1
$\begingroup$

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 $K^2*\sigma^2$?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

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 $a$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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
    Apr 30, 2019 at 10:29
  • 1
    $\begingroup$ Indeed -- wlog. $\endgroup$
    – user44143
    Apr 30, 2019 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.