May I know how to exclude tied rank, and interpret the sum of the Poisson random variables?

Question

for the first part of the question, my interpretation is that:

So, there are two tied ranks, so I should delete them from the data set?

For the second part of the question, I want to ask how do I interpret the fact that the sum of independent Poisson random variables has a Poisson distribution. my interpretation is that : If $X \sim P_o(\lambda_1) , and Y \sim P_o(\lambda_2)$. We can conclude that $ (X+Y)\sim P_0(\lambda_1 + \lambda_2) $.

Thank you so much for your reply, guys. And sorry for the fact that I don't know how to insert columns in the question, so I can only attach picture for illustration.

Answer 1

The complete proof that $X+Y\sim\mathrm{Pois}(\lambda_1+\lambda_2)$ when $X$ and $Y$ are independent is as follows: let $n\geqslant 0$ be a nonnegative integer. Then \begin{align} \mathbb P(X+Y=n) &=\sum_{k=0}^n \mathbb P(X=k)(Y=n-k)\\ &= \sum_{k=0}^ne^{-\lambda_1}\frac{\lambda_1^k}{k!}e^{-\lambda_2}\frac{\lambda_2^{n-k}}{(n-k)!}\\ &= \sum_{k=0}^n \frac1{k!(n-k)!}e^{-\lambda _1k}\lambda_1^ke^{-\lambda_2(n-k)}\lambda_2^{n-k}\\ &= \sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{e^{-\lambda _1k}\lambda_1^ke^{-\lambda_2(n-k)}\lambda_2^{n-k}}{n!}\\ &= \sum_{k=0}^n \binom nk \frac{e^{-\lambda _1k}\lambda_1^ke^{-\lambda_2(n-k)}\lambda_2^{n-k}}{n!}\\ &= \frac{e^{-\lambda}}{n!}\sum_{k=0}^n \binom nk \lambda_1^j\lambda_2^{n-j}\\ &=\frac{e^{-\lambda}}{n!} (\lambda_1+\lambda_2)^n\\ &= \frac{e^{-\lambda}\lambda^n}{n!}, \end{align} so that $\mathbb P(X+Y=n) = \frac{e^{-\lambda}\lambda^n}{n!}$ where $\lambda = \lambda_1+\lambda_2$, where we have used independence of $X$ and $Y$, discrete convolution of $X$ and $Y$ and the binomial theorem.