If A and B have the same distribution, e.g. Poisson, and they are independent. Is $P(A + B = k) = P(A = k)P(B = k)$ ? If so, why? If not, what is the correct way to calculate $P(A + B = k)$?
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No. Having $A + B = k$ is not equivalent to $A = k$ and $B = k$, as in the latter case you would have $A + B = 2k$. To find when $A + B = k$ you have to consider all the ways that $A$ and $B$ could sum to $k$. This means that $$P(A + B = k) = \sum_j P(A = j) P(B = k-j),$$ where the sum is over all feasible values of $j$ (i.e., all values of $j$ where you could actually have $A = j$ and $B = k-j$). This operation is called a (discrete) convolution of the common probability mass function of $A$ and $B$ with itself. |
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