Is the variation of the mean 0, and the variance of standard deviation just the variance?

Question

Just trying to get some properties straight. Was looking at a problem online which reads:

Let X be a random variable with mean μ and variance σ2. What is the variance of X/σ+10μ?

The solution of which is:

$Var(X/σ+10μ)=Var(X/σ)=Var(X)/σ2=1$.

I assume this means $Var(X/σ+10μ)$ expands to $Var(X/σ)$ + $Var(10μ)$, then implying $Var(10μ)=0$.

Next, I assume $Var(X/σ)$ expands to $Var(X)/Var(σ)$, implying $Var(σ)=σ^2$.

Is this correct? Are these universal properties for all distributions? Or am I way off target?

Answer 1

That is correct, but it's not generally true that $V(X+Y)=VX+VY$ unless $X,Y$ are uncorrelated. The result you're after uses two elementary properties of variances. For any constant $c$,

$$V(X+c)=VX$$

and

$$V(cx)=c^2 VX$$

Answer 2

The basic properties used can be summed up as follows. Let $a$ and $b$ be constants. Then $$\text{Var}(aX+b)=a^2\text{Var}(X).$$ The proof goes like this. Recall that the variance of a random variable $W$ is $E((W-\mu_W)^2)$. Let $W=aX+B$. Then $\mu_W=a\mu_X+b$.

It follows that $W-\mu_W=(aX+b)-(a\mu_X+b)=a(X-\mu_X)$. Thus $$(W-\mu_W)^2=a^2(X-\mu_X)^2.$$ Now take the expectations of both sides. We get $\text{Var}(W)=a^2\text{Var}X$.

Remark: The variance of a random variable is a measure of its "wiggliness." Adding the constant $b$ to a random variable shifts the mean, but does not change the wiggliness.

Answer 3

Be aware that $Var \left(\dfrac X \sigma \right)$ does not equal to $\dfrac {Var(X)} {Var(\sigma)}$. What happens here is that $\sigma$ is interpreted as a constant. Therefore it comes out as $\dfrac 1 {\sigma^2}$, hence $Var \left(\dfrac X \sigma \right) = \dfrac {Var(X)} {\sigma^2} = \dfrac {\sigma^2} {\sigma^2} = 1$.