Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 2 / 4F (Probability)

For $t>0$, monotonicity gives $\{X\ge k\}=\{e^{tX}\ge e^{tk}\}$. Apply Markov inequality to the nonnegative variable $e^{tX}$:If the expectation is infinite, the bound remains true but uninformative. At $t=0$ the right side is one, so that endpoint follows directly. Minimizing over $t\ge 0$ gives the Chernoff bound.

The Poisson distribution with unit parameter has mass $e^{-1}/j!$. Its moment-generating function isFor an integer $k\ge 1$, multiply the exponential probability bound by $e$ to obtain ${\sum }_{j=k}^{\infty }1/j!\le \exp (e^t-kt)$. The exponent is minimized at $e^t=k$, hence $t=\log k\ge 0$, including $t=0$ when $k=1$. Consequently