Let $X_1,\dots,X_n$ be independent random variables, $X_i \sim \operatorname{Poisson}(\lambda_i),$ $i=1,\dots,n.$ Let $$S=n^{-1}\sum_{i=1}^n X_i, \quad\quad \lambda=n^{-1}\sum_{i=1}^n \lambda_i.$$ Find an upper bound for $P(S-\lambda>t)$. What $t$ do we need in order $P(S-\lambda>t)\leq n^{-\tau}$ for some $\tau>0$?

I am not sure how to go about finding an upper bound for $P(S-\lambda > t)$ in this problem. Any help would be appreciated.

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    $\begingroup$ Try Cantelli's inequality. $\endgroup$ – user158565 Nov 4 '18 at 0:07
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    $\begingroup$ This seems like a poorly-worded pair of questions; as is, an upper bound of $1$ works well for the first question, and we can always find a $\tau>0$ such that $n^{-\tau} \geq 1-P(S=0)$, so any $t>-\lambda$ works for the second question. (Note $P(S=0)$ is the probability of $x=0$ where $x \sim \text{Poisson}(\sum \lambda_i)$, and is always $>0$.) $\endgroup$ – jbowman Nov 4 '18 at 0:28
  • $\begingroup$ Is this a homework question? In that case add the self-study tag. $\endgroup$ – Martijn Weterings Nov 4 '18 at 7:45

$\sum_{i=1}^n X_i$ follows Poisson distribution with parameter $\sum_{i=1}^n \lambda_i$. So $S =n^{-1} \sum_{i=1}^n X_i$ has mean $n^{-1}\sum_{i=1}^n \lambda_i = \lambda$ and variance $n^{-2}\sum_{i=1}^n \lambda_i = n^{-1}\lambda$.

According to Cantelli's inequality, $$P(S-\lambda>t) \le \frac {n^{-1}\lambda}{n^{-1}\lambda + t^2}$$

To get $P(S-\lambda>t)\leq n^{-\tau}$, we need $$\frac {n^{-1}\lambda}{n^{-1}\lambda + t^2} = n^{-\tau}$$

So $t=\sqrt{\frac {\lambda(n^{\tau}-1)}n}$

  • $\begingroup$ Could we maybe calculate a harder bound using the following? $$P(S - \lambda > t) = P(n S > n(t+\lambda)) = 1- F_\lambda(n(t+\lambda)) = 1 - \sum_{i=1}^{\lfloor n(t+\lambda) \rfloor} \frac{\lambda^i}{i!} \leq n^{-\tau} $$ where $F_\lambda(x)$ is the cumulative distribution of the Poisson distribution with rate parameter $\lambda$ (also expressed by the regularized Gamma function) $\endgroup$ – Martijn Weterings Nov 4 '18 at 8:01
  • $\begingroup$ @MartijnWeterings I think the original problem is not good. The teacher want the student to use some kind of probability inequality, but the distribution is specified. Under this situation, we can calculate the exact probability, as you suggested. Because of "upper bound", I think my Answer is what the teacher wanted. $\endgroup$ – user158565 Nov 4 '18 at 16:23

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