How to compute the variances of quotients of normal variables? $x$ and $y$ are independent normal random variables.  $z1=x/(x+y)$ and $z2=y/(x+y).$
How to obtain the variances for $z1$ and $z2$? 
I understand through the delta method, $\operatorname{Var}(z1)$ can be approximated as
$$((Ey)^2 \operatorname{Var}(x)+(Ex)^2(\operatorname{Var}(x)+\operatorname{Var}(y)))/(E(x)+E(y))^4$$
and $\operatorname{Var}(z2)$ can be approximated as
$$((Ex)^2 \operatorname{Var}(y)+(Ey)^2(\operatorname{Var}(x)+\operatorname{Var}(y)))/(E(x)+E(y))^4.$$
Apparently, these two expressions would not be equal to each other but we know the variances of $z1$ and $z2$ are supposed to be equal based on their functional relationship with $x$ and $y.$  Please advise.
 A: Let's solve this first for the case where $X$ and $Y$ are iid standard Normal variables.  (The question appears to assume $X$ and $Y$ are identically distributed anyway, suggesting this is a good starting point.)  
Because $X$ and $Y$ are identically distributed, the $Z_i$ are exchangeable and therefore also must be identically distributed.  Let their common variance be $\sigma^2$ and suppose for a moment it is finite.  Since $Z_1+Z_2=1$ is constant it has zero variance, allowing us to compute
$$0 =\operatorname{Var}(Z_1+Z_2) = 2\sigma^2 + 2 \operatorname{Cov}(Z_1,Z_2),$$
thereby deducing
$$\operatorname{Cov}(Z_1,Z_2) = -\sigma^2.$$
But, writing $\mu=E[Z_1] =E[Z_2]$ (whose finiteness is assured by the assumed finiteness of $\sigma^2$), algebra shows us
$$\eqalign{
\sigma^2 = -\operatorname{Cov}(Z_1,Z_2) &= -E[Z_1 Z_2] + \mu^2 \\
&= -E\left[\frac{XY}{(X+Y)^2}\right]  + \mu^2\\
&= -\frac{1}{4} E\left[\frac{(X+Y)^2 - (X-Y)^2}{(X+Y)^2}\right]  + \mu^2\\
&= \frac{1}{4}\left(-1 + E\left[\frac{(X-Y)^2}{(X+Y)^2}\right]\right) + \mu^2.
}$$
The numerator and denominator in that final fraction are independent because $(X-Y, X+Y)$ are jointly Normally distributed with zero covariance.  Since a multiple of the denominator $(X+Y)^2$ must therefore have a $\chi^2(1)$ distribution, and the density of that distribution in a neighborhood of $0$ is positive, the ratio must have infinite variance.
We are compelled to reject the original assumption that $\sigma^2$ is finite; and there is the answer: $\operatorname{Var}(Z_1) = \operatorname{Var}(Z_2) = \infty.$

It is straightforward to generalize these arguments to arbitrary independent Normal variables--the algebra gets a little messier, but the same principle applies: the ratios are random variables with infinite variances.
BTW, because the $Z_i$ sum to unity their means must either sum to unity or be undefined, whence $\mu=1/2$ (returning to the original simplified setting) or else $\mu$ is undefined.  This is a very general result, holding for any bivariate random variable $(X,Y)$ where $X$ and $Y$ have identical expectations $\mu.$
A: Following $$f'(t) = \frac{g'(t)h(t) - g(t)h'(t)}{h^2(t)}$$
$$\frac {\partial z_1} {\partial x} = \frac {x+y - x}{(x+y)^2} = \frac {y}{(x+y)^2}$$
$$\frac {\partial z_1} {\partial y} = \frac {- x}{(x+y)^2}$$
$$\frac {\partial z_2} {\partial x} = \frac {-y}{(x+y)^2}$$
$$\frac {\partial z_2} {\partial y} = \frac {x}{(x+y)^2}$$
So their variance are the same.
$$\mathrm{Var}(z_1) = \mathrm{Var}(z_2) \approx \frac {(\mathrm{E}(y))^2 \mathrm{Var}(x)+(\mathrm{E}(x))^2\mathrm{Var}(y)}{[\mathrm{E}(x)+\mathrm{E}(y)]^4}$$
This method is applicable when $\mathrm{E}(x) + \mathrm{E}(y)$ have a certain distance from 0.
