Menu Close

Question-209027




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 ))]
f(x)=12k+1x=1,2,3,4,5,..nfromMomentgeneratingfunction(MGF)Mx(t)=E(etx)Mx(t)=nx=1etxf(x)Mx(t)=nx=1etx.12k+1Mx(t)=12k+1nx=1etxMx(t)=12k+1(et+e2t+e3t+e4t..ent)Thisisgeometricseriesfromnx=1arn1=a(1rn)1ra=etr=etet(1ent)1rMx(t)=12k+1[et(1ent)1r]1stderivative=mean=Mx(t)=12k+1[et(1ent)1et]Mx(t)=12k+1[et(1ent)etnet(1et)2]t=0Mean(x)=12k+1applyquotientruleMx(t)=ddt[12k+1et(1ent)etnet(1et)2]Mx(t)=d2dt2[12k+1.et(1ent)etnet(1et)2]

Leave a Reply

Your email address will not be published. Required fields are marked *