May I know how to exclude tied rank, and interpret the sum of the Poisson random variables?for the first part of the question, my interpretation is that: V a l u e R a n k 1 1 2 2 2 2 (Tie rank) 2 2 (Tie rank) 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 ∼ P o ( λ 1 ) , a n d Y ∼ P o ( λ 2 ). We can conclude that ( X + Y ) ∼ P 0 ( λ 1 + λ 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.

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May I know how to exclude tied rank, and interpret the sum of the Poisson random variables?for the first part of the question, my interpretation is that:                     V        a        l        u        e                    R        a        n        k                            1                    1                            2                    2                            2                    2                    (Tie rank)                            2                    2                    (Tie rank)            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  ∼      P    o    (      λ    1    )  ,  a  n  d  Y  ∼      P    o    (      λ    2    ). We can conclude that   (  X  +  Y  )  ∼      P    0    (      λ    1    +      λ    2    ).Thank you so much for your reply, guys. And sorry for the fact that I don&#039;t know how to insert columns in the question, so I can only attach picture for illustration.

zalitiaf · Accepted Answer

The complete proof that   X  +  Y  ∼      P    o    i    s    (      λ    1    +      λ    2    ) when X and Y are independent is as follows: let   n  ⩾  0 be a nonnegative integer. Then                              P                (        X        +        Y        =        n        )                            =                  ∑                      k            =            0                    n                          P                (        X        =        k        )        (        Y        =        n        −        k        )                                          =                  ∑                      k            =            0                    n                          e                      −                          λ              1                                                            λ            1            k                                k            !                                    e                      −                          λ              2                                                            λ            2                          n              −              k                                            (            n            −            k            )            !                                                            =                  ∑                      k            =            0                    n                          1                      k            !            (            n            −            k            )            !                                    e                      −                          λ              1                        k                                    λ          1          k                          e                      −                          λ              2                        (            n            −            k            )                                    λ          2                      n            −            k                                                            =                  ∑                      k            =            0                    n                                      n            !                                k            !            (            n            −            k            )            !                                                              e                              −                                  λ                  1                                k                                                    λ              1              k                                      e                              −                                  λ                  2                                (                n                −                k                )                                                    λ              2                              n                −                k                                                          n            !                                                            =                  ∑                      k            =            0                    n                                                    (                                            n            k                                              )                                                                          e                              −                                  λ                  1                                k                                                    λ              1              k                                      e                              −                                  λ                  2                                (                n                −                k                )                                                    λ              2                              n                −                k                                                          n            !                                                            =                              e                          −              λ                                            n            !                                    ∑                      k            =            0                    n                                                    (                                            n            k                                              )                                                λ          1          j                          λ          2                      n            −            j                                                            =                              e                          −              λ                                            n            !                          (                  λ          1                +                  λ          2                          )          n                                                  =                                            e                              −                λ                                                    λ              n                                            n            !                          ,            so that       P    (  X  +  Y  =  n  )  =                    e                  −          λ                            λ        n                    n      !       where   λ  =      λ    1    +      λ    2  , where we have used independence of X and Y, discrete convolution of X and Y and the binomial theorem.

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

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