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 $x$.

  • 3
    $\begingroup$ 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? $\endgroup$
    – Ian
    Feb 11, 2021 at 16:39
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    $\begingroup$ @Ian That's correct, that is what I intended to ask. Sorry if my question wasn't clear. $\endgroup$ Feb 11, 2021 at 16:40

1 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.


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