Standard deviation of $Z = 9 - 3Y^3$

Question

If Y is a random variable with probability mass function:

 Y | Pr(Y = y)
-1 | 0.4
 0 | 0.5
 1 | 0.1

I need to find the standard deviation of $Z = 9 - 3Y^3$.

Approach: since it's the variance I know that 9 should not affect it and I essentially have $3^2 Var(Y^3)$ using one of the identities. My understanding is that I should be able to use the delta method to find $Var(Y^3)$ but I am not sure how. I don't think I can directly use the identity $Var(X) = E[X^2] - E[X]^2$ can I?

Answer 1

Since $(-1)^3=-1$, $0^3=0$, and $1^3=1$, $Y^3$ has the same distribution as $Y$. Therefore $$\mathrm{Var}(9-3Y^3) = 3^2\mathrm{Var}(Y^3) = 9\mathrm{Var}(Y). $$ Since $$\mathrm{Var}(Y) = \mathbb E[Y^2] - \mathbb E[Y]^2 = \frac12 - \left(\frac3{10}\right)^2 = \frac{41}{100},$$ it follows that $$\mathrm{Var}(Z) = 9\cdot\frac{41}{100} = \frac{369}{100}.$$

Answer 2

Hint: You can transform the random variable.

  Z | Pr(Y = y)
    12 | 0.4
     9 | 0.5
     6 | 0.1

Now you can calculate $Var(Z)$ directly.

Update

The straightforward calculation is

$$Var(Z)=0.4\cdot (12-9.9)^2+0.5 \cdot (9-9.9)^2+0.1 \cdot (6-9.9)^2=3.69$$