dz asked:
> Hi, anybody knows how to caculate the variance of x/y? where x and y are
two
> independent variables with normal dis n(a1,b1) and
> n(a2,b2) respectively.
Yes. And no.
Sitting down?
Trick question! It is always undefined, as is the mean! Moreover,this
holds for any continuous distributions st the density of the denominator
function at y=0 is nonzero.
Imagine just the region y = (- epsilon, epsilon). Let A be the minimum
of the density of Y over that interval; for small enough epsilon this is
>0.
Now, the mean, E(X/Y), is defined to be \integral \integral x/y f_x f_y
dx dy evaluated over the whole (x,y) plane. But if you try to evaluate this
over regions
(0,infinity) x (epsilon/2, epsilon)
you get \integral x f_x dx \integral from epsilon/2 to epsilon 1/y
f(y) dy.
The first factor is always positive (or 0 iff X is always negative); the
second
is bounded below by A (ln(epsilon) -ln(epsilon/2)) = A ln 2.
So each strip from epsilon to epsilon/2, epsilon/2 to epsilon/4, etc,
makes a contribution greater than A ln 2 \integral x f_x dx; thus the
integral diverges.
If X is always negative, the same argument holds, but integrating over
(-infinity,0).
Now, this seems daft. Intuitively, if the heights of husbands and wives
were independent (they aren't, but let that pass - independence is not
really the issue here) and normally distributed, you would sureluy expect
the mean of the ratio to exist? Yes, but. If they were truly normally
distributed, even in the far tails, you would expect one person in a few
quadrillion or something to have negative height, or a height of a few
microns. It is those hypothetical microscopic people who would provide a
scattering of astronomically large ratios.
So, what do you do? Well, there are limiting techniques such as Cauchy
principle values, but that is a highly abstract and formal technique that
you had better understand _very_ well before trying it on a real world
problem. Alternatively, you will find that if the mean is several standard
deviations away from 0, you can trim anywhere from 1% to (say) 0.00001%
of the extreme values from your distributions and the trimmed means you get
will not vary within that range. That "plateau" before the weirdness starts
is a good practical value to use instead. For any reasonable-sized sample,
the sampling distribution of the mean will gather around that value.
If the mean of Y is close to 0, you cannot do that - you have something
like a Cauchy distribution that is genuinely heavy-tailed even for practical
purposes. In a case like that, the sampling distribution of the mean of a
small sample will not converge to anything, and you had better use the
median instead. The question is: is Y~0 a definite possibility for the sort
of sample sizes you envisage?
-Robert Dawson
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