Robert Dawson wrote:
>
> 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
---------------------------------------------------------------------------
If I may add a few comments ---
william knight
[EMAIL PROTECTED]
(1) Since the ratio, X/Y, is wanted, I doubt that X and Y
are normally distributed. One seldom takes ratios
when X and Y may occasionally be negative (See Robert
Dawson's heights example.)
(2) Is that ratio really a good idea? Ratios can be nasty:
Even if Y cannot actually be zero, a small Y inflates the
ratio badly. Consider this data:
Case 1 2 3 4 5
X 1 5 1/4 5 1/5
Y 1 4 1/5 1/5 5
X/Y 1 1.25 0.8 25 0.04
I question the meaning of the average of these ratios
(about 6), let alone the meaning of the variance.
(3) I am puzzled why the original question didn't ask about
the expected ratio as well as the variance.
(4) Possible approaches
(4.1) Transform the data to logarithms. Means and variances
follow from usual linear formulae. I deduce there are no
zeroes since the original question concerned ratios which
would fail with division by zero.
(4.2) Look at X/(X+Y)
This avoids small denominator problems. The variance
(and average which, surprisingly was not asked about) can
be approximated by methods found under "ratio estimates"
in any textbook on the statistical methods of survey
sampling.
(5) It would help to know what X and Y are: Heights?
Number of Clintonia borealis plants in a quadrat?
X=length of fibula, Y=length of tibia?
Strength of treated vs untreated concrete specimens?
(None of these can be negative, and hence none of
these can be normal.) Indeed, it is dangerous to
answer the question without knowing this.
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