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Short question: Why is this true??

Long question:

Very simply, I am trying to figure out what justifies this first equation. The author of the book I am reading, (context here if you want it, but not necessary), claims the following:

Due to the assumption of near-gaussianity, we can write:

$$ p_0(\xi) = A \; \phi(\xi) \; exp( a_{n+1}\xi + (a_{n+2} + \frac{1}{2})\xi^2 + \sum_{i=1}^{n} a_i G_i(\xi)) $$

Where $p_0(\xi)$ is the PDF of your observed data that has maximum entropy, given that you had only observed a series of expectations, (simple numbers) $c_i, i = 1 ... n$, where $c_i = \mathbb{E}\{G_i(\xi)\}$, and $\phi(\xi)$ is the PDF of a standardized gaussian variable, that is, 0 mean, and unit variance.

Where all this is going is that he uses the above equation as a starting point for making the PDF, $p_0(\xi)$ simpler, and I get how he does it, but I do not get how he justifies the above equation, ie, the starting point.

I have tried to keep it brief to as not to obfuscate anyone, but if you want additional details please let me know in the comments. Thanks!

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1 Answer 1

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(Note: I've changed your $\xi$ to $x$.)

For a random variable $X$ with density $p$, if you have constraints $$ \int G_i(x)\,p(x)\,dx=c_i \, , $$ for $i=1,\dots,n$, the maximum entropy density is $$ p_0(x)=A\exp\left(\sum_{i=1}^n a_iG_i(x)\right) \, , $$ where the $a_i$'s are determined from the $c_i$'s, and $A$ is a normalization constant.

In this context, the Gaussian approximation ("near-gaussianity") means two things:

1) You accept to introduce two new constraints: the mean of $X$ is $0$ and the variance is $1$ (say);

2) The corresponding $a_{n+2}$ (see bellow) is much bigger than the other $a_i$'s.

These additional constraints are represented as $$ G_{n+1}(x)=x \, , \qquad c_{n+1}=0 \, , $$ $$ G_{n+2}(x)=x^2 \, , \qquad c_{n+2}=1 \, , $$ yielding $$ p_0(x)=A\exp\left(a_{n+2}x^2 + a_{n+1}x + \sum_{i=1}^n a_iG_i(x)\right) \, , $$ which can be rewritten as (just "add zero" to the exponent) $$ p_0(x)=A\exp\left(\frac{x^2}{2} - \frac{x^2}{2} + a_{n+2}x^2 + a_{n+1}x + \sum_{i=1}^n a_iG_i(x)\right) \, , $$ leading to what you want: $$ p_0(x)=A'\,\phi(x)\exp\left(a_{n+1}x + \left(a_{n+2}+\frac{1}{2}\right)x^2 + \sum_{i=1}^n a_iG_i(x)\right) \, ; $$ ready to be Taylor expanded (using the second condition of the Gaussian approximation).

Doing the approximation like a Physicist (which means that we don't care about the order of the error term), using $\exp(t)\approx 1+t$, we have the approximate density $$ p_0(x) \approx A'\,\phi(x)\left(1+a_{n+1}x + \left(a_{n+2}+\frac{1}{2}\right)x^2 + \sum_{i=1}^n a_iG_i(x)\right) \, . $$ To finish, we have to determine $A'$ and the values of the $a_i$'s. This is done imposing the conditions $$ \int p_0(x)\,dx=1 \, , \qquad \int x \,p_0(x)\,dx=0 \, , \qquad \int x^2 \,p_0(x)\,dx=1 $$ $$ \int G_i(x)\, p_0(x)\,dx=c_i \, , \quad i=1,\dots,n \, , $$ to obtain a system of equations, whose solution gives $A'$ and the $a_i$'s.

Without imposing additional conditions on the $G_i$'s, I don't believe that there is a simple solution in closed form.

P.S. Mohammad clarified during a chat that with additional orthogonality conditions for the $G_i$'s we can solve the system.

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  • $\begingroup$ Zen, thanks very much. I (somewhat) understand now. What is not clear to me though, is when you say "In this context, the Gaussian approximation ("near-gaussianity") means that you accept to introduce two new constraints: that the mean of X is 0 and the variance is (say) 1." , I dont understand, why for something to be 'near gaussian', means for it to have $\mu=0$ and $\sigma^2=1$. What if it was just another r.v. that happened to have those same values? $\endgroup$
    – Spacey
    Commented Aug 29, 2012 at 14:39
  • $\begingroup$ Hi Mohammad. I've added more information to the answer. To get the former expression of $p_0(x)$ you use only what I've called the first condition of the Gaussian approximation. You will use the second condition when you do the Taylor expansion of this $p_0(x)$. I hope this helps. $\endgroup$
    – Zen
    Commented Aug 29, 2012 at 19:42
  • $\begingroup$ Would you mind posting as a comment the final expression for the $p_0(x)$ after you do the remaining computations? Thanks. $\endgroup$
    – Zen
    Commented Aug 29, 2012 at 20:07
  • $\begingroup$ yes, he is saying that the final expression is: $p_0(z) \approx \phi(z) (1 + \sum_{i=1}^{N} c_i F_i(z))$ $\endgroup$
    – Spacey
    Commented Aug 29, 2012 at 20:11
  • $\begingroup$ I think there is a typo in the last equation?... $a_{n+1}x$ is happening twice?... $\endgroup$
    – Spacey
    Commented Aug 29, 2012 at 20:20

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