I have a set of correlated truncated normal random variables $X_1, X_2, \cdots, X_n$ for which the mean $\mu_i$ and the standard deviation $\sigma_i$ are given. At the same the correlation matrix $\Sigma_{n \times n}$ is given. Each random variables $X_i$ is a truncated normal RV to a given interval $[a_i .. b_i]$.

I would like to run the PCA method to reduce the dimensionality. I was thinking that using the pcacov function in Matlab would solve the problem but I do not know if this would be ok since I work with truncated normal rvs.

[PC, variances, explained] = pcacov($\Sigma_{n \times n}$) would give the principal components. In other words one can rewrite the vector $\textbf{X}$ as a linear combination $\textbf{X} = PC \times \textbf{Y}$ where $\textbf{Y}$ represents a set of independent normal rvs.

If the usage of this function is correct I am curious to know what would be the mean and the deviation of each $Y_i$ and if these rvs are also truncated to the given interval $[a_i .. b_i]$.

I assume the standard deviation of each $Y_i$ is $\sqrt{variances_i}$ but what about the mean?

Thanks, Bogdan.


1 Answer 1


Recall that PCA only deals with second order moments; the $PC$ matrix allows us to write the original variable $X$ as an orthogonal transformation (say, an axis rotation) of another variable $Y$ which is uncorrelated (its covariance matrix is diagonal). Afterwards, PCA picks the components of $Y$ that have greater variance.

In the special case in which $X$ (and hence $Y$) is jointly normal, $Y$ components will be , not only uncorrelated but independent.

In your case, there's nothing special. You just apply the usual formulas and get the transformed correlation matrix, the same as it were jointly normal.

Small caveat: all this assumes that $X$ -and hence $Y$- has zero mean; in other words, this is assumed to be applied to the mean-centered variables. So, the final mean of the transformed variable is the same as the original.

Because the original variables are truncated, their support lies inside a (hyper)rectangle. The corresponding transformed support will be a rotated (hyper)rectangle, and hence they will not (in general) correspond to truncated normals.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .