2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ This seems helpful: en.wikipedia.org/wiki/… $\endgroup$ – tomka Jul 27 '15 at 17:07
1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.