# Bernoulli’s inequality and probability

Suppose we have independent events $E_1,E_2,\ldots,E_m$, each of which occur with probability $1-\varepsilon$. The event that all of the $E_i$ occur is $E = \bigcap_{i=1}^m E_i$. By using independence we can calculate the probability of $E$, $P(E) = P\left(\bigcap_{i=1}^m E_i\right) = \prod_{i=1}^m P(E_i) = (1-\varepsilon)^m.$

We could also get a lower bound on $P(E)$ by using the union bound. This gives, $P(E) = 1-P(E^c) = 1-P\left(\bigcup_{i=1}^m E_i^c\right) \ge `1-\sum_{i=1}^m P(E_i^c) = 1-m\varepsilon.$

We can therefore conclude that $(1-\varepsilon)^m \ge 1-m\varepsilon$. This is an instance of Bernoulli’s inequality. More generally, Bernoulli’s inequality states that $(1+x)^m \ge 1+mx,$

for all $x \ge -1$ and $m \ge 1$. This inequality states the red line is always underneath the black curve in the below picture. For an interactive version of this graph where you can change the value of $m$, click here.

Our probabilistic proof only applies to that case when $x = -\varepsilon$ is between $-1$ and $0$ and $m$ is an integer. The more general result can be proved by using convexity. For $m \ge 1$, the function $f(x) = (1+x)^m$ is convex on the set $x > -1$. The function $g(x) = 1+mx$ is the tangent line of this function at the point $x=0$. Convexity of $f(x)$ means that the graph of $f(x)$ is always above the tangent line $g(x)$. This tells us that $(1+x)^m \ge 1+mx$.

For $m$ between $0$ and $1$, the function $(1+x)^m$ is no longer convex but actually concave and the inequality reverses. For $m \le 0$, $(1+x)^m$ becomes concave again. These two cases are visualized below. In the first picture $m = 0.5$ and the red line is above the black one. In the second picture $m = -1$ and the black line is back on top.