Transcription of MATH246 | Probability and Random Processes
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MATH246 | Probability and Random Processes
MATH246 Probability and Random ProcessesSolution to Homework Four1. Note that for alln,Xn={1 if the outcome isH 1 if the outcome isT.(a) The only two sample paths:(b) Given that the coin is fair, we haveP[Xn= 1] =P[outcome isH] =12P[Xn= 1] =P[outcome isT] =12.(c)P[Xn= 1, Xn+k= 1] =P[Xn= 1] =12P[Xn= 1, Xn+k= 1] =P[Xn= 1] =12P[Xn= 1, Xn+k= 1] =P[ ] = 0P[Xn= 1, Xn+k= 1] =P[ ] = 0 Hence, the joint pmfP[Xn=i, Xn+k=j] ={12, i=j0, i6=j.(d)P[Xn] = (1)P[Xn= 1] + ( 1)P[Xn= 1] =12 12= (n1, n2) =E[{Xn1 E[Xn1]}{Xn2 E[Xn2]}]=E[Xn1Xn2]= (1)(1)P[Xn1= 1, Xn2= 1] + ( 1)( 1)P[Xn1= 1, Xn2= 1]=12+12= (a)E[Z(t)] =E[Xt+Y] =tmX+mYCZ(t1, t2) =E[{(Xt1+Y) (t1mX+mY)}{(Xt2+Y) (t2mX+mY)}]=E[{t1(X mX) + (Y mY)}{t2(X mX) + (Y mY)}]=t1t2E[{X mX}2] +t1E[(X mX)(Y mY)]+E[{Y mY}2] +t2E[(Y mY)(X mX)]=t1t2 2X+ (t1+t2) X Y XY+ 2Y.(b) For joint Gaussian Random variables (see Example , page 244 of textbook), ifXandYarejointly Gaussian Random variables, thenZ(t) =Xt+Yis also a Gaussian Random variable for part (a),mZ(t) =tmX+mYVAR[Z(t)] =CZ(t, t) =t2 2X+ 2t X Y XY+ 2 YHence, the pdf ofZ(t) isfZ(t)(z) =1 2 VAR[Z(t)]exp{ 12 VAR[Z(t)](z mZ(t))2}.}}
MATH246 | Probability and Random Processes Solution to Homework Four 1. Note that for all n, Xn = n 1 if the outcome is H ¡1 if the outcome is T (a) The only two …
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