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

	Answered by aleks041103 last updated on 30/Jun/24

	$m o m e n t g e n e r a t i n g f u n c t i o n i s$ $m (t) = \underset{k = 0}{\sum^{\infty}} \frac{m_{k}}{k!} t^{k} = m_{0} + m_{1} t + . . .$ $w h e r e m_{k} = \underset{- \infty}{\int^{\infty}} x^{k} f (x) d x a r e t h e m o m e n t s$ $o f t h e p r o b . d e n s i t y f u n c t i o n f (x) .$ $\Rightarrow m (t) = \frac{t}{1 - t} = 0 + t + t^{2} + t^{3} + . . . =$ $= m_{0} + m_{1} t + . . .$ $\Rightarrow m_{0} = 0$ $b u t t h i s i s i m p o s s i b l e, s i n c e$ $m_{0} = \int f (x) d x = 1$ $f o r a n y P D F f (x) .$
	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

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