How does one find the mean of a sum of dependent variables? I know that the mean of the sum of independent variables is the sum of the means of each independent variable. Does this apply to dependent variables as well?
 A: Expectation (taking the mean) is a linear operator.
This means that, amongst other things, $\mathbb{E}(X + Y) = \mathbb{E}(X) + \mathbb{E}(Y)$ for any two random variables $X$ and $Y$ (for which the expectations exist), regardless of whether they are independent or not.
We can generalise (e.g. by induction) so that $\mathbb{E}\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \mathbb{E}(X_i)$ so long as each expectation $\mathbb{E}(X_i)$ exists.
So yes, the mean of the sum is the same as the sum of the mean even if the variables are dependent. But note that this does not apply for the variance! So while $\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$ for independent variables, or even variables which are dependent but uncorrelated, the general formula is $\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) + 2\mathrm{Cov}(X, Y)$ where $\mathrm{Cov}$ is the covariance of the variables.
A: TL; DR:
Assuming it exists, the mean is an expected value, and the expected value is an integral, and the integrals have the linearity property with respect to sums.
TS; DR:
Since we are dealing with the sum of random variables $Y_n = \sum_{i=1}^n X_i$, i.e. of a function of many of them, the mean of the sum $E(Y_n)$ is with respect to their joint distribution (we assume that all means exist and are finite) Denoting $\mathbf X$ the multivariate vector of the $n$ r.v.'s, their joint density can be written as $f_{\mathbf X}(\mathbf x)= f_{X_1,...,X_n}(x_1,...,x_n)$ and their joint support 
$D = S_{X_1} \times ...\times S_{X_n}$
Using the Law of Unconcscious Statistician  we have the multiple integral
$$E[Y_n] = \int_D Y_nf_{\mathbf X}(\mathbf x)d\mathbf x$$.
Under some regularity conditions we can decompose the multiple integral into an $n$-iterative integral: 
$$E[Y_n] = \int_{S_{X_n}}...\int_{S_{X_1}}\Big[\sum_{i=1}^n X_i\Big]f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n $$
and using the linearity of integrals we can decompose into
$$ = \int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n \; + ...\\ ...+\int_{S_{X_n}}...\int_{S_{X_1}}x_nf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n $$
For each $n$-iterative integral we can re-arrange the order of integration so that, in each, the outer integration is with respect to the variable that is outside the joint density. Namely,
$$\int_{S_{X_n}}...\int_{S_{X_1}}x_1f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_n = \\\int_{S_{X_1}}x_1\int_{S_{X_n}}...\int_{S_{X_2}}f_{X_1,...,X_n}(x_1,...,x_n)dx_2...dx_ndx_1$$ 
and in general
$$\int_{S_{X_n}}...\int_{S_{X_j}}...\int_{S_{X_1}}x_jf_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_j...dx_n =$$
$$=\int_{S_{X_j}}x_j\int_{S_{X_n}}...\int_{S_{X_{j-1}}}\int_{S_{X_{j+1}}}...\int_{S_{X_1}}f_{X_1,...,X_n}(x_1,...,x_n)dx_1...dx_{j-1}dx_{j+1}......dx_ndx_j$$
As we calculate one-by-one the integral in each $n$-iterative integral (starting from the inside), we "integrate out" a variable and we obtain in each step the "joint-marginal" distribution of the other variables. Each $n$-iterative integral therefore will end up as $\int_{S_{X_j}}x_jf_{X_j}(x_j)dx_j$.
Bringing it all together we arrive at 
$$E[Y_n ] = E[\sum_{i=1}^n X_i] = \int_{S_{X_1}}x_1f_{X_1}(x_1)dx_1 +...+\int_{S_{X_n}}x_nf_{X_n}(x_n)dx_n $$
But now each simple integral is the expected value of each random variable separately, so 
$$ E[\sum_{i=1}^n X_i] = E(X_1) + ...+E(X_n) $$
$$= \sum_{i=1}^nE(X_i) $$
Note that we never invoked independence or non-independence of the random variables involved, but we worked solely with their joint distribution.
