Determining variance from sum of two random correlated variables

For any two random variables: $$\text{Var}(X+Y) =\text{Var}(X)+\text{Var}(Y)+2\text{Cov}(X,Y).$$ If the variables are uncorrelated (that is, $\text{Cov}(X,Y)=0$), then

$$\tag{1}\text{Var}(X+Y) =\text{Var}(X)+\text{Var}(Y).$$ In particular, if $X$ and $Y$ are independent, then equation $(1)$ holds.

In general $$ \text{Var}\Bigl(\,\sum_{i=1}^n X_i\,\Bigr)= \sum_{i=1}^n\text{Var}( X_i)+ 2\sum_{i< j} \text{Cov}(X_i,X_j). $$ If for each $i\ne j$, $X_i$ and $X_j$ are uncorrelated, in particular if the $X_i$ are pairwise independent (that is, $X_i$ and $X_j$ are independent whenever $i\ne j$), then $$ \text{Var}\Bigl(\,\sum_{i=1}^n X_i\,\Bigr)= \sum_{i=1}^n\text{Var}( X_i) . $$

You can also think in vector form:

$$\text{Var}(a^T X) = a^T \text{Var}(X) a$$

where $a$ could be a vector or a matrix, $X = (X_1, X_2, \dots, X_n)^T$ is a vector of random variables. $\text{Var}(X)$ is the covariance matrix.

If $a = (1, 1, \dots, 1)^T$, then $a^T X$ is the sum of all the $x_i's$.

Let's work this out from the definitions. Let's say we have 2 random variables $x$ and $y$ with means $\mu_x$ and $\mu_y$. Then variances of $x$ and $y$ would be:

$${\sigma_x}^2 = \frac{\sum_i(\mu_x-x_i)(\mu_x-x_i)}{N}$$ $${\sigma_y}^2 = \frac{\sum_i(\mu_y-y_i)(\mu_y-y_i)}{N}$$

Covariance of $x$ and $y$ is:

$${\sigma_{xy}} = \frac{\sum_i(\mu_x-x_i)(\mu_y-y_i)}{N}$$

Now, let us consider the weighted sum $p$ of $x$ and $y$:

$$\mu_p = w_x\mu_x + w_y\mu_y$$

$${\sigma_p}^2 = \frac{\sum_i(\mu_p-p_i)(\mu_p-p_i)}{N} = \frac{\sum_i(w_x\mu_x + w_y\mu_y - w_xx_i - w_yy_i)^2}{N} = \frac{\sum_i(w_x(\mu_x - x_i) + w_y(\mu_y - w_yy_i))^2}{N} = \frac{\sum_i(w^2_x(\mu_x - x_i)^2 + w^2_y(\mu_y - y_i)^2 + 2w_xw_y(\mu_x - x_i)(\mu_y - y_i))}{N} = w^2_x\frac{\sum_i(\mu_x-x_i)^2}{N} + w^2_y\frac{\sum_i(\mu_y-y_i)^2}{N} + 2w_xw_y\frac{\sum_i(\mu_x-x_i)(\mu_y-y_i)}{N} = w^2_x\sigma^2_x + w^2_y\sigma^2_y + 2w_xw_y\sigma_{xy}$$