How do I find a mean and variance of ratio of two random variables for given mean, variance, and co-variance?

Question

Given two random variables $X$ and $Y$ with mean of $\mu_X$ and $\mu_Y$, variances of $\sigma_X^2$ and $\sigma_Y^2$, and covariance of $C_{XY}$, how to find the approximation of the mean and variance of $Z = \frac{Y}{X}$ in terms of $\mu_X$, $\mu_Y$, $\sigma_X^2$, $\sigma_Y^2$, and $C_{XY}$?

Answer 1

Let $f(X,Y)$ be the joint PDF of $X$ and $Y$. Taylor series of $f(X,Y) = \frac{X}{Y}$ around $\mu_X$ and $\mu_Y$ gives $$ \begin{split}f(x,y) &= f(\mu_X,\mu_Y) \\&+ (x-\mu_x)\frac{\partial f(x,y)}{\partial x}\Big\vert_{(x,y)=(\mu_X,\mu_Y)} \\&+ (y-\mu_y)\frac{\partial f(x,y)}{\partial y}\Big\vert_{(x,y)=(\mu_X,\mu_Y)} \\& +\frac{1}{2}(x-\mu_x)^2\frac{\partial^2 f(x,y)}{\partial x^2}\Big\vert_{(x,y)=(\mu_X,\mu_Y)} \\&+\frac{1}{2}(y-\mu_y)^2\frac{\partial^2 f(x,y)}{\partial y^2}\Big\vert_{(x,y)=(\mu_X,\mu_Y)} \\&+ (x-\mu_x)(y-\mu_y)\frac{\partial^2 f(x,y)}{\partial x\partial y}\Big\vert_{(x,y)=(\mu_X,\mu_Y)} \end{split}$$ First let us take the expectation of $f(x,y)$, and using the derivatives of $f(x,y) = \frac{x}{y}$, you get $$E(\frac{X}{Y}) = E(f(x,y)) \simeq \frac{\mu_X}{\mu_Y} - \frac{C_{XY}}{\mu_Y^2} + \frac{\sigma_Y^2 \mu_X}{\mu_Y^3} + \ldots $$

You can do the same to compute the variance of $\frac{X}{Y}$/ P.S: This is an approximation of ratio means for any distribution, given that the ratio exists with probablity 1. Now, there are some known distriubtions that arise from ratios of densities, such as the Cauchy distribution. The random variable associated with this distribution comes about as the ratio of two Gaussian (normal) distributed variables with zero mean.