I am trying to understand how I should approach the problem of a Taylor approximation to the expectation of the ratio of two random variables. In my particular problem I am concerned with the following ratio estimated using a sample of size $n$

$$\hat{\gamma_i}=\frac{x_i\sum_{i=1}^{n} y_i}{\sum_{i=1}^{n} x_i}=\frac{x_i\bar{y}}{\bar{x}}$$

We may assume for simplicity $E(x_i)=\mu_x$ and $E(y_i)=\mu_y$, but we may not have $E(x_iy_i) \ne E(x_i)E(y_i)$.

I try to find $E(\hat{\gamma})$. How should I approach this problem?

  • $\begingroup$ You should check out web.stanford.edu/class/cme308/OldWebsite/notes/…. It has a similar problem (a ratio) that is worked out. $\endgroup$ – StatsStudent Jul 27 '15 at 16:37
  • $\begingroup$ Are the $y_i$'s independent of the $x_i$'s? $\endgroup$ – Alecos Papadopoulos Jul 27 '15 at 16:42
  • $\begingroup$ Something seems strange about that material. I do not think it is legitimate (i.e. precise enough) to make a first order Taylor approximation to a ratio of two random variables. As discussed in that document, this is, namely, simply the ratio of expectations. I am looking for a second or third order approximation of my ratio. $\endgroup$ – tomka Jul 27 '15 at 16:55
  • $\begingroup$ @AlecosPapadopoulos No, not necessarily, $E(x_i y_i) \ne E(x_i)E(y_i)$. $\endgroup$ – tomka Jul 27 '15 at 16:56
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    $\begingroup$ This seems helpful: en.wikipedia.org/wiki/… $\endgroup$ – tomka Jul 27 '15 at 17:07

The difficulty in your expression comes from the $1/\bar{x}$ term, which is the term we will need to expand.

$$ \frac{1}{\bar{x}} = \frac{1}{E(\bar{x})+\bar{x}-E(\bar{x})}$$

$$ \frac{1}{\bar{x}} = \frac{1}{E(\bar{x})} - \frac{\bar{x}-E(\bar{x})}{E(\bar{x})^2} + \dots $$

So, for example, we find that:

$$ E(\frac{1}{\bar{x}}) \approx \frac{1}{E(\bar{x})} - 0 + \frac{var(\bar{x})}{E(\bar{x})^3} $$

Your case is more complex because you also have $x_i \bar{y}$ but you should be able to finish from there

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    $\begingroup$ In the question, the OP verifies that we cannot assume the $y_i$ are independent of the $x_i$. That assumption, however, seems to be the basis of your focus on $1/\bar x$ alone. $\endgroup$ – whuber Jul 29 '15 at 11:59
  • $\begingroup$ The 2nd order Taylor expension for a ratio with two dependent random variables is given here. The difficulty is in writing it out. en.wikipedia.org/wiki/… $\endgroup$ – tomka Jul 29 '15 at 13:11

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