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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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Hint: The event on the right hand side seems to deal with a situation where $A+B = k+k = 2k$, which is unrelated to the left hand side. How could you change the $k$'s on the rhs so that it has some relationship with the lhs? – whuber Apr 13 '11 at 23:10
up vote 4 down vote accepted

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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