Question Number 209027 by Spillover last updated on 30/Jun/24
Answered by Spillover last updated on 05/Jul/24
![f(x)=(1/2^(k+1) ) x=1,2,3,4,5,.....n from Moment generating function(MGF) M_x (t)=E(e^(tx) ) M_x (t)=Σ_(x=1) ^n e^(tx) f(x) M_x (t)=Σ_(x=1) ^n e^(tx) .(1/2^(k+1) ) M_x (t)=(1/2^(k+1) ) Σ_(x=1) ^n e^(tx) M_x (t)=(1/2^(k+1) ) (e^t +e^(2t) +e^(3t) +e^(4t) .....e^(nt) ) This is geometric series from Σ_(x=1) ^n ar^(n−1) =((a(1−r^n ))/(1−r)) a=e^t r=e^t ((e^t (1−e^(nt) ))/(1−r)) M_x (t)=(1/2^(k+1) ) [ ((e^t (1−e^(nt) ))/(1−r))] 1^(st) derivative =mean =M_x (t)=(1/2^(k+1) ) [ ((e^t (1−e^(nt) ))/(1−e^t ))] M_x ^′ (t)=(1/2^(k+1) ) [ ((e^t (1−e^(nt) )−e^(tn) e^t )/((1−e^t )^2 ))] t=0 Mean(x^− )=−(1/2^(k+1) ) apply quotient rule M_x ^′ (t)=(d/dt)[(1/2^(k+1) ) ((e^t (1−e^(nt) )−e^(tn) e^t )/((1−e^t )^2 ))] M_x ^(′′) (t)=(d^2 /dt^2 )[(1/2^(k+1) ) .((e^t (1−e^(nt) )−e^(tn) e^t )/((1−e^t )^2 ))]](https://www.tinkutara.com/question/Q209227.png)
$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},…..{n} \\ $$$${from}\:{Moment}\:{generating}\:{function}\left({MGF}\right) \\ $$$${M}_{{x}} \left({t}\right)={E}\left({e}^{{tx}} \right) \\ $$$${M}_{{x}} \left({t}\right)=\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{tx}} {f}\left({x}\right) \\ $$$${M}_{{x}} \left({t}\right)=\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{tx}} .\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\: \\ $$$${M}_{{x}} \left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{tx}} \\ $$$${M}_{{x}} \left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\left({e}^{{t}} +{e}^{\mathrm{2}{t}} +{e}^{\mathrm{3}{t}} +{e}^{\mathrm{4}{t}} …..{e}^{{nt}} \right) \\ $$$${This}\:{is}\:{geometric}\:{series}\:\: \\ $$$${from}\:\:\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{ar}^{{n}−\mathrm{1}} =\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}} \\ $$$${a}={e}^{{t}} \:\:\:\:\:\:\:\:\:{r}={e}^{{t}} \:\:\:\:\:\:\:\:\:\frac{{e}^{{t}} \left(\mathrm{1}−{e}^{{nt}} \right)}{\mathrm{1}−{r}} \\ $$$${M}_{{x}} \left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\left[\:\frac{{e}^{{t}} \left(\mathrm{1}−{e}^{{nt}} \right)}{\mathrm{1}−{r}}\right] \\ $$$$\mathrm{1}^{{st}} \:{derivative}\:={mean}\: \\ $$$$={M}_{{x}} \left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\left[\:\frac{{e}^{{t}} \left(\mathrm{1}−{e}^{{nt}} \right)}{\mathrm{1}−{e}^{{t}} }\right] \\ $$$${M}_{{x}} ^{'} \left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\left[\:\frac{{e}^{{t}} \left(\mathrm{1}−{e}^{{nt}} \right)−{e}^{{tn}} {e}^{{t}} }{\left(\mathrm{1}−{e}^{{t}} \right)^{\mathrm{2}} }\right] \\ $$$${t}=\mathrm{0}\:\:\:\: \\ $$$${Mean}\left(\overset{−} {{x}}\right)=−\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} } \\ $$$${apply}\:{quotient}\:{rule} \\ $$$${M}_{{x}} ^{'} \left({t}\right)=\frac{{d}}{{dt}}\left[\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\:\frac{{e}^{{t}} \left(\mathrm{1}−{e}^{{nt}} \right)−{e}^{{tn}} {e}^{{t}} }{\left(\mathrm{1}−{e}^{{t}} \right)^{\mathrm{2}} }\right] \\ $$$${M}_{{x}} ^{''} \left({t}\right)=\frac{{d}^{\mathrm{2}} }{{dt}^{\mathrm{2}} }\left[\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\:.\frac{{e}^{{t}} \left(\mathrm{1}−{e}^{{nt}} \right)−{e}^{{tn}} {e}^{{t}} }{\left(\mathrm{1}−{e}^{{t}} \right)^{\mathrm{2}} }\right] \\ $$$$ \\ $$