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Chapter 3, Section 9 Discrete Random Variables

Chapter 3, Section 9 Discrete Random Variables. Moment-Generating Functions.  John J Currano, 12/15/2008. æ. ö. k. (  c ) k  j E [ Y j ]. E [ ( Y – c ) k ]. k. =. å. ç. ÷. j. è. ø. =. j. 0. E [ ( Y – E ( Y ) ) k ] = E [ ( Y –  ) k ]. m. =. k.

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Chapter 3, Section 9 Discrete Random Variables

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  1. Chapter 3, Section 9Discrete Random Variables Moment-Generating Functions  John J Currano,12/15/2008

  2. æ ö k (c)kj E [Y j] E[(Y – c)k] k = å ç ÷ j è ø = j 0 E[(Y – E(Y))k] = E[(Y – )k] m = k Definitions. Let Y be a discrete random variable with probability function, p(y), and let k = 1, 2, 3, . Then: is the kth moment of Y (about the origin). is the kth moment of Y about c. is the kth central moment of Y. Notes.=E(Y) = the first moment of Y ( =1)  2=V(Y) = the second central moment of Y ( =2 ) =E(Y2) – [E(Y)]2=2 – (1)2. What is the first central moment of Y ?

  3. Definition. If Y is a discrete random variable, its moment-generating function (mgf) is the function provided this function of t exists (converges) in some interval around 0. Note. The mgf, when it exists, gives a better characterization of the distribution than the mean and variance – it completely determines the distribution in the sense that if two random variables have the same mgf, then they have the same distribution. So to find the distribution of a random variable, we can find its mgf and compare it to a list of known mgfs; if the mgf is in the list, we have found the distribution. We shall also see that the mgf gives us an easy way to find the distribution’s mean and variance.

  4. ( ) tY ty å = = m ( t ) E e e p ( y ) y k ¥ ( ty ) å å = p ( y ) k ! = y k 0 k ¥ ( ty ) å å = p ( y ) k ! = k 0 y k k ¥ t t k å å = factoring out y p ( y ) k ! k ! = k 0 y ( ) k ¥ t k å = E Y k ! = k 0 Why it is called the moment-generating function: Using the Maclaurin Series for ex at the right, we derive by Theorem 3.2 using the Maclaurin series for ety interchanging the summations by Theorem 3.2

  5. - k 2 ¥ ( ) ( ) ( ) - × k ( k 1 ) t 2 1 k 2 2 ¢ ¢ ¢ ¢ = = = å m ( t ) E Y ; m ( 0 ) E Y E Y ; k ! 2 ! = k 2 Why it is called the moment-generating function: Thus, et cetera.

  6. t 2 t 3 t 3 1 2 = + + m ( t ) e e e 6 6 6 t 2 t 3 t ¢ 9 1 4 14 7 ¢ = + + = = = m ( t ) e e e , so E ( Y ) m ( 0 ) . 6 6 6 6 3 ¢ ¢ = m ( t ) ( ) t 2 t 3 t 2 8 36 1 27 ¢ ¢ + + = = = e e e , so E Y m ( 0 ) 6 . 6 6 6 6 ( ) 2 2 = = 5 7 - 6 3 9 Theorem. for k = 1, 2, 3, . . . where mY(t) is the mgf of Y (provided it exists). Also, mY(0) = E(e0Y) = E(1) = 1 for any RV, Y. Example(p. 142 #3.155). Given that find: (a) E(Y); (b) V(Y); (c) the distribution of Y. (a) (b) Thus, V(Y) = E(Y2) [E(Y)]

  7. t 2 t 3 t 3 1 2 = + + m ( t ) e e e 6 6 6 ( ) tY ty t 2 t 3 t 3 1 2 = = = + + å m ( t ) E e e p ( y ) e e e 6 6 6 y 3 1 2 , , , 6 6 6 Theorem. for k = 1, 2, 3, . . . where mY(t) is the mgf of Y (provided it exists). Also, mY(0) = E(e0Y) = E(1) = 1 for any RV, Y. Example(p. 142 #3.155). Given that find: (a) E(Y); (b) V(Y); (c) the distribution of Y. (c) Since and the support of Y consists of the coefficients of t in the exponents of the powers of e in the nonzero terms of the power series, Y has support {1, 2, 3}. The probability that Y assumes each of these values is the coefficient of the exponential in the corresponding term, so respectively. Y = 1, 2, 3, with probabilities

  8. æ ö n n - ty y n y ç ÷ å e p q ç ÷ y è ø = y 0 ( ) æ ö n n y - t n y ç ÷ å = pe q ç ÷ y è ø = y 0 Example. Find the moment-generating function of Y ~ bin(n, p) and use it to find E(Y) and V(Y). Solution. Then, by Theorem 3.2 = (pet+ q)nby the Binomial Theorem

  9. Theorem. If Y ~ bin(n, p), then its mgf is ) ( - n 1 t t + × n pe q pe - - n 1 n 1 ¢ Þ = = + × × = × × = E ( Y ) m ( 0 ) n ( p q ) p 1 n ( 1 ) p 1 np ( ) ( ) - - n 1 n 2 é ù t t t t t + × - + × × + n pe q pe n ( n 1 ) pe q pe pe ¢ ¢ = m ( t ) ë û ( ) 2 ¢ ¢ Þ = = E Y m ( 0 ) - - n 2 n 1 - × × × × + × × n ( n 1 ) ( 1 ) p 1 p 1 n ( 1 ) p 1 2 2 2 2 = - + = - + n ( n 1 ) p np n p np np 2 2 2 2 2 Þ = - + - V(Y) = E(Y2)  [E(Y)] n p np np ( np ) 2 = - + = - = np np np ( 1 p ) npq . Now differentiate to find E(Y) and V(Y): (a) (b)

  10. ( ) 2 5, bin 3 2 3 3 2 æ ö æ ö 5 5 æ ö æ ö æ ö æ ö 2 1 2 1 ç ÷ ç ÷ + ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ ç ÷ 3 3 3 3 è ø è ø è ø è ø 2 3 è ø è ø Example. If the moment-generating function of a random variable, Y, , find Pr(Y = 2 or 3). Solution. From the form of the moment-generating function, , so that we know that Y ~ P(Y = 2 or 3) = P(Y = 2) + P(Y = 3) =

  11. Homework Problem Results (pp. 142-143) 3.147 If Y ~ Geom(p), then 3.158 If Y is a random variable with moment-generating function, mY(t), and W = aY + b where a and b are constants, then mW (t) = maY+b (t) = et bmY (at).

  12. ( ) t l - e 1 = m ( t ) e . Y Other Moment-Generating Functions 1. If Y ~ NegBin(r, p), then This is most easily proved using results in Chapter 6. Exercise. Use mY(t) to find E(Y) and V(Y). 2. If Y ~ Poisson(), then This is Example 3.23 on p. 140, where it is proved and then used (in Example 3.24) to find E(Y) and V(Y).

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