If $X_i$ is poisson. I know that the mgf of $X_i$ is $e^{\lambda(e^t-1)}$.

What would the distribution of $Y$ look like if $Y$ = the sum of all $X_i$?

is it Poisson itself?

  • 1
    $\begingroup$ Without knowing their joint distribution, we cannot answer anything. On one extreme, where $X_i$'s are independent, then their sum again has Poisson distribution. On the other extreme, if $X_1 = \cdots = X_n$, then their sum is just $nX_1$, which is no longer Poisson. $\endgroup$ – Sangchul Lee Feb 19 at 1:29

It is true that if $X_1, \dots, X_n$ are independent Poisson variables with mean $\lambda$, then $Y := X_1 + \dots + X_n$ is Poisson with mean $n\lambda$.

To see this, observe that the moment generating function of $Y$ is $$ \left( e^{\lambda (e^t - 1)} \right)^n = e^{n\lambda (e^t - 1)},$$ which is precisely the moment generating function of a Poisson variable with mean $n\lambda$.

This should feel intuitive. Imagine you have a Poisson process, where the expected number of events per unit time is $\lambda$. If for each $i \in \{1, \dots, n\}$, $X_i$ represents the number of events observed in time interval $[i - 1 , i)$, then $Y = X_1 + \dots + X_n$ represents the number of events observed in time interval $[0, n)$, and this is Poisson-distributed with mean $n\lambda$.

  • $\begingroup$ how would it be different if it was Gamma assuming t he $$X_i $$ are iid? How would that look? $\endgroup$ – George Harrison Feb 19 at 1:33
  • $\begingroup$ @GeorgeHarrison How about trying the same method? The mgf for the $\rm{Gamma}(\alpha, \beta)$ is $(1 - t / \beta)^{-\alpha}$. So if you have $n$ such variables, the mgf for the sum is $(1 - t /\beta)^{-n\alpha}$, which is the mgf for $\rm{Gamma}(n\alpha, \beta)$. $\endgroup$ – Kenny Wong Feb 19 at 1:37

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