I have a random variable $M>1$ whose variance and expected value I know (as well as any other moments). So for example, $\text{var}(M) = c$ and $\mathrm{E}(M) = d$. I was wondering whether there was a way to provide a bound (either lower or upper -- both would be useful) on $\text{var}(\frac{1}{M})$?

The simplest bound I have (just from the definition of variance) is,

$\text{var}(\frac{1}{M}) \leq \mathrm{E}(\frac{1}{M^2})$.

However this isn't all that useful to me since I can't calculate the RHS. I have tried as a next step using Jensen's inequality, but that also doesn't seem to help because my next step:

$\mathrm{E}(\frac{1}{M^2}) \geq \frac{1}{\mathrm{E} (M^2)}$,

so I get two oppositely signed inequalities! Therefore I can't seem to conclude anything useful.

I'm guessing there is an inequality like Jensen's, but for variances (or higher central moments) but I can't seem to find one.

Does anyone know how I can estimate bounds here?

Edit: suppose I also know that the random variable has only has support for $M>>1$.

  • $\begingroup$ You need to ensure $\hat{M}$ does not get close to 0. Else, $E[1/\hat{M}]$ can be infinite, as can the variance, or they may not even exist. A trivial example is when $\hat{M}=0$ with prob 1. So then with prob 1 $1/\hat{M}$ does not even exist. Or if $\hat{M}$ is uniform over $[-1,1]$, so then $1/\hat{M}$ exists with prob 1, but $E[1/\hat{M}]$ does not exist. $\endgroup$ – Michael Dec 22 '16 at 0:04
  • $\begingroup$ To make the problem more meaningful, add the condition that the expected value and the variance of the estimator are both required to exist (as finite quantities). $\endgroup$ – Mark Fischler Dec 22 '16 at 0:18
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    $\begingroup$ @MarkFischler : The problem is not that the estimator $\hat{M}$ moments exist. We can have $\hat{M} \in [-1,1]$ always, so all moments exist and are bounded by 1 in magnitude, but $Var(1/\hat{M})$ can be an arbitrarily large real number, and can even be $\infty$. $\endgroup$ – Michael Dec 22 '16 at 0:25
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    $\begingroup$ If you know that $M \ge k \gt 0$ then you can conclude that $\text{var}(\frac{1}{M}) \le \frac{1}{4k^2}$ $\endgroup$ – Henry Dec 22 '16 at 1:01
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    $\begingroup$ Even simpler (for $M\geq k$) is $Var(1/M) = E[1/M^2]-E[1/M]^2 \leq 1/k^2 - 1/E[M]^2$, which is better than Henry's bound when $E[M] < 2k/\sqrt{3}$. $\endgroup$ – Michael Dec 22 '16 at 1:43

Suppose that $X \geq k$ and that $E[X]=m$ for some given real numbers $m, k$ such that $m\geq k>0$. We want to bound $Var(1/X)$. We know by Jensen’s inequality that $1/m \leq E[1/X] \leq 1/k$. Thus:

\begin{align} Var(1/X) &= E\left[\frac{1}{X^2}\right] - E\left[\frac{1}{X}\right]^2\\ &\leq E\left[\frac{1}{X}\frac{1}{k}\right] - E\left[\frac{1}{X}\right]^2 \\ &= E\left[\frac{1}{X}\right]\frac{1}{k} - E\left[\frac{1}{X}\right]^2 \\ &\overset{(a)}{\leq} \max_{c \in [1/m, 1/k]} \left[\frac{c}{k} - c^2\right] \\ &= \left\{ \begin{array}{ll} \frac{1}{4k^2} &\mbox{ if $m \geq 2k$} \\ \frac{1}{mk} - \frac{1}{m^2}& \mbox{ if $m \in [k, 2k]$} \end{array} \right. \end{align} where (a) is justified because $E[1/X] \in [1/m, 1/k]$. Notice that this bound is identical to Henry’s bound when $m \geq 2k$, but is better when $m \in [k, 2k)$.

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