A joint density function is given by
f(x,y) ={kx, for 0<x<1, 0<y<1; 0, otherwise}
where k is a constant. What is Cov (X,Y)?
Solution: 0
ASM Study Manual by K. Ostaszewski offers this explanation:
The joint density is a product of functions of x, fX(x)=k(sub 1)*x, and y, fY(y)=k(sub 2)*1, where k(sub 1)*k(sub 2) = k, on a rectangular region. Hence, X and Y are independent, and Cov (X,Y) = 0.
Is it a fair assumption that if a joint equation has a rectangular boundary, in this case (0-1 on both x and y) and there is only one variable listed, (x in this case), that this joint distro is uniform and therefore independent? Can I make a snap evaluation and say, "Okay, only one variable in the pdf; ignore the constant k; rectangle limits; This is independent, no COV"?
f(x,y) ={kx, for 0<x<1, 0<y<1; 0, otherwise}
where k is a constant. What is Cov (X,Y)?
Solution: 0
ASM Study Manual by K. Ostaszewski offers this explanation:
The joint density is a product of functions of x, fX(x)=k(sub 1)*x, and y, fY(y)=k(sub 2)*1, where k(sub 1)*k(sub 2) = k, on a rectangular region. Hence, X and Y are independent, and Cov (X,Y) = 0.
Is it a fair assumption that if a joint equation has a rectangular boundary, in this case (0-1 on both x and y) and there is only one variable listed, (x in this case), that this joint distro is uniform and therefore independent? Can I make a snap evaluation and say, "Okay, only one variable in the pdf; ignore the constant k; rectangle limits; This is independent, no COV"?
Sample Problem # 7 May 2001