how to show that $ \frac{\theta e^t(1-\theta)}{(1-\theta+\theta e^t)^2} \leq\frac{1}{4}$?

Question

let $0 \leq \theta \leq 1$ , then how to show that $\forall t\in R$ $$ \frac{\theta e^t(1-\theta)}{(1-\theta+\theta e^t)^2} \leq\frac{1}{4}?$$

This is a step of a proof of hoffeding's lemma.

Answer 1

Maybe start with $4\theta e^t(1-\theta)\leq(1-\theta+\theta e^t)^2$ from the inequality $$4xy\leq(x+y)^2.$$

Answer 2

With $$\left|\alpha+\dfrac{1}{\alpha}\right|\geq2$$ then let $$\alpha=\sqrt{\dfrac{\theta}{1-\theta}}e^{\frac12t}$$ gives the desire inequality after simplification.

Answer 3

Just using algebra $$f(t)=\frac{\theta e^t(1-\theta)}{(1-\theta+\theta e^t)^2}$$ $$f'(t)=\frac{(\theta -1) \theta e^t \left(\theta +\theta e^t-1\right)}{\left(\theta \left(e^t-1\right)+1\right)^3}$$ So, the first derivative cancels for $$\theta +\theta e^{_*}t-1=0 \implies t_*=\log \left(\frac{1-\theta }{\theta }\right)\implies f(t_*)=\frac 14$$

Computing the second derivative, $f''(t_*)=-\frac 18$ confirms that $t_*$ corresponds to a maximum.

Answer 4

Observe $p(1-p) \le \frac 1 4$ (That's probably where the $\frac 1 4$ comes from). We will be finished if we can prove

$$ \frac{e^t}{(1-p+p e^t)^2} \le 1$$

Just modify proof here then.

Answer 5

Pf: Observe that the numerator can be written as a product of 2 terms whose sum is inside the $()^2$ of the denominator, so by AM-GM inequality, we have that for $x=1-p>0$ and $y=pe^t > 0$

$$\frac{xy}{(x+y)^2} \le \frac 1 4$$

QED