Hi, Shaofeng ,

Denote

E(X)=mX , E(Y)=my, variance of X is V(X)= E(X-mX)2 , variance of Y is V(Y)=E(Y-mY)2.

We have

V(X)=E(X2-2XmX+mX2)= E(X2)- mX2, аналогично V(Y)= E(Y2)- mY2.

From here

E(X2)= V(X)+ mX2, E(Y2)= V(Y)+ mY2.

E(X2)- E(Y2)= V(X)- V(Y) if . mX= mY.

Then E(X2)- E(Y2)=0, if V(X)- V(Y)=0.

Therefore, from E(X)= E(Y) does not always follow E(X2)= E(Y2).

Arnold S. Korkhin

thanks for your answer, but will the E( |X| )= E( |Y| )? that is the mean of the absolute value of X and Y

Hi, Shaofeng ,

The answer to your question is in the application.

From this and previous answers it follows that E(X) and E( |X| ) depend on the variance V(X). Therefore, the equations you propose are incorrect generally.

The answer to your question is in the application.

From this and previous answers it follows that E(X2) and E( |X| ) depend on the varianceV(X). Therefore, the equations you propose are incorrect generally.

A pair of discrete random variables X and Y that not satisfy the property in the question can be easily constructed as follows. Let X be the discrete uniform random variable with values 1 and 3, i.e., Prob(X =1) = Prob(X =3) =1/2, and let Y be the constant random variable equal to √5, i.e., Prob(Y =√5) = 1. Then we have E(X) =E(|X|) = 2, E(X^2) = 5, E(|Y|) = E(Y) = √5 and E(Y^2) = 5. Hence, E(X^2) = E(Y^2) , but E(|X|) ≠ E(|Y|).