# Variance of X/Y

Suppose you have two normally distributed, independent random varianbles X and Y, where X has mean $\mu _x$ and variance $\sigma^2 _x$ and Y has mean $\mu _y$ and variance $\sigma^2 _y$.

For each, $\mu \gg \sigma^2$ and $\mu \gg 0$.

We know from sampling that the distribution of $X/Y$ is approximately normal with mean $\mu _x/\mu _ y$.

The question is: what is the approximate variance of this distribution?

Below is the sampling histogram:

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Are you saying that both the numerator and denominator means are much greater than zero? If the probability near 0 in the denominator is at all appreciable the distribution would be heavytailed. If the denominator mean is 0 the ratio would not have a finite mean. The ratio of two independent standard normals is in fact Cauchy. Also the ratio would not have a mean exactly mu$_x$/mu$_y$. –  Michael Chernick Jul 19 '12 at 22:07
Yes, the mean of X is much larger than 0 (so that the probability of X=0 is ~0), and likewise the mean of Y is much larger than 0. –  lrAndroid Jul 19 '12 at 22:10
But why do you say that the mean of the ratio is the ratio of is the ratio of the means? By Jensen's inequality there is some bias. –  Michael Chernick Jul 19 '12 at 22:13
The mean of the ratio is approximately the ratio of the means. When X and Y are both distributed with mean 5000 and variance 115, the mean was estimated at 1.0000050310725317 which is about 1. –  lrAndroid Jul 19 '12 at 22:14
I think this question may be of interest. –  Macro Jul 20 '12 at 0:30
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Technically, the variance is infinite because you are dividing by a variable with a positive density around $0$ while the numerator has a positive density away from $0$, and this forces the tails to be too large for the variance to exist. In fact, the expected value doesn't exist, either. In practice you may be able to ignore this because it is a very rare event for the denominator to be close to $0$.

Let $\Delta X = X-\mu_X$ and $\Delta Y = Y-\mu_Y.$

Let $r(X,Y) = \frac{X}{Y}$

$r(X,Y) = r(\mu_x + \Delta X, \mu_y + \Delta Y)$

$\approx r(\mu_X,\mu_y) + \frac{\partial r}{\partial X}(\mu_X,\mu_Y)\Delta X + \frac{\partial r}{\partial Y}(\mu_X,\mu_Y)\Delta Y + O((\Delta X)^2+(\Delta Y)^2)$.

If the variances of $X$ and $Y$ are small enough, then we can ignore the higher order terms and compute the variance of the linear approximation.

$\text{Var}\bigg(r(\mu_X,\mu_y) + \frac{\partial r}{\partial X}(\mu_X,\mu_Y)\Delta X + \frac{\partial r}{\partial Y}(\mu_X,\mu_Y)\Delta Y)\bigg)$

$=(\frac{\partial r}{\partial X}(\mu_X,\mu_Y))^2 \text{Var}(X) + (\frac{\partial r}{\partial Y}(\mu_X,\mu_Y))^2\text{Var}(Y)$

since this is just a constant plus a linear combination of $X$ and $Y$.

$\frac {\partial r}{\partial X}(x,y) = \frac 1 y$ so at $(\mu_x,\mu_y)$ it is $\frac{1}{\mu_y}$.

$\frac {\partial r}{\partial Y}(x,y) = \frac {-x} {y^2}$ so at $(\mu_x,\mu_y)$ it is $\frac{-\mu_x}{\mu_y^2}$.

Therefore, the variance of $\frac XY$ will appear to be about

$$\frac{\sigma^2_X}{\mu_Y^2} + \frac{\mu_X^2 \sigma^2_Y}{\mu_Y^4} = \frac{\mu_Y^2\sigma_X^2 + \mu_X^2 \sigma^2_Y}{\mu_Y^4}.$$

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Actually, this answer is quite appropriate (+1), given that $X$ and $Y$ are resistance values and are therefore, in reality, $>0$. –  jbowman Jul 20 '12 at 3:16
@jbowman That means the assumption of normality is technically wrong for both the numerator and the denominator. –  Michael Chernick Jul 20 '12 at 5:02
(+1) I have seen this interesting approximation in "Introduction to the Theory of Statistics". The distribution of $X/Y$ can be misleading because, for certain values of the parameters, it looks like a normal distribution but it can also be bimodal. The tails are always heavy but this approximation can be used for approximating the cumbersome ratio distribution. These "closeness to normality" is observed when the coefficient of variation of $Y$ is small. –  user10525 Jul 20 '12 at 8:40