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Suppose a random vector $X\in\mathbb{R}^n$ follows a centered multivariate Gaussian distribution with zero mean and covariance $\Sigma$. We know that a linear combination of every elements in $X$ follows a Gaussian distribution. Suppose we fix a vector $a\in\mathbb{R}^n$, then the linear combination should be: $a^TX$. Then $a^TX$ follows a Gaussian distribution with zero mean, and variance $a^T\Sigma a$.

My question is what is the distribution of $(a^TX)^2$? Is it a simple chi-square distribution (with a little extra scaling to make $a^TX$ have unit variance) with degree of freedom 1?

Or Is it a generalized chi-square distribution? Since $(a^TX)^2=X^Taa^TX$, according to https://en.wikipedia.org/wiki/Generalized_chi-squared_distribution, if we take $aa^T$ as a single matrix, then $(a^TX)^2=X^Taa^TX$ is a quadratic form derived from a multivariate Gaussian.

Thank you!

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$XX^T$ is Wishart $W_n(\Sigma,1),$ so by a property of the Wishart distribution $(a^TX)^2=a^TXX^Ta$ is $W_1(a^T\Sigma a,1)$. Comparing the pdf of this distribution to that of a chi-square, you find that they are the same up to a scaling.

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My question is what is the distribution of $(a^TX)^2$? Is it a simple chi-square >distribution (with a little extra scaling to make aTX have unit variance) with >degree of freedom 1?

That's the right way, you can treat $a'X$ like any $\mathcal{N}(0,\sigma^2)$.

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