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Suppose I have a random variable $\mathbf{x} \sim \mathcal{N}(0, \sigma^{2} \mathbf{I})$, where $\mathbf{x} \in \mathbf{R}^{N}$, where $N$ is an even number. Let $\mathbf{x}_{r} = \mathbf{A}_{rev} \mathbf{x}$, where $\mathbf{A}_{rev}$ is the matrix that reverses the elements. I want to know what the distribution of $y = \mathbf{x}_{r}^{T} \mathbf{x}$ will be.

It seems that since $x_{i}$, the elements of $\mathbf{x}$, are i.i.d., then since $N$ is even, $x_{r_{i}}$ should be independent of $x_{i}$ $\forall i$. Therefore, according to the book "Products of Random Variables" (https://link.springer.com/chapter/10.1007/978-0-387-47694-0_7), the pdf of $y$ should be:

$$ f(y) = \frac{e^{-\frac{|y|}{\sigma^{2}}}}{\sigma^{2}(M-1)!} \sum_{k=0}^{M-1} \frac{(M+k-1)!}{2^{(M+k)} k! (M-k-1)!} \left(\frac{|y|}{\sigma^{2}}\right)^{M-1-k} $$

where $M = \frac{N}{2}$.

However, when I code this up in python like this:

    length = 2**6
    sigma_2 = 2.0

    #===Generate RV===#
    x = numpy.random.normal(0, numpy.sqrt(sigma_2), length)
    x_r = x[::-1]
    y = sum(x_r*x)

I get the following:PDF Comparison

The left hand side is the corresponding histogram of the computed $y$ values, and the right hand side is the analytic pdf. It is clear they do not match.

When I do:

    length = 2**6
    sigma_2 = 2.0

    #===Generate RV===#
    x = numpy.random.normal(0, numpy.sqrt(sigma_2), length)
    x_r = numpy.random.normal(0, numpy.sqrt(sigma_2), length)
    y = sum(x_r*x)

I get:

PDF Comparison 2

which actually do match.


Question:

Why are these plots different? Is it because of the correlation induced by the random variable generation process, or am I missing something fundamental about what the pdf of $y$ should actually be?

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