2
$\begingroup$

Given that $1-x\leq \exp{(-x)}$, prove that

$$\prod_{t=1}^{T}\sqrt{1-4x_t^{2}}\leq \exp\left(-2\sum_{t=1}^{T}x_t^{2}\right)$$

How would one approach this problem. Without the square root I know that

$$1-2\sum_{t=1}^{T}x_t^{2}\leq \exp(-2\sum_{t=1}^{T}x_t^{2})$$

But I'm not sure what to do after that.

$\endgroup$
1
$\begingroup$

For all $t$, using the hint, we have $$1-(2x_t)^2\leq \exp\bigg(-(2x_t)^2\bigg)$$ Thus $$1-4x_t^2\leq \exp(-4x^2_t)$$ Take the square root on each side $$\sqrt{1-4x_t^2}\leq \exp(-2x^2_t)$$ Multiplying all these inequalities for $1\leq t\leq T$ yields

$$\prod_{t=1}^T\sqrt{1-4x_t^2}\leq \prod_{t=1}^T\exp(-2x^2_t)$$

In other words $$\prod_{t=1}^T\sqrt{1-4x_t^2}\leq \exp(-2 \sum_{t=1}^T x^2_t)$$

$\endgroup$
  • $\begingroup$ wow..brilliant. i forgot how the square root can get pushed into the exponent $\endgroup$ – user1436508 Nov 8 '18 at 4:56
  • $\begingroup$ can you explain this $\sqrt{1-4x_t^2}\leq \exp(-2x^2_t)$ How did you get $-2x^2_t$ Shouldnt x^2 become x $\endgroup$ – user1436508 Nov 8 '18 at 5:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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