Let X be a r.v. has the probability function given by:
Find the expectation of the random variable
Y=2X+1 & Z=X^2+3X
Since E(Y)=∑〖yP(x)〗∴E(Y)=E(2X+1)=2E(X)+1ButE(X)=∑_x x P(x)=(-3)(1/3)-1(2/5)+1(1/5)+3(1/15)=-1∴E(Y)=2(-1)+1=-1To obtain the expectation of Z we evaluate:E(X^2 )=9(1/3)+(2/5)+(1/5)+9(1/15)=21/5∴E(Z)=E(X^2 )+3E(X)=21/5+3(-1)=6/5=1.2
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need an explanation for this answer? contact us directly to get an explanation for this answerE(Y)=∑〖yP(x)〗
∴E(Y)=E(2X+1)=2E(X)+1
But
E(X)=∑_x x P(x)=(-3)(1/3)-1(2/5)+1(1/5)+3(1/15)=-1
∴E(Y)=2(-1)+1=-1
To obtain the expectation of Z we evaluate:
E(X^2 )=9(1/3)+(2/5)+(1/5)+9(1/15)=21/5
∴E(Z)=E(X^2 )+3E(X)
=21/5+3(-1)=6/5=1.2