Simplify this statistical average

Question

I have a quantity $\tau$ given by: $$ \frac{1}{\tau} = \frac{1}{\tau_1}+\frac{1}{\tau_2}+\frac{1}{\tau_3} $$

where $\tau_1$, $\tau_2$ and $\tau_3$ are some constituent quantities. Now these $\tau$'s are function of a variable $x$, but the ensemble averages of $\tau_i$ are known, given by:

$$ \frac{\langle \tau_i^2\rangle}{\langle \tau_i\rangle^2} = \alpha_i $$

Is there a way to express or simplify the ensemble average of total $\tau$ in terms of $\alpha_i$'s: $$ \frac{\langle \tau^2\rangle}{\langle \tau\rangle^2} $$

If not, what would be the necessary information needed w.r.t. the actual distribution of $\tau_i$ with $x$.

Answer 1

First things first, as we know no relations between the $\tau_i$'s a priori, it's best to split this up via linearity of expectation.

Second, there is not a way in terms of your $\alpha_i$'s. This can be shown by considering the exponential distribution, where the $\alpha_i$'s will always be $2$, but the expectation of the inverse varies.

As for what would be proper information,I believe the only real way is to find $\mathbb{E}(1/X_i)$. One trick people I've seen used to compute this is to write it as $$\mathbb{E}(1/X_i) = \mathbb{E}\left(\int_0^\infty e^{-tX_i}dt \right) = \int_0^\infty \mathbb{E}(e^{-tX_i})dt$$ So maybe that will help. Best of luck.