so the formal approach is quite heavy in algebra.....vl try to explain
covariace is E(X*Y)-E(X)*E(Y)
E(X*Y)=integration(-1,1).integration(y= - underoot(1-x^2), y=underoot(1-x^2)(x*y)*1/pi.dy.dx
or
E(X*Y)=integration(-1,1).integration(x= - underoot(1-y^2), x=underoot(1-y^2)(x*y)*1/pi.dx.dy
u vl get the same result
and
E(X)=0=E(Y)
the logic lies in the domain of the PDF wich is nthing but a circle. so we have to use limits (derived from eqn of circle x^2+y^2=1) y= - underoot(1-x^2), y=underoot(1-x^2) as limits to convert the inner integral into a function of x alone and then the outside integral will give values to x which ranges from -1 to 1

[Dreyfus]

Thanks Akshay......

[Akshay Jain]

its very difficult to explain maths in writing.....

[The Villain]

Thanxxx  akshay really appreciate :))