Let Ω = N 0 2 , A = P o t ( Ω ) and P the product of two poisson distributions with paremeter λ 1 , λ &gt; 0, i.e P ( { ( n 1 , n 2 ) } ) = λ 1 n 1 λ 2 n 2 n 1 ! n 2 ! e − λ 1 − λ 2 Define then X : Ω → N 0 , ( n 1 , n 2 ) ↦ n 1 + n 2 . Prove that X is poisson distributed with parameter λ 1 + λ 2

Question

Let   Ω  =            N        0    2    ,      A    =  P  o  t  (  Ω  ) and P the product of two poisson distributions with paremeter       λ    1    ,      λ    &amp;gt;    0, i.e      P    (  {  (      n    1    ,      n    2    )  }  )  =                    λ        1                              n            1                                      λ        2                              n            2                                              n        1            !              n        2            !            e          −              λ        1            −              λ        2            Define then   X  :  Ω  →            N        0    ,     (      n    1    ,      n    2    )  ↦      n    1    +      n    2  . Prove that X is poisson distributed with parameter       λ    1    +      λ    2

erkvisin7s · Accepted Answer

Step 1Your probability space is explicitly the product space   (            N        0    ×            N        0    ,      P    (            N        0    )  ⊗      P    (            N        0    )  ,            P                      λ        0              ×            P                      λ        1              ), where             P              λ       is the Poisson probability measure on   (            N        0    ,      P    (            N        0    )  ). Since you have explicitly defined probabilities on the non-negative integers, you don&#039;t need a random variable to map from another underlying probability space: you already have             P                      λ        0              (  {      n    1    }  )  =            λ      1                        n          1                                    n        1            !            e          −              λ        1             for all       n    1    ∈            N        0  .Step 2Technical point: Remember that a probability measure is defined on sets in the sigma-algebra. For a discrete measure, it is sufficient to define it on the singletons   {  {  n  }  :  n  ∈            N        0    }.)Thus, to &quot;fix&quot; the notation in your proof, we can write                              P                (        X        =        n        )                            =                  P                (        {        (                  n          1                ,                  n          2                )        :        X        (                  n          1                ,                  n          2                )        =        n        }        )                                          =                  P                (        {        (                  n          1                ,                  n          2                )        :                  n          1                +                  n          2                =        n        }        )                                          =                  P                          (                      ⋃                          k              =              0                        n                    {          (                      n            1                    ,                      n            2                    )          :                      n            1                    =          k          ,                      n            2                    =          n          −          k          }          )                =        …            The first line simply clarifies what the usual notation       P    (  X  =  n  ) means: it is the probability of the event where the random variable takes on a certain value (i.e. set of 2-tuples   (      n    1    ,      n    2    ) that add to n). The rest continues as you have written. After splitting into a sum (by additivity), for each   0  ≤  k  ≤  n we will have       P    (  {  (      n    1    ,      n    2    )  :      n    1    =  k  ,      n    2    =  n  −  k  }  )  =            P                      λ        1              (  {  k  }  )            P                      λ        2              (  {  n  −  k  }  ) since we have a product measure.This notation is indeed not as easy to understand as the usual random variable notation (which allows for a more &quot;probabilistic&quot; way of thinking). By abstracting away the underlying probability space, we can simply say that       X    1    ∼      P    o    i    s    s    o    n    (      λ    1    ) and       X    2    ∼      P    o    i    s    s    o    n    (      λ    2    ) are two independent random variables on the same probability space, so that for   X  =      X    1    +      X    2   we can simply write       P    (  X  =  n  )  =      P    (      X    1    +      X    2    =  n  )  =      ∑          k      =      0        n        P        (          X      1        =    k    ,          X      2        =    n    −    k    )    =  …. The key takeaway is that product measures are essentially the same as independent random variables.

Gisselle Hodges · Accepted Answer

Step 1                                    P                (        X        =        n        )                            =                              P                (        {        (                  n          1                ,                  n          2                )        :                  n          1                +                  n          2                =        n        }        )                            (1)                    =                              P                          (          {          (                      n            1                    ,                      n            2                    )          :                      n            1                    +                      n            2                    =          n          }          ⋂                      ⋃                          k              =              0                        ∞                    {          (                      n            1                    ,                      n            2                    )          :                      n            1                    =          k          }          )                                    (2)                    =                              P                          (                      ⋃                          k              =              0                        ∞                    (          {          (                      n            1                    ,                      n            2                    )          :                      n            1                    +                      n            2                    =          n          }          ∩          {          (                      n            1                    ,                      n            2                    )          :                      n            1                    =          k          }          )          )                                    (3)                    =                              P                          (                      ⋃                          k              =              0                        n                    (          {          (                      n            1                    ,                      n            2                    )          :                      n            1                    +                      n            2                    =          n          }          ∩          {          (                      n            1                    ,                      n            2                    )          :                      n            1                    =          k          }          )          )                                    =                              P                          (                      ⋃                          k              =              0                        n                    {          (                      n            1                    ,                      n            2                    )          :                      n            1                    =          k          ,                      n            2                    =          n          −          k          }          )                                    (4)                    =                                      ∑                      k            =            0                    n                          P                (        {        (                  n          1                ,                  n          2                )        :                  n          1                =        k        ,                  n          2                =        n        −        k        }        )                            =                                      ∑                      k            =            0                    n                          P                (        {        (        k        ,        n        −        k        )        }        )                            (5)                    =                                      ∑                      k            =            0                    n                                                    λ              1              k                                      λ              2                              n                −                k                                                          k            !            (            n            −            k            )            !                                    e                      −            (                          λ              1                        +                          λ              2                        )                                              =                                          (                          λ              1                        +                          λ              2                                      )              n                                            n            !                                    e                      −            (                          λ              1                        +                          λ              2                        )                          .            This shows   X  ∼  Poisson  (      λ    1    +      λ    2    )Step 2Details explanation:(1): because   Ω  =      ⋃          k      =      0        ∞    {  (      n    1    ,      n    2    )  :      n    1    =  k  }  )(2): set operation.(3): when   k  &amp;gt;  n, the intersection is   ∅.(4): These n sets are disjoint.(5): Use the specification of P directly.

Let Omega=N_0^2, A=Pot(Omega) and P the product of two poisson distributions with paremeter lambda_1, λ>0, i.e

Answered question

Answer & Explanation

New Questions in High school probability