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If $N = \sum_{i=0}^{n} I_i$ where $I_i$ is the indicator variable of event $A_i$, with $P(A_i)=p_i$, how can one show that

$$ P(N=0) \leq \frac{\operatorname{Var}(N)}{E(N)^2} $$

using Chebyshev's inequality or otherwise?

I don't know how to apply Chebyshev's inequality here – it doesn't seem like there's anything that has the right form.

Edit: The first part of the question is to calculate the expectation and variance of $N$ in terms of $p_i$ and $p_{ij} = P(A_i \cap A_i)$ - perhaps this could be useful for the second part?


Edit: Full question:

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  • $\begingroup$ How do you know $p_{ij}$ without independence of events $A_i$'s? Chebyshev/Markov holds for all rv w/o looking into their characteristics, so in most cases, those bounds are no good. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 27 '20 at 13:39
  • $\begingroup$ We don't know what the $p_{ij}$ are, we just have to find the variance of N in terms of $p_i$ and $p_{ij}$. $\endgroup$ – asfjbkjabf Nov 27 '20 at 13:40
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not viewable to some, such as those who use screen readers. Scanned pages from books are discouraged on SE network. Questions should contain sufficient context so that it is answerable with the text alone. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 27 '20 at 13:42
  • $\begingroup$ The calculations for $var(N)$ and $E(N)$ are irrelevant. Chebyshev is just Markov's inequality applied to "squared deviation from average" $(X-E[X])^2$. Markov's inequality's proof is incredibly simple: get a whole $E[X]$, truncate it with $1_{\{X \ge a\}}$, write out the inequality and bounds, and wrap out things. You may replace $E[X]$ with $var(X)$ and rework the proof. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 27 '20 at 13:48
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I think I have an answer to this question but would appreciate any feedback! :)


Firstly note that $$ P(N=0) \leq P(|N-E(N)| = E(N)) $$

We know that

$$ P(|N-E(N)| \geq E(N)) \leq \frac{Var(N)}{E(N)^2} $$ using Chebyshev.

We want to connect these two expressions. We can do this by noting that $$ P(|N-E(N)| \geq E(N)) = P(|N-E(N)| = E(N)) + P(|N-E(N)| > E(N)) \geq P(|N-E(N)| = E(N)) $$

Therefore, putting this all together, we have

$$ P(N = 0) \leq P(|N-E(N)| = E(N)) \leq P(|N-E(N)| \geq E(N))\leq \frac{Var(N)}{E(N)^2} $$

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  • $\begingroup$ The first equality is wrong: what about the event $\{N = 2 E[N] = 2 \sum_i p_i \}$? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Nov 27 '20 at 14:24
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    $\begingroup$ Good point thanks. I have edited it $\endgroup$ – asfjbkjabf Nov 27 '20 at 14:25

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