Let N be a nonnegative, integer-valued random variable. Show that
P{N >0} ≥\frac{(E[N])²}{E[N²]}
and explain how this inequality can be used to derive additional bounds on a reliability function.
Hint:
E[N²] = E[N² |N >0]P{N >0} (Why?)
≥ (E[N |N >0])²P{N >0} (Why?)
Now multiply both sides by P{N >0}.
E\left[N^2\right]=E\left[N^2 \mid N>0\right] P\{N>0\}
\geqslant(E[N \mid N>0])^2 P\{N>0\}, \quad \text { since } E\left[X^2\right] \geqslant(E[X])^2
Thus,
\begin{aligned}E\left[N^2\right] P\{N>0\} & \geqslant(E[N \mid N>0] P[N>0])^2 \\& =(E[N])^2\end{aligned}
Let N denote the number of minimal path sets having all of its components functioning. Then r({p})=P\{N>0\}. Similarly, if we define N as the number of minimal cut sets having all of its components failed, then 1 – r({p})=P\{N>0\}. In both cases we can compute expressions for E[N] and E\left[N^2\right] by writing N as the sum of indicator (i.e., Bernoulli) random variables. Then we can use the inequality to derive bounds on r({p}).