Say that X ~ Bin(2,1/2) is a Binomial random variable corresponding to flipping

Question

Say that X ~ Bin(2,1/2) is a Binomial random variable corresponding to flipping a fair coin twice and counting the heads up. Construct the random variable Y in the following way: when X comes up zero, let Y ~ Bin(2.1/2) again (i.e. repeat the experiment), but when X comes up one or two, let Y ~ Unif(3), the discrete uniform random variable that takes on values 1,2,3. (a) What is the joint probability mass function for X and Y? Which is to say, compute P{X = k.Y = j) for every possible value of k and j. (b) Use the joint probability function from above to compute the covariance between these two random variables.

Explanation / Answer

X~Bin(2,0.5)

so the pmf of X is P[X=x]=2Cx(0.5)2   x=0,1,2

so X can take 3 values 0,1 or 2.

now when X=0 then Y~Bin(2,0.5)

so the pmf of X is P[Y=y]=2Cy(0.5)2   y=0,1,2

when X=1 or 2 Y~Unif(3)

so the pmf of Y is P[Y=y]=1/3 y=1,2,3

so when X=0 Y can take 0,1,2

when X=1 Y can take 1,2,3

when X=2 Y can take 1,2,3

now P[X=0]=2C0(0.5)2=0.25

P[X=1]=2C1(0.5)2=0.5

P[X=2]=2C2(0.5)2=0.25

now P[Y=0|X=0]=2C0(0.5)2=0.25

P[Y=1|X=0]=2C1(0.5)2=0.5

P[Y=2|X=0]=2C2(0.5)2=0.25

P[Y=1|X=1]=1/3

P[Y=2|X=1]=1/3

P[Y=3|X=1]=1/3

P[Y=1|X=2]=1/3

P[Y=2|X=2]=1/3

P[Y=3|X=2]=1/3

a) so the joint probability is as follows

when X=0 Y=0,1,2

P[X=0,Y=0]=P[Y=0|X=0]*P[X=0]=0.25*0.25=0.0625

P[X=0,Y=1]=P[Y=1|X=0]*P[X=0]=0.5*0.25=0.125

P[X=0,Y=2]=P[Y=2|X=0]*P[X=0]=0.25*0.25=0.0625

when X=1   Y=1,2,3

P[X=1,Y=1]=P[Y=1|X=1]*P[X=1]=1/3*0.5=0.167

P[X=1,Y=2]=P[Y=2|X=1]*P[X=1]=1/3*0.5=0.167

P[X=1,Y=3]=P[Y=3|X=1]*P[X=1]=1/3*0.5=0.167

when X=2 Y=1,2,3

P[X=2,Y=1]=P[Y=1|X=2]*P[X=2]=1/3*0.25=0.083

P[X=2,Y=2]=P[Y=2|X=2]*P[X=2]=1/3*0.25=0.083

P[X=2,Y=3]=P[Y=3|X=2]*P[X=2]=1/3*0.25=0.083

hence the joint probability mass function is

b) so X~Bin(2,0.5) so E[X]=2*0.5=1

now E[Y]=0*0.0625+1*0.375+2*0.3125+3*0.25=1.75

now E[XY]=0*0*0.625+0*1*0+0*2*0+1*0*0.125+1*1*0.167+1*2*0.083+2*0*0.0625+2*1*0.167+2*2*0.083+3*0*0+3*1*0.167+3*2*0.083

         =1.998

so covariance between X and Y is

cov(X,Y)=E[XY]-E[X]*E[Y]=1.998-1*1.75=0.248   [answer]

X=0 X=1 X=2 total Y=0 0.0625 0 0 0.0625 Y=1 0.125 0.167 0.083 0.375 Y=2 0.0625 0.167 0.083 0.3125 Y=3 0 0.167 0.083 0.25 total 0.25 0.5 0.25 1

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