I understand that the variance of the sum of two independent normally distributed random variables is the sum of the variances, but how does this change when the two random variables are correlated?

  • 2
    $\begingroup$ You have to add twice the covariance. $\endgroup$ Mar 2, 2012 at 1:55
  • $\begingroup$ There is also a good (and simple) explanation on Insight Things. $\endgroup$ Mar 7, 2016 at 14:17

4 Answers 4


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 $$ \text{Var}\Bigl(\,\sum_{i=1}^n X_i\,\Bigr)= \sum_{i=1}^n\text{Var}( X_i) . $$

  • 1
    $\begingroup$ I am unfamiliar with the summation(i<j). Can you explain what this notation means? $\endgroup$
    – Soo
    Mar 2, 2012 at 2:13
  • 4
    $\begingroup$ @soo You calculate all covariances $\text{Cov}(X_i,X_j)$ with $i<j$ and sum them up. Another way to write $2\sum_{i<j}$ in this case is to write $\sum_{i\ne j}$. (The 2 is there in the first sum because in the second sum you calculate, e.g., $\text{Cov}(X_1,X_2)$ and $\text{Cov}(X_2,X_1)$, but these are equal. $\endgroup$ Mar 2, 2012 at 2:17
  • $\begingroup$ David, excellent explanation, the 2 in the 2*cov(...) makes more sense now. Also, can you explain why you wouldn't define an upper limit "n" in the summation(i<j)? Just for my personal curiosity. Thanks $\endgroup$
    – Soo
    Mar 2, 2012 at 2:21
  • $\begingroup$ @soo For your first comment, that's correct. I'll just let your comment be the addendum, if that's ok. $\endgroup$ Mar 2, 2012 at 2:23
  • $\begingroup$ @soo To be rigorous, I should have written $\sum\limits_{i<j\atop 1\le j\le n }$ or something like that. No upper limit though. The lower limit perfectly describes what the index set is. $\endgroup$ Mar 2, 2012 at 2:26

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 $y$:

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

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


Consider a function of two variables, $ z = f(x, y) $. Then the variation of z, $\delta z$, is $$\tag{1} \delta z = \frac{df}{dx} \ \delta x $$ where $$ \frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{ dy}{dx}. $$ Squaring equation (1) we get $$ (\delta z)^2 = \Big[ \left( \frac{\partial f}{\partial x} \right)^2 + 2 \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \frac{dy}{dx} + \left( \frac{\partial f}{\partial y}\right)^2 \left( \frac{dy}{dx} \right)^2 \Big] (\delta x)^2. $$ Multiplying this out we get $$ (\delta z)^2 = \left( \frac{\partial f}{\partial x} \right)^2 (\delta x)^2+ 2 \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \delta x \delta y + \left( \frac{\partial f}{\partial y}\right)^2 (\delta y)^2, $$ where we have used that $\delta y = \frac{dy}{dx} \delta x$. Now we can identify the quadratic variation terms with the variances and covariance of random variables: $$ \text{Var}(z) = \left( \frac{\partial f}{\partial x} \right)^2 \text{Var}(x) + 2 \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \text{Cov}(x,y) + \left( \frac{\partial f}{\partial y}\right)^2 \text{Var}(y). $$ When the function $f$ is just a sum of $x$ and $y$ then the partial derivative terms are all equal to one, giving $$\text{Var}(z) = \text{Var}(x) + 2\ \text{Cov}(x,y) + \text{Var}(y). $$

  • $\begingroup$ This was brilliant. Thank you. $\endgroup$
    – user96265
    Oct 16, 2021 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.