Gaussian vector and variance

**akbar** · December 17th 2009, 07:19 AM

Let $(\epsilon_1,...,\epsilon_{2n-1})$ be a random vector with the density:
$p(x_1,...,x_{2n-1})=c_n\exp(-\frac{1}{2}(x_1^2+\Sigma_{i=1}^{2n-2}(x_{i+1}-x_i)^2+x_{2n-1}^2))$

One can check this is a Gaussian vector with mean vector zero and the $2n-1\times 2n-1$ correlation matrix has an inverse given by the tridiagonal form:

$\left(\begin{array}{ccccc} 2 & -1 & 0 & ... & 0 \\ -1 & 2 & -1 & .& : \\ 0 & -1 & . & & 0 \\ : & .& & 2 & -1 \\ 0 & ...& 0 & -1 & 2 \end{array}\right)=M_{2n-1}$

By induction, one can also check that for all $n\geq 1$ , $\det(M_n)=n+1$ which enables to obtain the normalizing constant $c_n=\frac{\sqrt{2n}}{(2\pi)^{\frac{2n-1}{2}}}$ using the usual formula for the Gaussian density (see for example Grimmett & Stirzaker).

Now for the question: how do you prove that there is a constant $a$ which does not depend on $n$ such that $Var(\epsilon_n)\geq a.n$ for all $n\geq 1$ ?

By advance, thanks for your help.

**Laurent** · December 19th 2009, 01:22 PM

Originally Posted by akbar

Let $(\epsilon_1,...,\epsilon_{2n-1})$ be a random vector with the density:
$p(x_1,...,x_{2n-1})=c_n\exp(-\frac{1}{2}(x_1^2+\Sigma_{i=1}^{2n-2}(x_{i+1}-x_i)^2+x_{2n-1}^2))$

One can check this is a Gaussian vector with mean vector zero and $2n-1\times 2n-1$ tridiagonal correlation matrix of the form:

$\left( \begin{array}{ccccc} 2 & -1 & 0 & ... & 0 \\ -1 & 2 & -1 & .& : \\ 0 & -1 & . & & 0 \\ : & .& & 2 & -1 \\ 0 & ...& 0 & -1 & 2 \end{array}\right)=M_{2n-1}$
(note: for some obscure reason, large parenthesis \left(, \right(, are not recognized by the forum editor...) (works for me ?!; otherwise you can use the environment "pmatrix", it's lighter to use)

No, this is not the covariance matrix but its inverse. If it was the covariance matrix, you could read the variance from the diagonal coefficients...

The following way works: integrate the marginals one after another (from both sides toward the middle variable $x_n$ ; by symmetry you just have to do one side), do it by hand for the first ones in order to get a pattern suitable for a proof by induction. This way you can resume to studying a real-valued sequence defined by induction (something like $u_{n+1}=2-\frac{1}{u_n}$ where I don't specify what I'm dealing with...), and you need an asymptotic expansion for this sequence. This becomes a calculus question, where various methods apply (you should be able to even get an asymptotic equivalence for the variance).

I let you try to perform this computation; tell me if you don't succeed.

**akbar** · December 19th 2009, 02:04 PM

Yes it is the inverse, sorry about that. So the coefficient is actually $c_n=\frac{\sqrt{2n}}{(2\pi)^\frac{2n-1}{2}}$
I wrote my post too quickly, there is also no syntax issue with parenthesis:
$\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)$
Will have a go at the asymptotic expansion. Will keep you posted (so to speak).

**akbar** · December 29th 2009, 10:59 AM

Having carried the calculation, the strange thing is that I actually end up with an explicit formula for the variance $V(\epsilon_n)$ , without the need of a sequence expansion. Here are the details of my calculations (sorry for the trivialities):

First we isolate each marginal variable in the quadratic form:
$x_1^2+(x_2-x_1)^2=2(x_1-\frac{1}{2}x_2)^2+\frac{1}{2}x_2^2$
We get a similar expression for $x_{2n-1}$ , by symmetry. At each step $p$ , on each side, we get a residual term of the form $\frac{1}{p+1}x_{p+1}^2$ (resp. $\frac{1}{p+1}x_{2n-(p+1)}^2$ ), used for the next step of integration.

Then, by integrating the marginals and using the symmetry:

$p(x_2,...,x_{2n-2})=\int_{x_1,x_{2n-1}}p(x_1,...,x_{2n-1})dx_1dx_{2n-1}$

$=c_n\exp(-\frac{1}{2}(\frac{1}{2}x_2^2+\Sigma_{i=2}^{2n-3}(x_{i+1}-x_i)^2+\frac{1}{2}x_{2n-2}^2)).(\sqrt{2\pi}\sqrt{\frac{1}{2}})^2$

Each (pair of) integration gives us $n-1$ extra terms of the form:
$2\pi\frac{p}{p+1}$ for $1\leq p \leq n-1$ . Their product gives $(2\pi)^{n-1}\frac{1}{n}$ .

We then get $p(x_n)=\frac{\sqrt{2n}}{(2\pi)^\frac{2n-1}{2}}(2\pi)^{n-1}\frac{1}{n}\exp(-\frac{1}{2}(\frac{1}{n}x_n^2+\frac{1}{n}x_n^2))=\f rac{1}{\sqrt{n\pi}}\exp(-\frac{1}{n}x_n^2)$

$V(\epsilon_n)=\frac{1}{\sqrt{n\pi}}\int_{x_n}x_n^2 e^{-\frac{1}{n}x_n^2}dx_n=\frac{n}{\sqrt{\pi}}\int u^2 e^{-u^2}du=\frac{n}{\sqrt<br /> {\pi}}\frac{\sqrt{\pi}}{2}=\frac{n}{2}$

Using a change of variable. Hence the result.
Too good to be true? Maybe your method is more general, in that case I would appreciate some details.

**Laurent** · December 29th 2009, 01:41 PM

Originally Posted by akbar

Hence the result.
Too good to be true? Maybe your method is more general, in that case I would appreciate some details.

This is highly probably correct. I can't remember what I got, but I'm not surprised since actually I had an explicit formula for my sequence as well. Maybe my method would work in a few other cases, but it is not very common to get asymptotic estimates for a sequence defined by induction, so it must not be very important. Your way is way simpler to explain. I did not believe in an explicit formula at first, hence what I did.

(for the computation of the variance, you could also simply recognize the pdf of a Gaussian of variance $\frac{n}{2}$ )

**akbar** · December 29th 2009, 02:16 PM

Thanks for the confirmation.
Best wishes for the new year.

**Laurent** · December 30th 2009, 02:15 AM

Thanks, and happy (upcoming) new year to you too!

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