Toss 4 fair coins and consider the following two r.v.'s: X = number of Head's in...

Toss 4 fair coins and consider the following two r.v.'s:

X = number of Head's in the first 3 coin tosses, Y = number of Head's in the last 2 coin tosses,

what is the covariance Cov(3+2X,4+Y )?

Expert Solution

so COV(3+2X,4+Y)=2COV(X,Y)

now COV(X,Y)=E[XY]-E[X]*E[Y]

now from the joint distribution,

E[XY]=0*0*P[X=0,Y=0]+1*0*P[X=1,Y=0]+2*0*P[X=2,Y=0]+3*0*P[X=3,Y=0]+

0*1*P[X=0,Y=1]+1*1*P[X=1,Y=1]+2*1*P[X=2,Y=1]+3*1*P[X=3,Y=1]+

0*2*P[X=0,Y=2]+1*2*P[X=1,Y=2]+2*2*P[X=2,Y=2]+3*2*P[X=3,Y=2]

=1*3/16+2*3/16+3*1/16+2*1/16+4*2/16+6*1/16

(3+6+3+2+8+6)16=28/16=1.75

E[X]=0*P[X=0]+1*P[X=1]*2*P[X=2]+3*P[X=3]=6/16+12/16+6/16=24/16=1.5

E[Y]=0*P[Y=0]+1*P[Y=1]+2*P[Y=2]=8/16+8/16=1

hence COV(X,Y)=1.75-1.5*1=0.25

so COV(3+2X,4+Y)=2COV(X,Y)=2*0.25=0.5 [answer]

orchestra answered 2 years ago

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