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There are $N$ random variables $X_1,\dots X_N$ and $Pr(X_i=1)=p$ $\forall i\in N$. Can we upper bound the probability that all random variables are $1$, i.e., $Pr(X_i=1,\forall i\in N)$. Note that the random variables are not independent.

Edit: How about a lower bound? Looking for answers other than $0,1$.

My attempt: I am thinking the product (as if independent) is a upperbound, but not sure.

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  • $\begingroup$ Well, 1 is an upper bound (containing very little information). $\endgroup$ Mar 20, 2014 at 18:21
  • $\begingroup$ @EricTowers you supposedly want something converging to $0$ as $N \to \infty$, like OP's product, which results in $p^N$... $\endgroup$
    – gt6989b
    Mar 20, 2014 at 18:22
  • $\begingroup$ without additional assumptions you can't do better than $p$, it's optimal for all $X_i$ representing the same r.v. $\endgroup$
    – mm-aops
    Mar 20, 2014 at 18:23
  • $\begingroup$ @mm-aops $p$ sound good, thanks. Could you also suggestion a lower bound please ? will $p^N$ work for a lower bound? $\endgroup$
    – triomphe
    Mar 20, 2014 at 18:53
  • $\begingroup$ it won't, if $p$ is small enough (with respect to $N$) you can get $0$, for example for $p < 1/2$ you can choose disjoint events. if $p$ is larger you can still get $0$, but this will require $N$ to be sufficiently big $\endgroup$
    – mm-aops
    Mar 20, 2014 at 19:01

2 Answers 2

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You are asking about the intersection of $N$ events. The simplest bounds are the Frechet inequalities: $$ P(A_1)+...+P(A_N)-(N-1) \leq P(A_1 \cap... \cap A_N) \leq \min\left[P(A_1), ..., P(A_N) \right] $$ There are also some more advanced lower bounds in this paper: http://www.jstor.org/discover/10.2307/1426084?uid=3739696&uid=2&uid=4&uid=3739256&sid=21106787384053

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I guess a lower bound on the probability of the intersection can be obtained by the union-bound:

\begin{align} \textrm{Pr}[\cap_i X_i] &= 1-\textrm{Pr}[\cup_i \overline{X_i}]\\ &\geq1-N\cdot\textrm{Pr}[\overline{X_1}]\\ &=1-N\cdot(1-p)\enspace, \end{align} which gives a nontrivial bound for $p>1-\frac{1}{N}$.

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