2
$\begingroup$

I would greatly appreciate any of you who could help me with this challenge. I am going to state the problem in sequential order, so as to make it clear:

  1. I have $n$ normally distributed random variables.

  2. I have statistical data for each random variable (thus, their mean and variance).

  3. I have historical data for each of the random variables, and from these data I can calculate the correlation between the random variables.

Now, assuming I know the value of one of the variables, how do I select the values of the other variables, such that all the $n$ values in the resulting vector are correlated accordingly. If that is possible, how do I extend this to a situation where I have two or more known values? For example, I know two or more values out of the $n$ variables (they satisfy the correlation constraint), and want to pick the others so that they also satisfy the correlation and their respective distribution functions.

I am not well-versed in statistics, so I would appreciate explanations that are as simple to understand as possible.

$\endgroup$

1 Answer 1

0
$\begingroup$

We're going to have to explain this using vectors and matricies.

Take a look at the 'Conditional distributions' section of the wikipedia page: https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Correlations_and_independence

We have a multivariate normal distribution over the variable $x$ with mean $\mu$ and covariance $\Sigma$. We want to condition it on some elements of $x$ being known. Let's say we can break $x$ up into two parts

$$ x = \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right) $$

where we know $x_2 = a$.

We'll break up our mean and covariance similarly,

$$ \mu = \left( \begin{array}{c} \mu_1 \\ \mu_2 \end{array} \right) $$

$$ \Sigma = \left( \begin{array}{cc} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{array}\right) $$

In this case, the new distribution over $x_1$ given that $x_2 = a$ has mean $$ \mu_{new} = \mu_1+\Sigma_{12}\Sigma_{22}^{-1}(a - \mu_2) $$ and covariance $$ \Sigma_{new} = \Sigma_{11} - \Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21} $$

That's the standard answer. However if you're not comfortable with matricies and vectors then this may need more explanation...

$\endgroup$
1
  • $\begingroup$ Thanks! I think that explanation is good enough ... I will try it out and see how it goes. By the way is there a some code that implements these kind of calculation in python? Thanks $\endgroup$
    – Bobby
    Nov 16, 2012 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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