Question
Let $X$ have the $Gamma(\alpha, \beta)$ density.
I.e. $$f_X(x) = \frac{1}{\gamma(\alpha)\beta^\alpha}x^{\alpha-1}e^{-\frac{x}{\beta}}$$
when $x >0$ and $0$ elsewhere.
The moment generating function of $X$ is $M(t) = \frac{1}{(1-\beta t)^\alpha}, t < \frac{1}{\beta}.$
(*) Accept that $P(X \ge r) \le e^{-rt}M(t)$ for every $r$ and all t, $0 < t < \frac{1}{\beta}$.
Derive a lower bound (depending only on $\alpha$) for $P(X \ge 2\alpha\beta)$
Answer
Using (*) and letting $r = 2\alpha\beta$ we have
$$P(X \ge 2\alpha\beta) \le e^{-2\alpha\beta t} \frac{1}{(1-\beta t)^\alpha}$$
$$\iff P(X \ge 2\alpha\beta) \le \frac{(e^{-2\beta t})^\alpha}{(1-\beta t)^\alpha}$$
$$\iff P(X \ge 2\alpha\beta) \le \bigg(\frac{e^{-2\beta t}}{(1-\beta t)}\bigg)^\alpha$$
But all I have now is an upper bound, and not only that, it is dependent on $\beta$ and $t$ as well as $\alpha$. Am I completely off track here? Anyone know how to solve this problem?
Directly using the distribution function:
$$P(X \geq 2\alpha \beta)= 1 - P(X < 2\alpha \beta)=1 - F(2\alpha \beta)\\=1-\frac{1}{\Gamma(\alpha)\beta^\alpha}\int_0^{2\alpha\beta}s^{\alpha-1}e^{-\frac{s}{\beta}}\, ds.$$
Let $u = s/\beta$.
Then
$$P(X \geq 2\alpha \beta)= 1- \frac{1}{\Gamma(\alpha)}\int_0^{2\alpha}u^{\alpha-1}e^{-u}\, du>0.$$ So equality here gives a lower bound that depends only on $\alpha$.
For an upper bound, you can use Markov's inequality.
$$P(X \geq 2\alpha \beta) \leq \frac{E(X)}{2\alpha\beta} = \frac{\alpha \beta}{2\alpha\beta}=\frac1{2}.$$