Finding variance of a point estimator I'm having trouble understanding the solution for an example of finding the variance of a point estimator. Here is the question

I'm confused about the whole last line of the explanation. How do we go from $V(2Y)$ to $4V(Y)$ and then divide that by n?
 A: I've been reading too many stats books. This is from Wackerly, correct?
Remember, $\bar{Y} = \dfrac{Y_1 + Y_2+ \cdots+ Y_n}{n}$, where the $\{Y_i\}_{i=1}^{n}$ are random variables.
$\bar{Y}$ is a random variable too.
So since $\bar{Y}$ is a random variable, recall the theorem
$$\text{Var}[aX] = a^2\text{Var}[X]$$
and hence $\text{Var}[2\bar{Y}] = 2^2\text{Var}[\bar{Y}] = 4\text{Var}[\bar{Y}]$.
Now for the next part. Do the same thing except write out what $\bar{Y}$ is now:
$$4\text{Var}[\bar{Y}] = 4\text{Var}\left[\dfrac{Y_1 + Y_2 +\cdots+ Y_n}{n} \right] = \dfrac{4}{n^2}\text{Var}\left[Y_1 + Y_2 + \cdots + Y_n\right]\text{.}$$
We have that $\{Y_i\}_{i=1}^{n}$ is a random sample. Whenever you see "random sample," you should think independent and identically distributed. Since these are independent, you have
$$\dfrac{4}{n^2}\text{Var}\left[Y_1 + Y_2 + \cdots + Y_n\right] = \dfrac{4}{n^2}\left\{\text{Var}[Y_1]+\text{Var}[Y_2] + \cdots + \text{Var}[Y_n] \right\}\text{.}$$
(This ONLY holds in independence. If you don't have independence, you have to worry about covariance terms.) Now the $Y_i$ are identically distributed, so they have the same variance. So you can just write
$$\dfrac{4}{n^2}\left\{\text{Var}[Y_1]+\text{Var}[Y_2] + \cdots + \text{Var}[Y_n] \right\} = \dfrac{4}{n^2}\left\{\text{Var}[Y_i]+\text{Var}[Y_i] + \cdots + \text{Var}[Y_i] \right\}\text{.}$$
Note that there are $n$ $\text{Var}[Y_i]$ terms, so
$$\dfrac{4}{n^2}\left\{\text{Var}[Y_i]+\text{Var}[Y_i] + \cdots + \text{Var}[Y_i] \right\} = \dfrac{4}{n^2}\cdot n\text{Var}[Y_i] = \dfrac{4}{n}\text{Var}[Y_i]\text{.}$$
Lastly, $Y_i$ is uniform on $(0, \theta)$, so $\text{Var}[Y_i]=\dfrac{(\theta-0)^2}{12} = \dfrac{\theta^2}{12}$.
A: Comment [see my Comment and the excellent Answer by @Clarinetist, (+1)]: 
Here is a simulation that may help you visualize the distributions of the two estimators and their relative dispersions. 
In my simulation
I take $\theta = 5$ and $n = 9.$ The simulated distributions are
based on a million such samples from $Unif(0, \theta).$
 m = 10^6;  th=5;  n = 9.
 x = runif(m*n, 0, th)
 DTA = matrix(x, nrow=m) # m x n matrix; each row a sample
 a = rowMeans(DTA)       # vector of m sample means
 w = apply(DTA, 1, max)  # vector on m sample maximums
 est.1 = 2*a
 est.2 = ((n+1)/n)* w
 mean(est.1);  sd(est.1)
 ## 4.999734   # est of E(est.1) = 5 (unbiased)
 ## 0.962236   # compare with 0.9622504 (from example)
 mean(est.2);  sd(est.2)
 ## 4.999626   # est of E(est.2) = 5 (unbiased)
 ## 0.502947   # markedly smaller than for est.1

The figure below clearly shows that the second estimator
(based on the maximum) has the smaller variance.

