Why are negative constants removed from variance?

Question

Let $X_1$ and $X_2$ be random variables such that $X_i \sim N(1, 1) $. Why is the constant removed in the case of the variance $$ \mathrm{V}(X_1 + X_2 - 2) = 1 + 1 = 2 $$ but not in the case of the expectation $$ \mathrm{E}(X_1 + X_2 - 2) = 1 + 1 - 2 = 0 \;? $$

Answer 1

In very non math/statistical terms, variance measures how spread out the data is. Therefore, shifting the data by a constant term does not change how spread out the data is. However, the shift will change the the expected value of the data, because the expected value is the weighted average of the data values.

Answer 2

The variance of a constant is equal to zero.

Assuming $X_1$ and $X_2$ are independent: $$V(X_1 + X_2 - 2) = V(X_1) + V(X_2) + V(2) = 1+1+0 = 2 $$