Question

The moment-generating function of the random variable x having a chi-squared distribution with v degrees of freedom is

Answer 1

Demonstrate how the moments of a random variable x may be obtained
from its moment generating function by showing that the rth derivative of
E(ext) with respect to t gives the value of E(xr) at the point where t = 0.
Show that the moment generating function of the Poisson p.d.f. f(x) =
e−µµx/x!; x ∈ {0, 1, 2,...} is given by M(x, t) = exp{−µ} exp{µet
}, and
thence find the mean and the variance.
2. Demonstrate how the moments of a random variable may be obtained from
the derivatives in respect of t of the function M(t) = E{exp(xt)}.
If x = 1, 2, 3,... has the geometric distribution f(x) = pqx−1, where q =
1 − p, show that the moment generating function is
M(t) = pet
1 − qet .
Find E(x).
Answer: The moment generating function of x is
M(t) = X∞
x=1
extpqx−1 = p
q
X∞
x=1
¡
qet
¢x
= pet
X∞
x=0
¡
qet
¢x = pet
1 − qet .
To find E(x), we may use the quotient rule to differentiate the expression
M(t) with respect to t. This gives
dM(t)
dt = (1 − qet
)pet − pet
(−qet
)
(1 − qet)2 .
Setting t = 0 gives E(x)=1/p.
3. Let xi;i = 1,...,n be a set of independent and identically distributed
random variables. If the moment generating function of xi is M(xi, t) =
E{exp(xit)} for all i, find the moment generating function for y = Pxi.
Find the moment generating function of a random variable xi = 0, 1
whose probability density function if f(xi) = (1 − p)1−xi pxi , and thence
find the moment generating function of y = Pxi. Find E(y) and V (y).