Question-209030

Question Number 209030 by Spillover last updated on 30/Jun/24

Answered by aleks041103 last updated on 30/Jun/24

m o m e n t g e n e r a t i n g f u n c t i o n i s

m (t) = \underset{k = 0}{\sum^{\infty}} \frac{m_{k}}{k!} t^{k} = m_{0} + m_{1} t + \dots

w h e r e m_{k} = \underset{- \infty}{\int^{\infty}} x^{k} f (x) d x a r e t h e m o m e n t s

o f t h e p r o b . d e n s i t y f u n c t i o n f (x) .

\Rightarrow m (t) = \frac{t}{1 - t} = 0 + t + t^{2} + t^{3} + \dots =

= m_{0} + m_{1} t + \dots

\Rightarrow m_{0} = 0

b u t t h i s i s i m p o s s i b l e, s i n c e

m_{0} = \int f (x) d x = 1

f o r a n y P D F f (x) .

Answered by Spillover last updated on 04/Jul/24

s i n c e m (t) = \frac{t}{1 - t} i s n o t d e f i n e d f o r t h e t = 1

t h i s i m p l i e s t h a t t h e r e i s n o p r o b a b i l i t y d i s t r i b u t i o n

a s s o c i a t e d w i t h t h i s m o m e n t g e n e r a t i n g f u n c t i o n

Commented by aleks041103 last updated on 04/Jul/24

m a y b e .

b y d e f i n i t i o n

m (t) = E [e^{x t}]

m (0) = E [e^{x . 0}] = E [1] = 1 \neq \frac{0}{1 - 0} = 0

b u t

m (1) = E [e^{x}]

i f t h i s {d o e s n}^{'} t e x i s t t h e n

\int_{- \infty}^{+ \infty} e^{x} f (x) d x {d o e s n}^{'} t c o n v e r g e .

T h i s i m p l i e s a l m o s t n o t h i n g o r a t l e a s t I

{d o n}^{'} t s e e h o w t h i s p r o v e s t h a t f {d o e s n}^{'} t e x i s t .

F o r e x a m p l e f (x) = \frac{1}{2} e^{- x / 2} H (x), w h e r e

H (x) = 1 f o r x ⩾ 0 a n d H (x) = 0 f o r x < 0 .

T h i s i s a p d f, s i n c e i t i s l o c a l l y i n t e g r a b l e

a n d o b v . \int_{- \infty}^{\infty} f (x) d x = \int_{0}^{\infty} e^{- (x / 2)} d (x / 2) = 1

b u t \int_{- \infty}^{\infty} e^{x} f (x) d x = \int_{0}^{\infty} e^{(x / 2)} d (x / 2), w h i c h

d i v e r g e s .

Commented by Spillover last updated on 05/Jul/24

t h a n k y o u f o r c l a r i f i c a t i o n