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Example 4.12.
Let X be a random variable whose moment m generating
function is M(t) and n be any natural number. What is the
nth derivative of M(t) at t = 0?
Answer:
𝑑
𝑑𝑡
𝑑
M(t) =
𝑑𝑡
𝑑
= E( 𝑒 𝑡𝑥 )
𝑑𝑡
= E(x 𝑒 𝑡𝑥 )
E(𝑒 𝑡𝑥 )
Similarly,
𝑑2
𝑑𝑡 2
=
=
𝑑2
M(t) = 2
𝑑𝑡
2
𝑑
E( 2 𝑒 𝑡𝑥 )
𝑑𝑡
E(𝑥 2 𝑒 𝑡𝑥 )
E(𝑒 𝑡𝑥 )
Hence, in general we get
𝑑𝑛
𝑑𝑡 𝑛
𝑑𝑛
M(t) = 𝑛 E(𝑒 𝑡𝑥 )
𝑑𝑡
𝑛
𝑑
E( 𝑛 𝑒 𝑡𝑥 )
𝑑𝑡
𝑛 𝑡𝑥
=
= E(𝑥 𝑒 )
If we set t = 0 in the nth derivative, we get
𝑑𝑛
𝑑𝑡 𝑛
M(t)
= E(𝑥 𝑛 𝑒 𝑡𝑥 )
𝑡=0
= E(𝑥 𝑛 )
𝑡=0
Hence the nth derivative of the moment generating function of X
evaluated
at t = 0 is the nth moment of X about the origin.
𝑒 −𝑥 𝐟𝐨𝐫 𝐱 > 𝟎
f 𝐱 =
𝟎
𝐨𝐭𝐡𝐞𝐫𝐰𝐢𝐬𝐞
Answer:
What are the mean and variance of X?
The moment generating function of X is
M(t) = E(𝑒 𝑡𝑥 )
=
∞ 𝑡𝑥
𝑒 𝑓
0
𝑥 𝑑𝑥
∞
𝑒 𝑡𝑥 𝑒 −𝑥 𝑑𝑥 =
0
Hence
the nth derivative of the moment generating function of X
evaluated
at t = 0 is the nth moment of X about the origin.
∞ 𝑡𝑥 −𝑥
𝑒 𝑒 𝑑𝑥
0
=
∞ −(1−𝑡)𝑥
𝑒
𝑑𝑥
0
=
1
1−𝑡
if
=
1
1−𝑡
𝑒 −(1−𝑡)𝑥
∞
0
𝟏−𝐭 >𝟎
The expected value of X can be computed from M(t) as
𝐸 𝑥 =
𝑑
𝑑𝑡
M(t)
= (1 − 𝑡)−2
=
𝑡=0
𝑡=0
1
(1−𝑡)2
=
𝑡=0
= 1
𝑑
𝑑𝑡
(1 − 𝑡)−1
𝑡=0
Similarly
𝐸
𝑑2
𝑑𝑡 2
𝑥2
=
𝑑2
𝑑𝑡 2
(1 − 𝑡)−1
= 2 (1 − 𝑡)−3
=
M(t)
𝑡=0
𝑡=0
2
(1−𝑡)3
𝑡=0
𝑡=0
= 2
=
Therefore, the variance of X is
Var(x)= = 𝐄 𝐱 𝟐 − ( 𝝁)𝟐
= 2 – 1 =1
Example 4.16.
Let the random variable X have moment generating function
M(t) = (1 − 𝑡)−2 for t < 1. What is the third moment of X about the
origin?
Answer:
To compute the third moment E(X3) of X about the origin, we
need to compute the third derivative of M(t) at t = 0.
SOME SPECIAL
DISCRETE
DISTRIBUTIONS
5.2. Binomial Distribution
Definition 5.2.
The random variable X is called the binomial random variable
with parameters p and n if its probability density function is of the
form
 n  x n x
b (x ; n , p )    p q , x  0,1,
x 
, n.
Theorem 5.2.
If X is binomial random variable with parameters p and n , then
the mean, variance and moment generating functions are
respectively given by
5.6. Poisson Distribution
Definition 5.6.
A random variable X is said to have a Poisson distribution
if its probability density function is given by
Example 5.23.
If the moment generating function of a random variable X
𝑡−1)
4.6(𝑒
𝑒
is M(t) =
, then what are the mean and variance of X?
What is the probability that X is between 3 and 6,
that is P(3 < X < 6)?
Answer:
Since the moment generating function of X is given by
M(t) =
𝑡−1)
4.6(𝑒
𝑒
we conclude that X 1 POI(𝜆) with 𝜆 = 4.6. Thus, by
Theorem 5.8, we get
E(x) = 4.6 = Var(x)
𝑃 3<𝑥 <6 =𝑓 4 +𝑓 5
= F 6 − F(3)
= 0.686 − 0.326
= 0.36.
SOME SPECIAL CONTINUOUS DISTRIBUTIONS
Exponential Distribution
Definition 6.3.
A continuous random variable is said to be an exponential
random variable with parameter 𝜃 if its probability density
function is of the form
where 𝜃 > 0. If a random variable X has an exponential
density function
with parameter 𝜃, then we denote it by writing X ~𝐸𝑥𝑝 (𝜃 ).
Mean and variance of exponential distribution
1
𝐸 𝑥 = 𝜆 =
𝜃
𝑉𝑎𝑟 𝑥 = 𝜆2 =
1
𝜃2
Moment generating function:
Example 6.12.
What is the cumulative density function of a random variable
which has an exponential distribution with variance 25?
Answer:
6.4. Normal Distribution
Definition 6.7.
A random variable X is said to have a normal distribution
if its probability density function is given by
𝑓 𝑥 =
−1 𝑥−𝜇 2
1
𝑒2( 𝜎 )
𝜎 2𝜋
−∞ < X < ∞
where −∞ < μ < ∞ and 0 < 𝜎 2 < ∞ are arbitrary
parameters. If X has a
normal distribution with parameters μ and 𝜎 2 , then we
write X ~ 𝑁( 𝜇, 𝜎 2 ).
Theorem 6.6.
If 𝑋~ N(𝜇,𝜎 2 ), then
Definition 6.8.
A normal random variable is said to be standard normal, if its
mean is zero and variance is one. We denote a standard normal
random variable X by X ~ 𝑁 (0,1 )
The probability density function of standard normal distribution
is the following
2
𝑓 𝑥 =
𝑥
1
−2
𝑒
2𝜋
−∞ < X < ∞
Example 6.22.
If X 1 N(0, 1), what is the probability of the random variable X
less than or equal to −1.72?
Answer
P(X ≤−1.72) = 1 − P(X ≤1.72)
= 1 − 0.9573
(from table)
= 0.0427.
Example 6.24.
If X ~ N(3, 16), then what is P(4 ≤X ≤ 8)?
Answer:
P (4 ≤ X ≤ 8) =
=
1
P(
4
≤𝑍≤
4−3
P(
4
≤
𝑋−3
4
≤
5
)
4
= P (Z ≤ 1.25) − P (Z ≤ 0.25)
= 0.8944 − 0.5987
= 0.2957.
8−3
)
4