Moment Generating Functions in Probability Theory

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: ? ?? M(t) = ? = E(? = E(x ???) ?? E(???) ?????)

Similarly, ?2 ??2 M(t) = ?2 = E(?2 = E(?2???) Hence, in general we get ?? ??? M(t) =?? = E(?? = E(?????) ??2 E(???) ??2???) ???E(???) ??????)

If we set t = 0 in the nth derivative, we get ?? ??? M(t) = E(??) ?=0 = E(?????)?=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.

?? ??? ? > ? ????????? f ? = ? Answer: What are the mean and variance of X? The moment generating function of X is

M(t) = E(???) ???? ? ?? = 0 ???? ??? = 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.

??????? =0 ? (1 ?)??? 1 1 ?? (1 ?)? = 0 0 1 = if ? ? > ? 1 ? The expected value of X can be computed from M(t) as ? ?? M(t) (1 ?) 2 ? ?? (1 ?) 1 ?=0= ? ? = ?=0 = ?=0 1 = (1 ?)2 ?=0 = 1

Similarly ?2 ??2 M(t) ?=0= ? ?2= ?2 ??2 (1 ?) 1 ?=0 = 2 (1 ?) 3 ?=0 2 = (1 ?)3 ?=0 = 2

Therefore, the variance of X is Var(x)= = ? ?? (?)? = 2 1 = 1 Example 4.16. Let the random variable X have moment generating function M(t) = (1 ?) 2for 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 = = x ( ; , ) b x n p p q x , . n , 0,1,

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 is M(t) = ?4.6(?? 1) , 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) = ?4.6(?? 1)

weconclude that X 1 POI(?) with ? = 4.6. Thus, by Theorem 5.8, we get E(x) = 4.6 = Var(x) ? 3 < ? < 6 = ? 4 + ? 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 ? if its probability density function is of the form where ? > 0. If a random variable X has an exponential density function with parameter ?, then we denote it by writing X ~??? (? ).

Mean and variance of exponential distribution 1 ? ? ? = ? = 1 ??? ? = ?2= ?2 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 1 2(? ? ?)2 1 < X < ? ? = ? 2?? where < < and 0 < ?2< are arbitrary parameters. If X has a normal distribution with parameters and ?2, then we write X ~ ?( ?, ?2).

Theorem 6.6. If ?~ N(?,?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 ~ ? (0,1 )

The probability density function of standard normal distribution is the following 2?? ? 2 1 < X < ? ? = 2 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) = P(43 ? 3 8 3 4) 4 4 = P(1 4 ? 5 4) = P (Z 1.25) P (Z 0.25) = 0.8944 0.5987 = 0.2957.