# Question on Principal Component Analysis

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.

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.
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.