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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: π ππ‘ π M(t) = ππ‘ π = E( π π‘π₯ ) ππ‘ = E(x π π‘π₯ ) E(π π‘π₯ ) Similarly, π2 ππ‘ 2 = = π2 M(t) = 2 ππ‘ 2 π E( 2 π π‘π₯ ) ππ‘ E(π₯ 2 π π‘π₯ ) E(π π‘π₯ ) Hence, in general we get ππ ππ‘ π ππ M(t) = π E(π π‘π₯ ) ππ‘ π π E( π π π‘π₯ ) ππ‘ π π‘π₯ = = E(π₯ π ) If we set t = 0 in the nth derivative, we get ππ ππ‘ π M(t) = E(π₯ π π π‘π₯ ) π‘=0 = E(π₯ π ) π‘=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. π βπ₯ ππ¨π« π± > π f π± = π π¨ππ‘ππ«π°π’π¬π Answer: What are the mean and variance of X? The moment generating function of X is M(t) = E(π π‘π₯ ) = β π‘π₯ π π 0 π₯ ππ₯ β π π‘π₯ π βπ₯ ππ₯ = 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. β π‘π₯ βπ₯ π π ππ₯ 0 = β β(1βπ‘)π₯ π ππ₯ 0 = 1 1βπ‘ if = 1 1βπ‘ π β(1βπ‘)π₯ β 0 πβπ >π The expected value of X can be computed from M(t) as πΈ π₯ = π ππ‘ M(t) = (1 β π‘)β2 = π‘=0 π‘=0 1 (1βπ‘)2 = π‘=0 = 1 π ππ‘ (1 β π‘)β1 π‘=0 Similarly πΈ π2 ππ‘ 2 π₯2 = π2 ππ‘ 2 (1 β π‘)β1 = 2 (1 β π‘)β3 = M(t) π‘=0 π‘=0 2 (1βπ‘)3 π‘=0 π‘=0 = 2 = Therefore, the variance of X is Var(x)= = π π± π β ( π)π = 2 β 1 =1 Example 4.16. Let the random variable X have moment generating function M(t) = (1 β π‘)β2 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 ο¦ n οΆ x n οx b (x ; n , p ) ο½ ο§ ο· p q , x ο½ 0,1, ο¨x οΈ , 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 π‘β1) 4.6(π π 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) = π‘β1) 4.6(π π we conclude that X 1 POI(π) with π = 4.6. Thus, by Theorem 5.8, we get E(x) = 4.6 = Var(x) π 3<π₯ <6 =π 4 +π 5 = F 6 β F(3) = 0.686 β 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 π if its probability density function is of the form where π > 0. If a random variable X has an exponential density function with parameter π, then we denote it by writing X ~πΈπ₯π (π ). Mean and variance of exponential distribution 1 πΈ π₯ = π = π πππ π₯ = π2 = 1 π2 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 π π₯ = β1 π₯βπ 2 1 π2( π ) π 2π ββ < X < β where ββ < ΞΌ < β and 0 < π 2 < β are arbitrary parameters. If X has a normal distribution with parameters ΞΌ and π 2 , then we write X ~ π( π, π 2 ). Theorem 6.6. If π~ N(π,π 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 ~ π (0,1 ) The probability density function of standard normal distribution is the following 2 π π₯ = π₯ 1 β2 π 2π ββ < X < β Example 6.22. If X 1 N(0, 1), what is the probability of the random variable X less than or equal to β1.72? Answer P(X β€β1.72) = 1 β P(X β€1.72) = 1 β 0.9573 (from table) = 0.0427. Example 6.24. If X ~ N(3, 16), then what is P(4 β€X β€ 8)? Answer: P (4 β€ X β€ 8) = = 1 P( 4 β€πβ€ 4β3 P( 4 β€ πβ3 4 β€ 5 ) 4 = P (Z β€ 1.25) β P (Z β€ 0.25) = 0.8944 β 0.5987 = 0.2957. 8β3 ) 4