Question

6. Show that for any sequence of events (F)y-1

Answer 1

We will use mathematical induction to prove the inequality.

$\mathbb{P}\left ( \bigcap_{j=1}^{n} F_j\right )\ge \sum_{j=1}^{n}\mathbb{P}(F_j) - (n-1)$

For n = 1,

$LHS = \mathbb{P}\left ( \bigcap_{j=1}^{1} F_j\right ) = \mathbb{P}(F_1)$

$RHS = \sum_{j=1}^{1}\mathbb{P}(F_j) - (1-1) = \mathbb{P}(F_j)$

Thus,  $\mathbb{P}\left ( \bigcap_{j=1}^{n} F_j\right ) = \sum_{j=1}^{n}\mathbb{P}(F_j) - (n-1)$ for n = 1

and,

$\mathbb{P}\left ( \bigcap_{j=1}^{n} F_j\right )\ge \sum_{j=1}^{n}\mathbb{P}(F_j) - (n-1)$ holds for n = 1

Let the inequality condition hold for n = k, where k $\ge$ 1

$\mathbb{P}\left ( \bigcap_{j=1}^{k} F_j\right )\ge \sum_{j=1}^{k}\mathbb{P}(F_j) - (k-1)$ --- (1)

For n = k + 1

$\mathbb{P}\left ( \bigcap_{j=1}^{k+1} F_j\right ) = \mathbb{P}\left ( \bigcap_{j=1}^{k} F_j\right ) \bigcap \mathbb{P}(F_{k+1})$

$= \mathbb{P}\left ( \bigcap_{j=1}^{k} F_j\right ) + \mathbb{P}(F_{k+1}) - \mathbb{P}\left ( \bigcap_{j=1}^{k} F_j \bigcup F_{k+1} \right )$

$\ge \sum_{j=1}^{k}\mathbb{P}(F_j) - (k-1) + \mathbb{P}(F_{k+1}) - \mathbb{P}\left ( \bigcap_{j=1}^{k} F_j \bigcup F_{k+1} \right )$

$= \sum_{j=1}^{k}\mathbb{P}(F_j) + \mathbb{P}(F_{k+1}) - k+1 - \mathbb{P}\left ( \bigcap_{j=1}^{k} F_j \bigcup F_{k+1} \right )$

$= \sum_{j=1}^{k+1}\mathbb{P}(F_j) - (k+1-1) + \left [ 1 - \mathbb{P}\left ( \bigcap_{j=1}^{k} F_j \bigcup F_{k+1} \right ) \right ]$

As, $\mathbb{P}\left ( \bigcap_{j=1}^{k} F_j \bigcup F_{k+1} \right )$ is a probability value and it should lie between 0 and 1,

$1 - \mathbb{P}\left ( \bigcap_{j=1}^{k} F_j \bigcup F_{k+1} \right ) \ge 0$. So,

$\mathbb{P}\left ( \bigcap_{j=1}^{k+1} F_j\right ) \ge \sum_{j=1}^{k+1}\mathbb{P}(F_j) - (k+1-1)$

and hence the inequality holds for n = k + 1

Hence by mathematical induction,

$\mathbb{P}\left ( \bigcap_{j=1}^{n} F_j\right )\ge \sum_{j=1}^{n}\mathbb{P}(F_j) - (n-1)$