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Given I have a random variable $a$ that can be realised in the domain $D$ and has a finite variance $\sigma^2$. Furthermore I have a function $f$ which is differentiable(hence continuous) with an absolute derivative in the domain $D$ always less than $K$.

Is there then a way to prove that the variance of $f(a)$ is bounded by $K^2*\sigma^2$?

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Assume wlog that $E[a]=0$. Then $$\big|f(a)-f(0)\big|\le \left|\int_0^a|f'(x)|dx\right| \le \left|\int_0^a Kdx\right| \le K|a|\ \ \ \ \ \ \ \ \ \ $$ So $$E[(f(a)-f(0))^2]<K^2 E[a^2]$$ On the left, we can subtract $(E[f(a)]-f(0))^2$, which is non-negative; on the right, we can subtract $K^2E[a]^2$, which is zero. This gives $$E[f(a)^2]-E[f(a)]^2 \le K^2(E[a^2]-E[a]^2)$$ which indeed shows that the variance of $f(a)$ is at most $K^2$ times the variance of $a$.

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  • $\begingroup$ This proof can be extended to random variables with non zero mean, since variance is independent by addition and subtaction. So if $b=a-E(a)$ then $b$ and $a$ will have the same variance. $\endgroup$
    – Peter Mølgaard Pallesen
    Commented Apr 30, 2019 at 10:29
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    $\begingroup$ Indeed -- wlog. $\endgroup$
    – user44143
    Commented Apr 30, 2019 at 10:52

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