Thursday, June 15, 2017

Joint Distribution From Marginals

Consider two dependent random variables, \(y_1\) and \(y_2\), with a correlation coefficient \(\rho\).

Suppose you are given the marginal distributions \(\pi(y_1)\) and \(\pi(y_2)\) of the two random variables. Is it possible to construct the joint probability distribution \(\pi(y_1, y_2)\) from the marginals?

In general, the answer is no. There is no unique answer. The marginals are like shadows of a hill from two orthogonal angles. The shadows are not sufficient to specify the full 3D shape (joint distribution) of the hill.

Let us simplify the problem a little, so that we can seek a solution.

Let us assume \(y_1\) and \(y_2\) have zero mean and unit standard deviation. We can always generalize later by shifting (different mean) and scaling (different standard distribution). Let us also stack them into a single random vector \(Y = [y_1, y_2]\).

The covariance matrix of two such random variables is given by, \[C(Y) = \begin{bmatrix} E(y_1 y_1) - \mu_1 \mu_1 & E(y_1 y_2) - \mu_1 \mu_2 \\ E(y_2 y_1) - \mu_2 \mu_1 & E(y_2 y_2) - \mu_2 \mu_2 \end{bmatrix} = \begin{bmatrix} 1 & \rho \\ \rho  & 1 \end{bmatrix},\] where \(\mu\) and \(\sigma\) refer to the mean and standard deviation.


A particular method for sampling from the joint distribution of correlated random variables \(Y\) begins by drawing samples of independent random variables \(X = [x_1, x_2]\) which have the same distribution as the desired marginal distributions.

Note that the covariance matrix in this case is an identity matrix, because the correlation between independent variables is zero  \(C(X) = I\).

Now we recognize that the covariance matrix \(C(Y)\) is symmetric and positive definite. We can use Cholesky decomposition \(C(Y) = LL^T\) to find the lower triangular matrix \(L\).

The recipe then says that we can draw the correlated random variables with the desired marginal distribution by simply setting \(Y = L X\).


Suppose we seek two random variables whose marginals are normal distributions (zero mean, unit standard deviation) with a correlation coefficient 0.2.

The method above asks us to start with independent random variables \(X\) such as those below.

Cholesky decomposition with \(\rho\) = 0.2, gives us,  \[L = \begin{bmatrix} 1 & 0 \\ 0.1  & 0.9797 \end{bmatrix}.\] If we generate \(Y = LX\) using the same data-points used to create the scatterplot above, we get,

It has the same marginal distribution, and a non-zero correlation coefficient as is visible from the figure above.

1 comment:

Philips Huges said...

Its very useful to me. Wonderful blog.. Thanks for sharing informative Post.

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