Let’s say I have a parameterspace $\Omega$ with known probability density function $p(\omega), \omega \in \Omega$. I want to estimate the expectation $E[x]$ of a variable $x$ depending on $\omega \in \Omega$. I cannot calculate directly $x$ but I can make an estimation $\tilde{x}$ such that $\tilde{x}- x \leq \epsilon, \epsilon > 0$ and $x \leq \tilde{x}$.
It’s easy to prove that $|E[\tilde{x}] – E[x]| \leq \epsilon$. Is there a way to calculate a similar upper bound for $|Var[\tilde{x}] – Var[x] |$?
Kind regards, Koen
\begin{eqnarray*} \mathsf{Var}[\tilde x]-\mathsf{Var}[x] &=& \mathsf{Var}[x+(\tilde x-x)]-\mathsf{Var}[x] \\ &=& \mathsf{Var}[x]+\mathsf{Var}[\tilde x-x]+2\mathsf{Cov}[x,\tilde x-x]-\mathsf{Var}[x] \\ &=& \mathsf{Var}[\tilde x-x]+2\mathsf{Cov}[x,\tilde x-x] \\ &\le& 2\epsilon\sqrt{\mathsf{Var}[x]}+\epsilon^2\;. \end{eqnarray*}
A slightly more natural and perhaps more useful result might be
\begin{eqnarray*} \frac{\mathsf{Var}[\tilde x]}{\mathsf{Var}[x]} &=& \frac{\mathsf{Var}[x+(\tilde x-x)]}{\mathsf{Var}[x]} \\ &=& \frac{\mathsf{Var}[x]+\mathsf{Var}[\tilde x-x]+2\mathsf{Cov}[x,\tilde x-x]}{\mathsf{Var}[x]} \\ &\le& 1+2\frac{\epsilon}{\sqrt{\mathsf{Var}[x]}}+\left(\frac{\epsilon}{\sqrt{\mathsf{Var}[x]}}\right)^2\;. \end{eqnarray*}
(I left out all the absolute values signs for readability, but the inequality still holds if you add them.)