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Let $X_1,\cdots X_n$ i.i.d bernoulli(p) r'v's

Let $A = \sum_{i=1}^nX_i \geq l$, and $B= \sum_{i=1}^n X_i \geq k$

Then $A \circ B = \sum_{i=1}^nX_i \geq l+k$ (disjoint occurance of A, B)

Im trying to see why $\mathbb{P}(A\circ B)\leq \mathbb{P}(A)\mathbb{P}(B)$ holds without using the BK-inequality.

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possible duplicate of showing sum-inequality –  Did Nov 2 '12 at 7:14
    
You could show why $[S_n\geqslant\ell+k]$ is indeed $A\circ B$. –  Did Nov 2 '12 at 7:15

1 Answer 1

up vote 2 down vote accepted

Let $S_j = \sum \limits^j X_i, \tau_l = $ first hitting time to l. $\mathbb P (S_n > k+l) = \sum_i \mathbb P (\tau_l = i) \mathbb P(S_{n-i} > k)$ $\le \sum \mathbb P (\tau_l = i) \mathbb P(S_n > k) = $$ \mathbb P(S_n > l) \mathbb P(S_n > k)$

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