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I have random variables $X_1, X_2, \dots, X_n$ and $Y_1, Y_2, \dots, Y_n$, with $n$ a large integer. All pairs $(X_i, Y_i)$ are independent and identically distributed, but every $X_i$ and $Y_i$ within a pair are dependent. All $X_i$ and $Y_i$ yield positive real numbers.

I have a sample of each variable, I'll call the values $x_1, x_2, \dots, x_n$ and $y_1, y_2, \dots, y_n$. Then I can calculate $\mu_x = \frac{1}{n} \sum_{i=1}^n x_i$ and $\sigma_x = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu_x)^2$, and similar formulas for $\mu_y$ and $\sigma_y$.

The goal is to estimate $\mu = E[X_1] / E[Y_1]$ and to get a confidence interval with a given confidence (for example 95%). I'm not sure if I may do some assumptions, since the random variables are the output of a process that I do not fully understand. Maybe I may assume that all variables are almost normally distributed, but it would be better if the question can be answered with no assumptions or weaker assumptions.

To estimate $\mu$ I can simply calculate $\mu \approx \mu_x / \mu_y$, but the confidence interval gives me headaches. How can I calculate the confidence interval?

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  • $\begingroup$ I had a similar problem once and a solution I found in the book "Wolters - Introduction to variance estimation" was to use a jackknife approach for the variance, which can then be easily translated into confidence intervals. See the chapters there for explicit derivations. Bootstrapping as a natural extension would work too, but is not covered in the book. The third option is to use Taylor approximations - this is also covered in the book. In general, look for "ratio estimators". $\endgroup$
    – Nameless
    Sep 18, 2013 at 14:12
  • $\begingroup$ Error propagation is the way to go, I believe. $\endgroup$ Jan 1, 2016 at 19:56
  • $\begingroup$ What you are calling $\mu_x$ is a mean of your sample, not the mean of the population. So it is more commonly called $\bar{x}$. Similarly with $\sigma_x$ and $s_x$. Also either a square or square root is missing from your standard deviation equation, is that right? $\endgroup$
    – 2'5 9'2
    Jul 20, 2016 at 6:59
  • $\begingroup$ When you say $E[X_1]/E[Y_1]$, do you mean $E[X]/E[Y]$? (Side note: just noticing how old this question is...) $\endgroup$
    – 2'5 9'2
    Jul 20, 2016 at 7:04

2 Answers 2

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If $n$ is large, then you may use the asymptotic confidence interval. Let $\hat{\mu}_x=n^{-1}\sum X_i$, $\hat{\mu}_y=n^{-1}\sum Y_i$ and $\mu_x=\mathbb{E}X_1$, $\mu_y=\mathbb{E}Y_1$. Then using Delta method (we assume that $\mu_y\ne 0$), \begin{align} \sqrt{n}\left(\frac{\hat{\mu}_x}{\hat{\mu}_y}-\frac{\mu_x}{\mu_y}\right)&=\left[\left(\frac{1}{\mu_y},-\frac{\mu_x}{\mu_y^2}\right)+o_p(1)\right]\sqrt{n}\left( \begin{matrix} \hat{\mu}_x-\mu_x \\ \hat{\mu}_y-\mu_y \end{matrix} \right) \\ &\xrightarrow{d}\left(\frac{1}{\mu_y},-\frac{\mu_x}{\mu_y^2}\right)\mathcal{N}(0,V), \end{align} where $V=\operatorname{Var}\left(\left(X_1,Y_1\right)'\right)$ and convergence in distribution follows by the CLT (if the second moments of $X_1$ and $Y_1$ are finite).

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I have not made whole exercise, but wouldn't it be possible to try to approximate the pdf of the quotient assuming:

  • that your estimators $\bar{x} = 1/n \sum x_i$ and $\bar{y}$ are approximately normally distributed because of CLT
  • that the ratio of two independent normaly distributed random variables is a cauchy distribution. The problem in your case is that $X$ and $Y$ are correlated.
  • but anyway, you could estimate the covariance matrix of $\bar{x}$ and $\bar{y}$ via $\operatorname{cov}(\bar{x},\bar{y}) = \operatorname{cov}(x,y)/n \simeq 1/n^2 \sum (x_i-\bar{x})(y_i-\bar{y})$ and then build the joint pdf of $f(\bar{x},\bar{y})$ as a bivariate normal distribution
  • finally you get the distribution of the quotient via the changement of variable $u=\bar{x}, v=\bar{x}/\bar{y}$ leading to $g(v) = \int g(u,v) \, du$ with $g(u,v) \, du \, dv = f(\bar{x},\bar{y}) \, d\bar{x} \, d\bar{y}$

If you succeed to get an analytic function for $g(v)$, maybe you can use it to compute your confidence interval ?

You can look also at the wikipedia article on the ratio distribution

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