# sum of correlated random sample

Suppose I have 1000 draws each of two random variables X and Y.

If I wanted to sample the sum of these variables, I would simply calculate 1000 samples, i.e.

$$S_{i}=X_{i}+Y_{i}, i=1,2,…,1000$$

And that would give me draws from the pdf of the sum of these two variables.

Now suppose I know that the correlation between these two variables is ρ=0.8

A higher value of X is now more likely to coincide with a higher value of Y, and the standard deviation of the sums will be higher.

Can I resample the sum of the two variables without adding a specific assumption regarding the specific destribution of X and Y?

For example.

Assume X and Y are population forecasts for male population in Washington and Oregon in 2050. I have two separate models for population in Oregon and Washington that give me pdf for population in Washington/Oregon in 2050. Because the models were run independently of each other, X+Y=S will underestimate the standard deviance of the sum, simply because of the fact that coefficients and/or the future development of independent variables in the model for Washington and Oregon are likely to be correlated.

Let us also assume that I only have those draws and I cannot account for the fact that X and Y are related by designing a smarter model. I can however observe the correlation of population numbers in Washington and Oregon.

• I don't understand the question. You have already described how to draw samples of the sum. – Tom Minka Oct 22 '14 at 11:13
• Thank you for your reply. I know how to draw samples from the sum if I assume the two variables to be independent, they are however correlated. This fact is not reflected in my draws because I obtained these two sets of draws independently from each other. – mzuba Oct 22 '14 at 12:34

You cannot sample the sum without making distributional assumptions, but you can compute its variance, using the formula $${\rm var}(X+Y) = {\rm var}(X) + {\rm var}(Y) + 2 {\rm cov}(X,Y)$$ since ${\rm cov}(X,Y) = \rho \sqrt{{\rm var}(X) {\rm var}(Y)}$.