Expected and Variance When Additional Operations Are Added

Question

Suppose $E(X) = 1$ and $V(X) = 0.95$. Let's say $C = 20+10X$. My textbook says:

$ E(C) = E(20+10X) = 20+10E(X) = 30$

$V(C) = V(20+10X) = 10^2V(X) = 95$

I understand how they did the expected value, but am a bit confused about the Variance. Can someone explain it to me?

Answer 1

There are some properties of variance that you should know (and also know how to prove). They should be mentioned in your textbook. For any random variable $Y$,

• $V(aY) = a^2 V(Y)$
• $V(Y+c) = V(Y)$

The second property is intuitive: variance is some measure of variability/spread, which does not change if we shift the random variable by a shift $c$. The first property can be read from the definition of variance; roughly, it is the average distance squared to the mean $\mathbb{E} Y$, so if you scale by $a$, this squared distance is scaled by $a^2$.

Applying this to your example, we have $$V(20+10X) = V(10X) = 10^2 V(X).$$

Answer 2

Note that expectation is linear: $$\mathbb{E}[aX+b] = a\mathbb{E}[X]+b$$ Once we have this, deriving the variance expression is easy: \begin{align*} \text{Var}[aX+b] & = \mathbb{E}[(aX+b)^2]-\mathbb{E}[aX+b]^2 \\ & = \mathbb{E}[a^2X^2+2abX+b^2] - (a^2\mathbb{E}[X]^2+2ab\mathbb{E}[X]+b^2) \\ & = a^2\mathbb{E}[X^2]-\mathbb{E}[X]^2 = a^2(\mathbb{E}[X^2]-\mathbb{E}[X]^2) \\ & = a^2\text{Var}[X] \end{align*} Substituting in your values, we get the result.

Answer 3

$$\begin{array}{rcl} V(C) &=& V(20+10X) \\ &=& E[(20+10X)^2] - E[20+10X]^2 \\ &=& E[400+400X+100X^2] - (20+10E[X])^2 \\ &=& 400+400E[X]+100E[X^2] - (400+400E[X]+100E[X]^2) \\ &=& 100E[X^2] - 100E[X]^2 \\ &=& 100V(X) \\ &=& 10^2 V(X) \\ \end{array}$$