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I want to find the normalization constant for the following pdf

$$ f(x_1,x_2,\ldots,x_n) = C\cdot\Pi_{m=1}^n\exp\left\{-\frac{1}{2a^2}(x_m-\sum_{l=1}^L\alpha_lx_{m-l})^2+y_m\sum_{l=0}^L\beta_lx_{m-l}\right\} $$ where $n>L$ and $x_l=0$ for $l\leq0$. Namely, I want to find $C$ for which

$$ \int_{\mathbb{R}^n}f(x_1,x_2,\ldots,x_n)dx_1\ldots dx_n = 1. $$

Any idea?

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I think you mean $x_l=0$ for $l\le0$? – joriki Dec 15 '12 at 7:11
@joriki: Yes, you right. Thx – Josh Dec 15 '12 at 12:02
An idea would be to write the argument of the exponential in a matrix form, complete the squares and diagonalize the matrix. But it won't be pretty. – Marek Dec 15 '12 at 12:10
@Marek: Yes, I have tried that, and it's really not pretty. I thought that maybe there is a nice trick that I misses. – Josh Dec 15 '12 at 13:05
Then I'd suggest to transform from the index $l$ to the index $k=m-l$ so you don't have to deal with arithmetic inside the indices. – joriki Dec 15 '12 at 18:46
up vote 1 down vote accepted

From your comment, it seems that you may not be aware that the outer product is equivalent to a sum in the exponent. Using $\exp(a)\exp(b)=\exp(a+b)$ and transforming the indices as suggested in a comment, you can rewrite your function as

$$ f(x_1,x_2,\ldots,x_n) = C\exp\left\{\sum_{m=1}^n\left(-\frac{1}{2a^2}\left(x_m-\sum_{k=m-L}^{m-1}\alpha_{m-k}x_k\right)^2+y_m\sum_{l=0}^L\beta_{m-k}x_k\right)\right\}\;. $$

You can see from this representation that the matrix of the quadratic form (whose determinant you need) is a band matrix with bandwidth $2L+1$. The diagonal entries are constant ($1+\sum_i\alpha_i^2$) up to $m=n-L$, but decrease for greater $m$ (because successive $\alpha_i^2$ go missing) down to $1$ for $m=n$. Because of this asymmetry, I doubt that you'll find a nice closed form for the determinant.

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Yes, my previous comment is not clear (I was aware that the outer product is equivalent to a sum in the exponent). Anyway, it really looks that I will not find a nice solution. Thx. – Josh Dec 15 '12 at 19:31

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