Consider $X_{1},…,X_{30}$ independent Poisson random variables with mean 1.
I need to find a lower bound for
$$
P(25 \le \sum_{i=1}^{30}X_{i} \le 35)
$$
My first thought was that:
$$
\bar{X}_{30} = \frac{1}{30}\sum_{i=1}^{30}X_{i}
$$
so
$$
P(25 \le \sum_{i=1}^{30}X_{i} \le 35) = P(\frac{25}{30} \le \bar{X}_{30} \le \frac{35}{30}) \ge 1 – \frac{V(\bar{X}_{30})}{(5/30)^{2}}
$$
by Chebyshev's inequality since the mean is 1 and $\frac{25}{30} = 1 – 5/30$ and $\frac{35}{30} = 1 + 5/30$. However, this inequality gives me a nonsense result since $V(\bar{X}_{30}) = \sigma^{2}/30$ with $\sigma = 1$ since it's a Poisson distribution.
Where am I going wrong…?
Best Answer
Revised solution
Ok, after some thought, I realized the following:
The complement of the event in Chebyshevs inequality $P[|Y-30|\geq 5] \leq \frac{30}{25}$ is not $P[|Y-30|\leq 5] \geq 1- \frac{30}{25}$ but $P[|Y-30|< 5] > 1- \frac{30}{25}$ (changes to strict inequality) The discreteness of the Poisson makes this distinction important. Knowing this we can apply Chebyshev directly to your problem. But...we must apply it to $P[|Y-30|\geq 6]$ with the upper bound being $\frac{30}{36} = \frac{5}{6}$. Now, due to the discreteness of the Poisson counts, $P[|Y-30|< 6] = P[|Y-30|\leq 5] >1-\frac{5}{6} = \frac{1}{6}$ Which gives $0.16$. Comparing this to the true value of 0.63, we see how loose/conservative Chebyshev's bound is.
However, Chebyshev's inequality is definitely not the tightest bound out there. Since your RVs are independent, I'd take a look at Chernoff Bounds (also)which are tighter. Note that the sum of independent Poisson RVs is also Poisson (in your case Poisson(30)), so it can be directly applied.