Generally speaking, the expectation and variance of random variables have the following relationship:

Let x 1, x2, x3...xn be random variables, and the mathematical expectation is:

E(x 1+X2+X3+……+Xn)= E(x 1)+E(X2)+E(X3)+……+E(Xn)

That is, the expectation of sum equals the expected sum.

For variance, there are some special similar relations, but the conditions that random variables are independent of each other must be met, otherwise it will not be established, that is:

D(x 1 X2 X3……Xn)= D(x 1)+D(X2)+D(X3)+……+D(Xn)，

Where x 1, x2, x3...xn are independent random variables and must be clear, otherwise there will be errors, such as:

D(X+X)=？ The random variable x and itself are certainly not independent of each other. If calculated according to the independent formula: D(X+X)=D(X)+D(X)=2D(X), this is obviously the wrong answer. In fact, D(X+X)=D(2X)=4D(X).

I wonder if the above formula can solve your problem?