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 $C_{XY}$?

  • $\begingroup$ Nice question +1 $\endgroup$ – Ahmad Bazzi Oct 19 '16 at 0:07
  • $\begingroup$ Do you know anything else about the distributions? You're not guaranteed the mean and variance even exist in general. $\endgroup$ – Paul Oct 19 '16 at 0:23
  • $\begingroup$ @Paul No. I do not have any information about distributions. I need to find the approximation of the mean and variance. ( We can make an assumption that it exists) $\endgroup$ – Mat_python Oct 19 '16 at 0:26

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.

  • $\begingroup$ Great! Thank you. Do I apply variance same way as you showed above like, Var(X/Y) = Var(f(x,y)) ? $\endgroup$ – Mat_python Oct 19 '16 at 1:03
  • $\begingroup$ How did you get the second term i.e, (-Cxy)/mu_y^2? I am getting (x-mu_x)*(1/mu_y) - (y - mu_y)*(mu_x/mu_y^2) $\endgroup$ – Mat_python Oct 19 '16 at 1:20
  • $\begingroup$ from the last term of the first equation $\endgroup$ – Ahmad Bazzi Oct 19 '16 at 8:57

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