I've got a probability exam tonight and I'm just curious about an answer from a practice exam. Any tips/help would be much appreciated!
Here's what I've got:
X is a random variable
Y = 4X + 1, is another random variable
P(X=1) = 1/2, P(X=2) = 1/4, P(X=4) = 1/4
So far I've solved the following (The solutions are provided, so every answer I have is correct until we get to where I have a question.)
E(X) = 2, E(X^2) = 5.5, var(X) = 1.5
E(Y) = 9, var(Y) = 24
E(X + Y) = E(X) + E(Y) = 11 due to linearity of expectation.
Now for where I'm stuck…
I am asked to find var(X + Y), which is equal to var(X + 4X + 1) = var(5X + 1). The solution provided says that var(5X + 1) = 25var(X). Can anyone explain why this is the case? I know linearity of variance does not apply in this situation because Y is dependent on X, but I'm not sure how I would go about computing this variance during an exam.
Best Answer
In general, the variance of $aX+b$, where $a$ and $b$ are constants, is $a^2\text{Var}(X)$. This is a fact that you would be expected to know. It is not hard to prove. For by the definition of variance, we have $$\text{Var}(aX+b)=E(((aX+b)-E(aX+b))^2)=E((aX-aE(X))^2)=a^2E((X-E(X))^2)=a^2\text{Var}(X).$$