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.
Howard Hoffman responded:
> 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.
"When I use a word" said Humpty Dumpty, "it means what I want it to
mean.
It all depends on who's going to be master, you or the word."
The fact is that this is not the accepted meaning of independence. The
accepted meaning of independence is that the conditional probability
distribution of U does not depend on V. This can be rephrased usefully as:
"there is no value of U that, if observed, would tell you anything about the
value of V".
As (this may have been a homework question but I presume it's past the
due date by now, so I can be explicit) U=12 implies V=0 whereas U=11 makes
this impossible, U and V are not independent. [End of proof.]
Independence is actually quite hard to see from a scatterplot, as it
is often hard to determine by eye if two conditional samples {U_i:V_i in
(a1,a2)}
and {U_i:V_i in (b1,b2)} have similar distributions or not when the numbers
of
data points differ significantly. For this purpose, I like to use an array
of
side-by-side boxplots of the dependent variable, one for each interval of
the
independent variable.
Another frequently-confused-with-independence property of joint
distributions is that the two marginal distributions are causally
unrelated. This implies independence but does not follow from it (the
canonical example involves the toss of two coins, with the events being
A: heads on cent, B:exactly one head.)
We have: causally independent => independent => regression slope = 0.
-Robert Dawson
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