Lec-12: Prove that E(a)=a, E(ax)=aE(x) & E(ax+b)=aE(x)+b | Important proves

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For any constant a and b, E[aX+b]=aE[X]+b and E[aX-b]=aE[X]-b (Discrete) For any constant a and b, E[aX+b]=aE[X]+b and E[aX-b]=aE[X]-b (Discrete)
Properties of Expected Value: aE(X)+b = E(aX + b) Properties of Expected Value: aE(X)+b = E(aX + b)
#15-2 E(aX+b)= aE(X)+b proof for X discrete part 2 of 2 #15-2 E(aX+b)= aE(X)+b proof for X discrete part 2 of 2
#15-1 E(aX+b)= aE(X)+b proof for X discrete part 1 of 2 #15-1 E(aX+b)= aE(X)+b proof for X discrete part 1 of 2
E(aX+b) | Expectation | Mean | Var(aX+b) | Variance | Statistical Analysis E(aX+b) | Expectation | Mean | Var(aX+b) | Variance | Statistical Analysis
E(Y)=aE(x) +b and V(Y) =a^2V(Y) where Y=aX+b / problems on continues random variables E(Y)=aE(x) +b and V(Y) =a^2V(Y) where Y=aX+b / problems on continues random variables
Second law of Expectation Prove E(aX+b) = aE(X) + b Second law of Expectation Prove E(aX+b) = aE(X) + b
E(aX+bY)=aE(X)+bE(Y) for X,Y discrete - Proof part 3 of 3 E(aX+bY)=aE(X)+bE(Y) for X,Y discrete - Proof part 3 of 3