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 $K^2\*\sigma^2$?



Proof of variance bounds for transformed random variables
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 $a$.


