# Simulating a Gaussian process with an exponentially decaying covariance function

I'm trying to generate many draws (i.e., realizations) of a Gaussian process $e_i(t)$, $1\leq t \leq T$ with mean 0 and covariance function $\gamma(s,t)=\exp(-|t-s|)$.

Is there an efficient way to do this that wouldn't involve computing the square root of a $T \times T$ covariance matrix? Alternatively can anyone recommend an R package to do this?

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It's a stationary process (looks close to a simple version of an OU process). Is it uniformly sampled? –  cardinal May 24 '12 at 13:53
The R package mvtnorm has rmvnorm(n, mean, sigma) where sigma is the covariance matrix; you'd have to construct the covariance matrix for your sampled / selected $t$s yourself, though. –  jbowman May 24 '12 at 14:06
@jb Presumably $T$ is huge, otherwise the OP wouldn't be asking to avoid the matrix decomposition (which is implicit in rmvnorm). –  whuber May 24 '12 at 14:31

Yes. There is a very efficient (linear time) algorithm, and the intuition for it comes directly from the uniformly-sampled case.

Suppose we have a partition of $[0,T]$ such that $0=t_0 < t_1 < t_2 < \cdots < t_n = T$.

Uniformly sampled case

In this case we have $t_i = i \Delta$ where $\Delta = T/n$. Let $X_i := X(t_i)$ denote the value of the discretely sampled process at time $t_i$.

It is easy to see that the $X_i$ form an AR(1) process with correlation $\rho = \exp(-\Delta)$. Hence, we can generate a sample path $\{X_t\}$ for the partition as follows $$X_{i+1} = \rho X_i + \sqrt{1-\rho^2} Z_{i+1} \>,$$ where $Z_i$ are iid $\mathcal N(0,1)$ and $X_0 = Z_0$.

General case

We might then imagine that it could be possible to do this for a general partition. In particular, let $\Delta_i = t_{i+1} - t_i$ and $\rho_i = \exp(-\Delta_i)$. We have that $$\gamma(t_i,t_{i+1}) = \rho_i \>,$$ and so we might guess that $$X_{i+1} = \rho_i X_i + \sqrt{1-\rho_i^2} Z_{i+1} \>.$$

Indeed, $\mathbb E X_{i+1} X_i = \rho_i$ and so we at least have the correlation with the neighboring term correct.

The result now follows by telescoping via the tower property of conditional expectation. Namely, $$\newcommand{\e}{\mathbb E} \e X_i X_{i-\ell} = \e( \e(X_i X_{i-\ell} \mid X_{i-1} )) = \rho_{i-1} \mathbb E X_{i-1} X_{i-\ell} = \cdots = \prod_{k=1}^\ell \rho_{i-k} \>,$$ and the product telescopes in the following way $$\prod_{k=1}^\ell \rho_{i-k} = \exp\Big(-\sum_{k=1}^\ell \Delta_{i-k}\Big) = \exp(t_{i-\ell} - t_i) = \gamma(t_{i-\ell},t_i) \>.$$

This proves the result. Hence the process can be generated on an arbitrary partition from a sequence of iid $\mathcal N(0,1)$ random variables in $O(n)$ time where $n$ is the size of the partition.

NB: This is an exact sampling technique in that it provides a sampled version of the desired process with the exactly correct finite-dimensional distributions. This is in contrast to Euler (and other) discretization schemes for more general SDEs, which incur a bias due to the approximation via discretization.

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Just a few more remarks. (1) To get a good idea of what the continuous time process looks like, $n$ and $T$ must be chosen so that $\Delta$ is small, say less than $0.1$. (2) The inverse covariance (precision) matrix for the timeseries vector is tri-diagonal, as is its Cholesky root. –  Yves May 24 '12 at 17:55
@Yves: Thanks for your comments. To be clear, the procedure I've outlined gives an exact realization of the continuous-time process sampled on the corresponding partition; in particular, there is no discretization error like there is in typical Euler-scheme approximation to more general SDEs. The inverse Cholesky, as shown by the construction in the answer has nonzero terms only on the diagonal and lower off-diagonal, so it's a little simpler than tridiagonal. –  cardinal May 24 '12 at 18:13
Can you give an explicit form of the Cholesky decomposition here? I think that the answer by cardinal achieves just that, if you think about it, by expressing $X_i$ as a function of the history. –  StasK May 25 '12 at 2:54