Question 2.37: (Sums of Independent Poisson Random Variables) Let X and Y b......
(Sums of Independent Poisson Random Variables) Let X and Y be independent Poisson random variables with respective means λ_{1} and λ_{2}. Calculate the distribution of X + Y.
Since the event {X + Y = n} may be written as the union of the disjoint events {X = k, Y = n − k}, 0 ≤ k ≤ n, we have
P\{X+Y=n\}=\sum_{k=0}^{n}P\{X=k,\ Y=n-k\}=\sum_{k=0}^{n}P\{X=k\}P\{Y=n-k\}
=\sum_{k=0}^{n}e^{-\lambda_{1}}{\frac{\lambda_{1}^{k}}{k!}}e^{-\lambda_{2}}{\frac{\lambda_{2}^{n-k}}{(n-k)!}}
=e^{-(\lambda_{1}+\lambda_{2})}\sum_{k=0}^{n}\frac{\lambda_{1}^{k}\lambda_{2}^{n-k}}{k!(n-k)!}
={\frac{e^{-(\lambda_{1}+\lambda_{2})}}{n!}}\sum_{k=0}^{n}{\frac{n!}{k!(n-k)!}}\lambda_{1}^{k}\lambda_{2}^{n-k}
={\frac{e^{-(\lambda_{1}+\lambda_{2})}}{n!}}(\lambda_{1}+\lambda_{2})^{n}
In words, X_{1}+X_{2} has a Poisson distribution with mean \lambda_{1}+\lambda_{2}.
Related Answered Questions
Question: 2.27
Verified Answer:
Recalling (see Example 2.22) that E[X] = μ, we hav...
Question: 2.53
Verified Answer:
Surprisingly enough, the probability that node i i...
Question: 2.52
Verified Answer:
If we let X_{i} denote the lifetime...
Question: 2.51
Verified Answer:
Since E[X_{i}]={\frac{1}{2}},\mathrm{Var}(X...
Question: 2.49
Verified Answer:
Let X be the number of items that will be produced...
Question: 2.44
Verified Answer:
The moment generating function of X + Y is given b...
Question: 2.39
Verified Answer:
The joint density of X and Y is given by
f_...
Question: 2.45
Verified Answer:
\phi_{X+Y}(t)=\phi_{X}(t)\,\phi_{Y}(t)[/lat...
Question: 2.36
Verified Answer:
From Equation (2.18),
f_{X+Y}(a)={\frac{d}{...
Question: 2.33
Verified Answer:
To show that f(x, y) is a joint density function w...