Some (maybe) basic estimate of expected values involving brownian motion

Question

I have the following estimation in a paper I'm reading about Brownian Motion on a Torus: $$ 1+\theta\ \mathbb E_x[\tau_A] + \sum_{i=2}^\infty \left(\theta\sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]\right)^i \leq \exp\left(\theta\mathbb E_x[\tau_A] + 2\theta^2 \sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]^2\right) $$ where $\mathbb T_2$ is a Torus, $A\subset\mathbb T_2$ closed, $x\in\mathbb T_2$ and $0<\theta<\frac 1 2 \sup_{x\in\mathbb T_2}\mathbb E_x[\tau_A]^{-1}$.

Ideas: To prove this, I have already tried the Taylor series of $e^x$, the geometric sum, the Jensen inequality and a few other things, however nothing worked so far.

I'd be really glad if you could give me some thoughts about what else to do.

Answer 1

This has nothing to do with probablitities.

Set $x = \theta \mathbb E_x[\tau_A]$ and $y = \theta \sup_y \mathbb E_y[\tau_A]$. Then $0 x \leq y < \frac{1}{2}$. By the geometric series we get $$ \sum_{i=2}^\infty y^i = \frac{1}{1-y} - 1 - y = \frac{y^2}{1-y} < 2 y^2.$$ Hence $$ 1 + x + \sum_{i=2}^\infty y \leq 1 + x + 2y^2 \leq \exp(x+2y^2).$$ The last step uses the inequality $$1+z \leq \exp(z), \forall z > 0,$$ which follows from the Taylor expansion $$ \exp(z) = 1 + z + \frac{z^2}{2} + \dotsb.$$

Answer 2

Assuming you are missing a $\theta$ in the exponential, this is just $$ 1+z\leq e^{z} $$ for $z\geq 0$.