RISK MATHEMATICS (ASC512)
KHULUMA ZEBRON
ASSIGNMENT 1
DEPARTMENT OF MATHEMATICS
EDWARD MASOAMBETA
10TH OCTOBER 2022
BSC/COM/08/17
Question 1
Argue that the probability generating function PSn (z), of the sum of n variables is simply the product of:
PSn (z) =
n
Y
PXj (z)
j=1
SOLUTION
Let X1 and X2 be independent discrete random variables
Let S be a discrete random variable such that, S = X1 + X2
Then it can be seen that:
PS (z) = PX1 (z) · PX2 (z)
PROOF
Recall the definition of a Probability Generating Function, for a discrete random variable S, its PGF can be
defined as:
PS (z) = E[Z S ]
If S = X1 + X2 , we then see that
PS (z) = E[Z X1 +X2 ]
= E[Z X1 Z X2 ]
= E[Z X1 ] · E[Z X2 ]
= PX1 (z) · PX2 (z)
We can therefore let Sn = X1 + X2 + ... + Xn . Where each X1 + X2 + X3 + ... + Xn are independent discrete
random variables with PGF’s PX1 (z), PX2 (z), ..., PXn (z). Thus,
PSn (z) =
n
Y
PXj (z)
j=1
Question 2
Recall a poisson distribution. Find the MGF and PGF.
Poisson Probability Generating function
Let X be a discrete random variable with the Poisson distribution with parameter λ. It is also known that a
PGF can also be defined as:
X
PX (z) =
PX (x)z x
x≥0
Where PX (x) is the Probability Mass Function. For a Poisson Distribution, PX (k)
PX (k) =
1
e−λ λk
k!
We then see from the definitions above that:
PX (z) =
X e−λ λk
k!
k≥0
= e−λ
zk
X (λz)k
k≥0
k!
Because of the Taylor Series Expansion for Exponential Function which shows that:
ex =
X xn
n!
x≥0
It leads us to,
PX (z) = e−λ eλz
PX (z) = e−λ+λz
Poisson Moment Generating function
Let X be a discrete random variable with a Poisson distribution with parameter λ
From the definition of the Poisson distribution, X has probability mass function:
P r(X = n) =
λn e−λ
n!
We define a moment generating function as,
MX (t) = E[etX ] =
∞
X
P r(X = n)etn
n=0
We therefore see,
MX (t) =
∞
X
λn e−λ tn
e
n!
n=0
= e−λ
∞
X
(λet )n
n!
n=0
t
= e−λ eλe
= eλ(e
2
t
−1)