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](https://www.tinkutara.com/question/Q209350.png)
$${f}\left({x}\right)=\begin{cases}{\frac{\mathrm{1}}{{n}}\:\:\:\:{x}=\mathrm{1},\mathrm{2},\mathrm{3},…..=\frac{\mathrm{1}}{{n}}\left[{e}^{{t}} +{e}^{\mathrm{2}{t}} +{e}^{\mathrm{3}{t}} +……..\right]}\\{\mathrm{0}\:\:{else}\:{where}}\end{cases} \\ $$$$\left({a}\right)\:{moment}\:{generating}\:{function} \\ $$$${M}_{{x}} \left({t}\right)={E}\left({e}^{{tx}} \right) \\ $$$${M}_{{x}} \left({t}\right)=\underset{\mathrm{3}=\mathrm{1}} {\sum}{e}^{{tx}} .=\frac{\mathrm{1}}{{n}}\left[\underset{\mathrm{3}=\mathrm{1}} {\sum}{e}^{{tx}} \right] \\ $$$$=\frac{\mathrm{1}}{{n}}\left[{e}^{{t}} +{e}^{\mathrm{2}{t}} +{e}^{\mathrm{3}{t}} +……..\right] \\ $$$$=\frac{\mathrm{1}}{{n}}\left[\frac{\left[{e}^{{nt}} −\mathrm{1}\right.}{{e}^{{t}} −\mathrm{1}}\right] \\ $$$${Probability}\:{generatingfunction} \\ $$$${G}_{{x}} \left({s}\right)={E}\left({s}^{{x}} \right) \\ $$$$=\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{s}^{{x}} .\frac{\mathrm{1}}{{n}}=\frac{\mathrm{1}}{{n}\:}\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{s}^{{x}} \\ $$$$\frac{\mathrm{1}}{{n}\:}\left[{s}+{s}^{\mathrm{2}} +{s}^{\mathrm{3}} +{s}^{\mathrm{4}} +….+{s}^{{n}} \right] \\ $$$$=\frac{\mathrm{1}}{{n}}\left[\frac{{s}\left(\mathrm{1}−{s}^{{n}} \right)}{\mathrm{1}−{s}}\right] \\ $$$${For}\:{the}\:{expection}\:\:\:\:\:\:\:{M}_{{x}} ^{'} \left(\mathrm{0}\right)={E}\left({x}\right) \\ $$$${for}\:{variance}\:\:\:\:\:{M}_{{x}} ''−\left[{M}_{{x}} \left(\mathrm{0}\right)\right]^{\mathrm{2}} \\ $$$$ \\ $$