From Algorithms to Z-Scores - matloff - University of California, Davis

19.4. TRANSFORM METHODS 389
transform approach.
From Section 19.4.3:
where ν = µ1 + µ2.
gN(t) = gN1 (t)gN2 (t) = e−ν+νt
(19.66)
But the last expression in (19.66) is the generating function for a Poisson distribution too! And
since there is a one-to-one correspondence between distributions and transforms, we can conclude
that N has a Poisson distribution with parameter ν. We of course knew that N would have mean
ν but did not know that N would have a Poisson distribution.
So: A sum of two independent Poisson variables itself has a Poisson distribution. By induction,
this is also true for sums of k independent Poisson variables.
19.4.5 Random Number of Bits in Packets on One Link
Consider just one of the two links now, and for convenience denote the number of packets on the
link by N, and its mean as µ. Continue to assume that N has a Poisson distribution.
Let B denote the number of bits in a packet, with B1, ..., BN denoting the bit counts in the N
packets. We assume the Bi are independent and identically distributed. The total number of bits
received during that time period is
T = B1 + ... + BN
(19.67)
Suppose the generating function of B is known to be h(s). Then what is the generating function of
T?
Here is how these steps were made:
gT (s) = E(s T ) (19.68)
= E[E(s T |N)] (19.69)
= E[E(s B1+...+BN |N)] (19.70)
= E[E(s B1 |N)...E(s BN |N)] (19.71)
= E[h(s) N ] (19.72)
= gN[h(s)] (19.73)
= e −µ+µh(s)
(19.74)