Variance of inverse weighted sum

Let $X$ be a random variable with variance $Var[X]=\sigma^2$. Given a number of (independent) realizations $X_i$, is it possible to estimate the variance of the inverse weighted sum: $$Var\left[\frac{1}{\sum_i a_iX_i}\right]$$ ? What I am actually looking for is an estimate of the variance of the above expression without having a particular set of realizations - just knowing the weights $a_i$ and $\overline X$ and $\sigma^2$.

I found the delta method, which is essentially a kind of Taylor expansion: $$Var[f(\vec X)] \approx Var[X]\left(\sum_i\left(\frac{\partial}{\partial X_i}f(\vec X)\right)^2\right)\\ =Var[X]\left(\sum_i\left(-\frac{a_i}{2\left(\sum_i a_iX_i\right)^2}\right)^2\right)$$ However, this expressions still has $X_i$ inside, which doesn't solve my problem.

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I think there is a problem here. You can evaluate the variance of a random variable and given its realizations (as you mentioned they are $X_i$s) you can estimate it. –  Seyhmus Güngören Sep 19 '12 at 12:45
Sorry, my initial question was mistaken. Please see my edit. –  Jakob S. Sep 19 '12 at 12:50
I think estimation is not necessary here. It seems perhaps more ambiguous at the moment. Your expression for the Taylor epproximation also seems problematic. I suggest $X_i$ i.i.d random variables $\sim X$.Check the Taylor apprx. again. You have an expectation operation out of which you will get only some values (results independent of $X_i$) –  Seyhmus Güngören Sep 19 '12 at 12:56
In the simplest possible case of just one variable, given that $X$ has finite variance $\sigma^2$, can we be sure that $Y = X^{-1}$ has finite variance? For example, if $X$ is a standard exponential random variable with density function $\exp(-x)\mathbf 1_{[0,\infty)}$, $$E[X^{-n}] = \int_0^{\infty}x^{-n} \exp(-x)\,\mathrm dx = \Gamma(-n+1)$$ which is undefined for positive integers $n$. –  Dilip Sarwate Sep 19 '12 at 13:52
@DilipSarwate: Thanks for this interesting comment! I did not think about this before. There might be a fundamental problem indeed. –  Jakob S. Sep 20 '12 at 9:05