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I need to use a Bernstein-type bound to a random variable $y^Ty$, where $y= Qx$.

$x$ is a Gaussian vector where each entry is independent. $Q$ is a kernel matrix. So $y$ is the linear transformation of independent variables $x$, where entries of $y$ might be correlated. $y$ is jointly Gaussian.

I want to apply a Bernstein-type bound to the second moment of $y$, which is $y^Ty$. So I think I need to first show that $(Qx)^T(Qx)$ is sub-exponential, right?

Can anybody tell me the steps to do this? Or point to a good reference to follow.

I am also referring the tail bounds in Proposition 2.2 in this chapter https://www.stat.berkeley.edu/~mjwain/stat210b/Chap2_TailBounds_Jan22_2015.pdf

user77005
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  • By "linear transformation" don't you mean the *result* of a transformation? The map $x\to Qx$ is the linear transformation; $Qx$ is its result. Assuming this is your intent, is your question about whether a "squared sum of Gaussian variables" may be sub-exponential or would it be about the effect of linear transformations on sub-exponentiality? – whuber Sep 19 '18 at 15:55
  • Actually @whuber what I want to check is whether sums of squares of Gaussians which are potentially correlated are sub-exponential. I think the problem reduces to showing a chi-square variable that has correlations being sub-exponential. – user77005 Sep 20 '18 at 03:11
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    You can diagonalize the covariance matrix to make all correlations zero. For zero-mean Gaussians this will result in a sum of scaled Gamma functions, so the answer can be found at https://stats.stackexchange.com/questions/72479/generic-sum-of-gamma-random-variables/72486#72486. For non zero-mean Gaussians these are non-central Chi-squared variables, but you can expect a similar answer because the tail behavior is not affected by the shift in means. – whuber Sep 20 '18 at 03:16

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