Solved: TEAM PROJECT. Moment Generating Function. The

Chapter 24, Problem 24.1.128

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TEAM PROJECT. Moment Generating Function.

The moment generating function G(t) is defined by

\(G(t)=E\left(e^{t X_{J}}\right)=\sum_{j} e^{t x_{j}} f\left(x_{j}\right)\)

or

\(G(t)=E\left(e^{t \mathrm{X}}\right)=\int_{-\infty}^{x} e^{t x} f(x) d x\)

where X is a discrete or continuous random variable, respectively.

(a) Assuming that termwise differentiation and differentiation under the integral sign are permissible, show that \(E(X^{k}) = d^{(k)}(O)\), \(G^{(k)} = d^{k}G/dt^{k}\) in particular, \(\mu = G’(0)\)

(b) Shov. that the binomial distribution has the moment generating function

\(G(t)=\sum_{x=0}^{n} e^{t x}\left(\begin{array}{l} n \\ x \end{array}\right) p^{x} q^{n-x}=\sum_{x=0}^{n}\left(\begin{array}{l} n \\ x \end{array}\right)\left(p e^{t}\right)^{x} q^{n-x} \)

\(=\left(p e^{t}+q\right)^{n}\)

(c) Using (b), prove (3).

(d) Prove (4).

(e) Show that the Poisson distribution has the moment generating function \(G(t)=e^{-\mu} e^{\mu e}’\) and prove (6).

(f) Prove x \(\left(\begin{array}{c} M \\ x \end{array}\right)=M\left(\begin{array}{r} M-1 \\ x-1 \end{array}\right) \). Using this. prove (9).

Text Transcription:

G(t) = E(e^tX_j) = sum_j e^tx_j f(x_j)

G(t) = E(e^tX) = int ^x_-x e^tx f(x) dx

E(X^k) = G^(k) (0)

G^(k) = d^k G / dt^k

mu = G’ (0)

G(t) = sum ^n _x = 0  e^tx (_x^n) p^x q^n-x = sum ^n_x=0 (_x^n) (pe^t)^x q^n-x

=(pe^t + q)^n

G(t) = e^-mu e^mue’

(_x^M) = M (_x-1 ^M - 1)