# Let $X$ have the gamma $(r, \lambda)$ distribution. Show that $P(X \ge 2E(X)) \le (2/e)^r$

Let $X$ have the gamma $(r, \lambda)$ distribution. Show that $P(X \ge 2E(X)) \le (2/e)^r$.

I do not know how to approach this. I am thinking of Chernoff Bounds, but what trips me up is how to minimize the bound.

• It looks to me like you could simplify to $\lambda=1$ (since changing the scale parameter won't change the probability, you can do the case $\lambda=1$ without loss of generality -- the general case follows immediately by a rescaling at the end). The Chernoff bound looks like it should work just fine with the right choice of $t$. I don't think you need to worry about choosing it to minimize something, just try to find the value of $t$ that makes it come out. – Glen_b Apr 19 '17 at 2:50
• @Glen_b provides a method with statistical insight. A more pedestrian method, but perhaps not without insight of its own, is (after ignoring the irrelevant scale parameter $\lambda$) to write the probability as an integral over a positive variable $x$, use the substitution $x=2y$, and apply the simple inequality $e^{r-2y}\le e^{-y}$ for all $y \ge r$. The result will fall right out. – whuber Apr 19 '17 at 13:49