Statistics: the covarience?
how would i go about solving this problem?
Suppose that X and Y are random variables with joint density:
fX;Y (x; y) = 1 such that -y < x < y, where y lies in the set (0,1)
0 elsewhere
.
Show that Cov (X; Y ) = 0, but X and Y are not independent.
please could you show me this step by step? Thanks
Answers:
I'm not sure precisely which way you were taught to do this, but heres how I'd do it:
Cov(X,Y) = E[XY]-E[X]E[Y]. Now E[XY] = (integral from y=0 to y=1) of (integral from x=-y to x=y) of xy*1. That inner integral turns into (0.5x^2y) from x=-y to x=y; substitute those in and you get 0, so the whole thing becomes 0. Also, E[X] is obviously 0 (you could do that by integrating again, but its symmetric about the y axis), so the covariance is 0.
However, the variables are definitely not independent - if y<=0 then P(X>0.5)=0, but if y<=1 P(X>0.5) is bigger than 0.Should the BBC take the lead in demanding action on climate change?
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Suppose that X and Y are random variables with joint density:
fX;Y (x; y) = 1 such that -y < x < y, where y lies in the set (0,1)
0 elsewhere
.
Show that Cov (X; Y ) = 0, but X and Y are not independent.
please could you show me this step by step? Thanks
Answers:
I'm not sure precisely which way you were taught to do this, but heres how I'd do it:
Cov(X,Y) = E[XY]-E[X]E[Y]. Now E[XY] = (integral from y=0 to y=1) of (integral from x=-y to x=y) of xy*1. That inner integral turns into (0.5x^2y) from x=-y to x=y; substitute those in and you get 0, so the whole thing becomes 0. Also, E[X] is obviously 0 (you could do that by integrating again, but its symmetric about the y axis), so the covariance is 0.
However, the variables are definitely not independent - if y<=0 then P(X>0.5)=0, but if y<=1 P(X>0.5) is bigger than 0.