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Question Number 209031 by Spillover last updated on 30/Jun/24

	Answered by Spillover last updated on 07/Jul/24
	$f(x)= { (((1/n) x=1,2,3,.....=(1/n)[e^t +e^(2t) +e^(3t) +........])),((0 else where)) :} (a) moment generating function M_x (t)=E(e^(tx) ) M_x (t)=Σ_(3=1) e^(tx) .=(1/n)[Σ_(3=1) e^(tx) ] =(1/n)[e^t +e^(2t) +e^(3t) +........] =(1/n)[(([e^(nt) −1)/(e^t −1))] Probability generatingfunction G_x (s)=E(s^x ) =Σ_(x=1) ^n s^x .(1/n)=(1/(n ))Σ_(x=1) ^n s^x (1/(n ))[s+s^2 +s^3 +s^4 +....+s^n ] =(1/n)[((s(1−s^n ))/(1−s))] For the expection M_x ^′ (0)=E(x) for variance M_x ′′−[M_x (0)]^2$
	$f (x) = {\begin{cases} \frac{1}{n} x = 1, 2, 3, . . . . . = \frac{1}{n} [e^{t} + e^{2 t} + e^{3 t} + . . . . . . . .] \\ 0 e l s e w h e r e \end{cases}$ $(a) m o m e n t g e n e r a t i n g f u n c t i o n$ $M_{x} (t) = E (e^{t x})$ $M_{x} (t) = \sum_{3 = 1} e^{t x} . = \frac{1}{n} [\sum_{3 = 1} e^{t x}]$ $= \frac{1}{n} [e^{t} + e^{2 t} + e^{3 t} + . . . . . . . .]$ $= \frac{1}{n} [\frac{[e^{n t} - 1}{e^{t} - 1}]$ $P r o b a b i l i t y g e n e r a t i n g f u n c t i o n$ $G_{x} (s) = E (s^{x})$ $= \underset{x = 1}{\sum^{n}} s^{x} . \frac{1}{n} = \frac{1}{n} \underset{x = 1}{\sum^{n}} s^{x}$ $\frac{1}{n} [s + s^{2} + s^{3} + s^{4} + . . . . + s^{n}]$ $= \frac{1}{n} [\frac{s (1 - s^{n})}{1 - s}]$ $F o r t h e e x p e c t i o n M_{x}^{^{'}} (0) = E (x)$ $f o r v a r i a n c e M_{x}^{″} - {[M_{x} (0)]}^{2}$
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