# Simplify this statistical average

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 $$x$$.

• So in math notation, you have dependent random variables $X_1,X_2,X_3$, and you know $\frac{E[X_i^2]}{E[X_i]^2}$ for each $i$, and you want to know $\frac{E[Y^2]}{E[Y]^2}$ where $\frac{1}{Y}=\sum_{i=1}^3 \frac{1}{X_i}$. Is that correct?
– Ian
Feb 11, 2021 at 16:39
• @Ian That's correct, that is what I intended to ask. Sorry if my question wasn't clear. Feb 11, 2021 at 16:40

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.