Question

The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event that the automobile is damaged, repair costs can be modeled by a uniform random variable on the interval (0, 1500)

In the above question, I am asked to find Var(Y) where Y = (X - 250). The solution actually involves finding E(Y) and E(Y^2) through intergration, then Var(Y) = E(Y^2) - E(Y)^2. I understand the solution.

My question is why can not I use the properties of variance: Var(Y) = Var(X) - Var(250)??

Answer 1

By property,

Var(X-a) = Var(X) if 'a' is a constant.                                                                                                                         

Thus,

Var(X-250) = Var(X)