I stand corrected. I had not considered the requirement of homogeniety of
varience. Sure, if U=2 V must be 0. hence, for certain values of U , V can be
predicted exactly.
Howard S. Hoffman
Radford Neal wrote:
> Guidi Chan wrote:
>
> >> A fair die is rolled 2 times. X1 and X2 is the # of points showing on
> >> 1st and 2nd rolls.
> >>
> >> U = X1 + X2; V = X1 - X2.
> >>
> >> Show that U and V are NOT independent.
>
> In article <[EMAIL PROTECTED]>,
> Howard S. Hoffman <[EMAIL PROTECTED]> wrote:
>
> >If you make a scatterplot of all possible values of U and V you will
> >discover that for every value of U the mean value of V is 0. In other
> >words, the slope of the regression of U on V is zero. This, for me is proof
> >that U and V are independent.
>
> For U and V to be independent, it is not enough that the mean of V
> conditional on any particular value for U is always the same. It must
> be that the conditional distribution for V given any particular value
> for U is always the same. It is quite possible for the mean to be the
> same while the conditional distribution is different in other ways,
> eg, the standard deviation might be different.
>
> Radford Neal
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