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Limit in the Mean Convergence

Suppose we have a stochastic process X(t). Assume it's zero mean has correlation function RX(t1,t2)= [X(t1)X(t2)]. What we'd LIKE is to be able to represent X(t) as a superposition of orthonormal basis functions, scaled by independent random variables. The independence is a long shot. So what we'll DO is settle for uncorrelated projections on to the orthonormal basis functions.

So, we define $X_N(t) = \sum_{n=1}^N a_n \phi_n(t)$ with $a_n = \int \phi_n(t) X(t) dt$ on some interval and require the following:

\begin{displaymath}\lim_{N\rightarrow \infty} E[ (X_N(t) - X(t))^2] = 0
\end{displaymath}

That is, we require that XN(t) converge to X(t) in the mean square sense on the interval of interest.

Note however, that we SEEM to have presupposed that any old set of $\phi_n(t)$ will do. NOT! (in general). We require $E[a_i a_j] = \lambda_i \delta_{ij}$; that is, the ai are uncorrelated and have variances $\lambda_i$(remember, that X(t) is zero mean therefore so are the ai).

\begin{displaymath}E[a_i a_j]
=
\int \int X(t)\phi_i(t) X(\tau) \phi_j(\tau) dt d\tau
=
\int \int R_X(t,\tau)\phi_i(t)\phi_j(\tau) dt d\tau
\end{displaymath}

Now remember that we want the $\phi_i$ to be mutually orthonormal AND we would also like them to SPAN the function space on the interval in question (I've left off the limits for generality, but we are talking about a finite interval). We first rewrite the equation

\begin{displaymath}\lambda_i \delta_{ij}
=
E[a_i a_j]
=
\int \left ( \int R_X(t,\tau)\phi_j(\tau) d\tau
\right ) \phi_i(t) dt
\end{displaymath}

and we see that we must have

\begin{displaymath}\left ( \int R_X(t,\tau)\phi_j(\tau) d\tau
\right ) = \lambda_j \phi_j(t)
\end{displaymath}

which is a standard integral equation. This form is VERY VERY important. Become intimate with it. We call the $\lambda_i$ eigenvalues of the random process X(t) and likewise the $\phi_i(t)$ are the corresponding eigenfunctions. The analogy to linear algebraic concepts is almost exact. We do, however, require that Rx() be square integrable since if not, we have unbounded variances and all hell breaks loose (in a mathematical sense).

Please also note that had the process not been zero mean, we could have defined the integral equation in terms of the covariance function E[(X(t1)-mX(t1))(X(t2)-mX(t2))] with the same result. However, since carrying the mean around often means (no pun intended) a headache, we assume ``centered'' random variables.

Here are some interesting factoids which we list without proof for the most part:

I'll continue this later, but want to get it out. Be wary of typos above. I'll provide the two examples we did in class in what follows.


next up previous
Next: About this document ... Up: No Title Previous: No Title
Christopher Rose
1999-02-23