Question 2.45: Calculate the distribution of X + Y when X and Y are indepen......
Calculate the distribution of X + Y when X and Y are independent Poisson random variables with means \lambda_{1}{\mathrm{~and~}}\lambda_{2}, respectively.
\phi_{X+Y}(t)=\phi_{X}(t)\,\phi_{Y}(t)
=e^{\lambda_{1}(e^{t}-1)}e^{\lambda_{2}(e^{t}-1)}
=e^{(\lambda_{1}+\lambda_{2})(e^{t}-1)}
Hence, X + Y is Poisson distributed with mean \lambda_{1}{\mathrm{~and~}}\lambda_{2}, verifying the result given in Example 2.37.
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