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Backstory: My research adviser and I are working on a project and we got to a nasty fraction that needed to be averaged (a time average of the function). He just went ahead and took the average of the numerator and divided it by the average of the denominator. This made the algebra a LOT easier and the results are about what we would expect, but we're kind of unsure about how we got there.

So, is there ever a case where the average of a quotient is the quotient of the averages? Or could this just be a one in a million coincidence that we got the correct results?

EDIT: What OP means is if we have two functions say $f$ and $g$, let $avg$ denote the average, then is it possible $$ avg \left( \frac{f}{g}\right) = \frac{avg(f)}{avg(g)}$$

Knight wants Loong back
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CSchulzke
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  • I, for one, am confused as to what you mean by the "average of a quotient." To me, a quotient is a single expression, and the average of it would simply be itself. – JMoravitz Jan 16 '18 at 21:24
  • I think it can mean expected value here. – Arnaud Mortier Jan 16 '18 at 21:26
  • It is certainly possible. If all the fractions are equal the calculation is correct. If the fractions are not all equal it might be true, but would probably not be. If all the fractions are close in magnitude the calculation is (almost certainly) incorrect, but it will be close. It might well be close enough that would wouldn't notice. – Ross Millikan Jan 16 '18 at 22:18
  • You are asking whether $$E\left(\frac XY\right)=\frac{E(X)}{E(Y)}$$ can hold, presumably for positive random variables $X$ and $Y$. Equivalently, this is asking whether $$E(UV)=E(U)E(V)$$ ever happens in nondegenerate cases. The answer is a resounding "yes", it holds true as soon as, for example, $U$ and $V$ are independent, that is, as soon as $Y$ and $\frac XY$ are independent. – Did Jan 16 '18 at 22:40

3 Answers3

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In general, there's no reason to expect this to be a valid manipulation. Consider that for tiny positive $\epsilon$, the average of $1/\epsilon$ and $1/1$ is dominated by the $1/\epsilon$ and thus tends to $\infty$ as $\epsilon \to 0$, while separately averaging the numerator and denominator gives us the fraction $1/(\frac{1}{2} + \frac{\epsilon}{2})$, which tends to $2$ as $\epsilon \to 0$.

Gregory J. Puleo
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If by average you mean expected value, then I would be very surprised at your story.

Take $X$ to be a constant random variable, taking value $1$ all the time. Take $Y$ to be equal to $1$ with probability $0.5$ and to $2$ with probability $0.5$. Then $E(\frac{X}{Y})= 0.75$ whereas $\frac{E(X)}{E(Y)}=\frac{2}{3}$.

I think this is hardly ever true.

Arnaud Mortier
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Well, that would occur well

For two terms that would be $\frac 12(\frac ab + \frac cd) = \frac {\frac {a+c}2}{\frac {b+d}2}$

$\frac ab + \frac cd = 2\frac {a+c}{b+d}$

$\frac {ad + cb}{bd}= 2\frac {a+c}{b+d}$

$(ad+ cb)(b+d) = 2bd(a+c)$

$abd + cb^2 + ad^2 +cbd = 2abd + 2bdc$

$ad^2 + cb^2 = abd +cbd$

Which of course can happen.

Wolog, assume $b \le d$ and $b = e-k$ and $d = e+k$

$a(e+k)^2 + c(e-k)^2 = (a+c)(e+k)(e-k)$

$ae^2 + ak^2 + ce^2+ck^2 + 2aek - 2cek = ae^2 - ak^2 + ce^2 - ck^2$

$k^2(a+c) = ek(c-a)$.

$k(a+c) = e(c-a)$

$a(k+1) = c(e-1)$

This obviously can be true but but if it is, it'd be pretty coincidental.

fleablood
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