Prove a $2$-dimensional function is nonnegative

Question

This question arose when I am considering the disturbed Generating Moment Function (GMF) of a Bernoulli random variable. Mathematically speaking, given a Bernoulli random variable $X$ with mean $p$, we define its disturbed GMF as $$ \varphi_\varepsilon (t) \;=\; \mathbb{E} (e^{t(X - p - \varepsilon)}) \;=\; p e^{t(1-p-\varepsilon)} + (1-p) e^{-t(p+\varepsilon)}. $$ I want to prove for any $p\in [0, 1]$ (since it is the probability), for any $\varepsilon > 0$, $$ \varphi_\varepsilon (4 \varepsilon) \le 1, $$ which is equivalent to $$ f(p, \varepsilon, \theta) \;=\; e^{\theta\varepsilon (p+\varepsilon)} - p e^{\theta\varepsilon} + p - 1 \;\ge\; 0 $$ with $\theta = 4$, $p\in [0,1]$, and $\varepsilon > 0$.

When I tried to solve this, I found that it seems for any $t\ge 0$, for any $x,y\in \mathbb{R}$, $f(p, \varepsilon, t) \ge 0$, and the equality holds when $x=0$. I wonder if there is any way to solve this problem, or the former specific case. Any comment helps.

Answer 1

First, we claim that if $\theta > 8$, there exist $p\in (0, 1)$ and $\varepsilon > 0$ such that $e^{\theta\varepsilon (p+\varepsilon)} - p e^{\theta\varepsilon} + p - 1 < 0$. Indeed, let $f(\varepsilon) := e^{\theta\varepsilon (p+\varepsilon)} - p e^{\theta\varepsilon} + p - 1$. We have $f(0) = 0$, $f'(0) = 0$, and $f''(0) = p^2\theta^2 - p\theta^2 + 2\theta = \theta^2(p - 1/2)^2 - \frac{\theta(\theta - 8)}{4}$. Thus, if $\theta > 8$, there exists $p \in (0, 1)$ such that $f''(0) < 0$. Thus, there exist $p \in (0, 1)$ and $\varepsilon > 0$ such that $f(\varepsilon) < 0$.

Second, we have the following result.

Fact 1. Let $\theta \in [0, 8]$, $p \in [0, 1]$, and $\varepsilon > 0$. Then $e^{\theta\varepsilon (p+\varepsilon)} - p e^{\theta\varepsilon} + p - 1 \ge 0$.

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Proof of Fact 1.

Taking logarithm, it suffices to prove that $$F(\varepsilon) := \ln\left(e^{\theta\varepsilon (p+\varepsilon)} + p - 1\right) - \ln p - \theta\varepsilon. \tag{1}$$

We have $$F'(\varepsilon) = \frac{\theta(p + 2\varepsilon)e^{\theta\varepsilon (p+\varepsilon)}}{e^{\theta\varepsilon (p+\varepsilon)} + p - 1} - \theta \ge 0. \tag{2}$$ (The proof of (2) is given at the end.)

Also, we have $F(0) = 0$. Thus, $F(\varepsilon) \ge 0$ for all $\varepsilon > 0$.

We are done.

Proof of (2).

It suffices to prove that $$(p + 2\varepsilon)e^{\theta\varepsilon (p+\varepsilon)} \ge e^{\theta\varepsilon (p+\varepsilon)} + p - 1$$ or $$1 - p \ge (1 - p - 2\varepsilon)e^{\theta\varepsilon (p+\varepsilon)}. $$

We only need to prove the case that $1 - p - 2\varepsilon > 0$. It suffices to prove that $$1 - p \ge (1 - p - 2\varepsilon)e^{8\varepsilon (p+\varepsilon)}$$ or $$G(\varepsilon) := 1 - p - (1 - p - 2\varepsilon)e^{8\varepsilon (p+\varepsilon)} \ge 0.$$

We have $$G'(\varepsilon) = 2e^{8\varepsilon (p+\varepsilon)} (2p + 4\varepsilon - 1)^2 \ge 0.$$ Also, we have $G(0) = 0$. Thus, $G(\varepsilon) \ge 0$ for all $\varepsilon > 0$.

We are done.