Expected Value and Variance with Substitution? In preparation for a statistics test, I came across this practice question that is causing me a lot of difficulty: 
X, Y, and Z are i.i.d random variables with mean = 4 and variance = 1. Let W = $\frac{1}{\sqrt3}$XY - 2Z - $\frac{16}{\sqrt3)}$. Compute E(W) and Var(W).
I'm not quite sure how to approach it. First, I tried simplifying E(W), and got -8, which I don't think is correct:
E(W) = $\frac{1}{\sqrt3}$(E(X)E(Y))- 2E(Z) - $\frac{16}{\sqrt3}$
E(W) = $\frac{1}{\sqrt3}$(4)(4) - 2(4) - $\frac{16}{\sqrt3}$ = - 8
I know that $\operatorname{Var}(W) = E(W^2)-[E(W)^2]$, but that's basically it. Is it just a matter of substituting 1 into the equation? I'm not quite sure how to proceed.
Thanks!
 A: You have correctly computed
\begin{align}
E[W] &= E\left[\frac{1}{\sqrt{3}}XY - 2Z -\frac{16}{\sqrt{3}}\right]\\
&= \frac{1}{\sqrt{3}}E[XY] - 2E[Z] -\frac{16}{\sqrt{3}} &\scriptstyle{\text{linearity of expectation}}\\
&= \frac{1}{\sqrt{3}}E[X]E[Y] - 2E[Z] -\frac{16}{\sqrt{3}}
&\scriptstyle{X~\text{and}~Y~\text{are independent}}\\
&= \frac{1}{\sqrt{3}}\times 4 \times 4 - 2\times 4 - \frac{16}{\sqrt{3}}\\
&= -8.
\end{align}
You made a correct start to finding the variance of $W$ as $\operatorname{Var}(W) = E(W^2)-(E[W])^2$. You know the value of $(E[W])^2$ already from the above. Now all that remains is to figure out
$$E[W^2] = E\left[\left(\frac{1}{\sqrt{3}}XY - 2Z -\frac{16}{\sqrt{3}}\right)^2\right]$$
which you can do by multiplying out the square. You will get six terms of which one will have $X^2Y^2$ in it, another $Z^2$ in it, yet another $XYZ$, and one with $XY$ in it. Use linearity of expectation to write $E[W^2]$ as the sum of six expectations, and then use independence of $X$, $Y$, and $Z$ to write $E[XYZ]$ as $E[X]E[Y]E[Z]$ etc. Don't forget that $E[X^2] = \operatorname{Var}(X) + (E[X])^2$ etc.  Good luck!
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Since writing the above, BruceET, in a comment that he regrettably declines to incorporate into his own answer, has come up with a much simpler computation for $\operatorname{Var}(W)$. Since comments are ephemeral and can de deleted at any time by the author, I am memorializing this method below.
Since $XY$ and $Z$ are independent random variables,
$$\operatorname{Var}(W) = \operatorname{Var}\left(\frac{1}{\sqrt{3}}XY - 2Z -\frac{16}{\sqrt{3}}\right) = \frac{1}{{3}}\operatorname{Var}(XY)+4\operatorname{Var}(Z).$$ 
We of course know that $\operatorname{Var}(Z) = 1$, while
\begin{align}
\operatorname{Var}(XY) &= E[(XY)^2] - \left(E[XY]\right)^2\\
&= E[X^2Y^2] - \left(E[XY]\right)^2\\
&= E[X^2]E[Y^2] - \left(E[X]E[Y]\right)^2\\
&= \left(\operatorname{Var}(X)+(E[X])^2\right)\left(\operatorname{Var}(Y)+(E[Y])^2\right)- \left(E[X]E[Y]\right)^2\\
&= 17^2-16^2\\
&= (17+16)(17-16)\\
&= 33
\end{align}
leading to
$$\operatorname{Var}(W) = \frac{1}{{3}}\operatorname{Var}(XY)+4\operatorname{Var}(Z) = 15.$$
And that's it. No fuss, no complicated calculations, no million simulations yielding a sample variance of $14.99188$ which R rounds off to $15$ as the approximate value of the variance, instead of the above direct mathematical computation giving the exact value of the variance as $15$. Ah well, de gustibus non est disputandam.....
A: You are on the right track. You've already seen what is perhaps the deepest part---that $E(XY) = E(X)E(Y)$ because $X$ and $Y$ are independent.
Here is a simulation, in R, of a million realizations of $W$ to the given specifications. It is reasonable to expect two or three significant digits of
accuracy.
set.seed(928)
m = 10^6;  mu = 4;  sg = 1
x = rnorm(m,mu,sg);  y = rnorm(m,mu,sg);  z = rnorm(m,mu,sg)
w = (1/sqrt(3))*x*y - 2*z - 16/sqrt(3)
mean(w); var(w)
[1] -8.000974    # aprx E(W) = -8
[1] 14.99188     # aprx Var(W) = 15

Beyond your question: $W$ is slightly right-skewed (because of the product
term), so that the simulated distribution of $W$ is not a really close match to
$\mathsf{Norm}(\mu=-8,\sigma=\sqrt{15}).$ 

