I am trying to draw random numbers

$$Z_{l,m} = \int_{-1}^1 dx \, P_l^m(x)W(x)$$

Here $P_l^m(x)$ are the associated Legendre polynomials with integers $l\geq0$ and $-l\leq m \leq l$. The variable $W(x)$ corresponds to white noise with zero mean and variance


Here $\delta(x-x')$ is the delta distribution.

I noticed already, that $P_l^m$ is proportional to $P_l^{-m}$, hence I only need to draw numbers for $m\geq0$.

However, I am not sure at all if the remaining random numbers $Z_l^m$ are independent. Does anyone have ideas on either how to show the independence or on how to draw the random numbers?


I think it is better to avoid thinking in terms of bona fide white noise, as physicists often think. You should instead think in terms of computing

$$I_l^m=\int_0^2 P_l^m(x-1) dW(x).$$

This is a stochastic integral. (I shift everything to start at zero for consistency with standard notation in math.)

Approach 1:

Rewrite the stochastic integral by "formal integration by parts":

$$I_l^m=P_l^m(1) W(2)-P_l^m(-1)W(0)-\int_0^2 W(x) (P_l^m)'(x-1) dx \\= P_l^m(1) W(2) - \int_0^2 W(x) (P^m_l)'(x-1) dx.$$

The second term is now a regular integral, not a stochastic integral anymore. You can now approximate this integral as:

$$\sum_{k=1}^N (P_l^m)'\left ( \frac{2k}{N}-1 \right ) \sum_{j=1}^k N_j$$

where $N_j$ are iid normal random variables with mean zero and variance $\frac{2}{N}$. This is essentially the rectangle rule for the regular integral, with the inner sum serving to approximate a sample path of the Wiener process.

  • $\begingroup$ Thanks a lot and sorry for the late answer. Before marking it as an accepted answer, a question to Approach 2. Calculating the mean and covariance will let me draw random numbers. However, there might be dependencies hidden to the covariance. Just as an example: Take the Fourier decomposition. Replace the $P_l^m$'s by $\exp(ikx)$ and $\exp(-ikx)$ with $x \in [0,2\pi]$. Both will have zero covariance, but are related by complex conjugation and thus dependent on each other. So, I don't see approach 2 being always valid, only once the independent indices $l$ and $m$ are determined. $\endgroup$ – physicsGuy Jul 18 '16 at 7:56
  • $\begingroup$ In particular, if $P_l^m(x)$ would be an overcomplete basis on $[-1,1]$, then I could reduce the $(l,m)$ indices such that I get a complete basis $(l_c,m_c)$. Next I could express the $P_l^m$'s as a sum of the complete basis, which yields a reduced number of random numbers, compared to Approach 2. $\endgroup$ – physicsGuy Jul 18 '16 at 8:07
  • $\begingroup$ Multivariate normal r.v.s are fully specified by the mean and covariance. If you have hidden nonlinear dependence then you don't have normal r.v.s but in fact one can show that you do. $\endgroup$ – Ian Jul 18 '16 at 10:17
  • $\begingroup$ It is possible that I made a mistake with approach 2, you can try both approaches and compare. Approach 1 is definitely correct (modulo approximation error). $\endgroup$ – Ian Jul 18 '16 at 10:38
  • $\begingroup$ I thought a bit more about approach 2 and would set your answer as the correct one, if you change or remove approach 2, with the following reason. The associated Legendre polynomials $P_l^0(x)$ already form a complete set, because they are the 'normal' Legendre polynomials. Hence I can express $P_l^m (x) = \sum_L c_{L}^{l,m} P_L^0(x)$. This means, that after drawing numbers for the random integrals $I_l^0$ that you defined above, I know the 'random' integrals for $m \neq 0$ by applying the same transformation. $\endgroup$ – physicsGuy Jul 19 '16 at 8:48

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