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We all know that a chi-square distributed random variable: $U \sim \chi_n^2$ results from:

$ U = \sum_{i=1}^n X_i^2 $, where $X_i \sim \mathcal{N}(0,1)$ and all $X_i$ are i.i.d.

So, my question is whether we can obtain a chi-square distribution if instead of squaring each $X_i$, we multiply it times another independent random variable $Y_i$ with similar distribution. That is:

$V = \sum_{i=1}^n Xi Yi$ for $X_i \sim \mathcal{N}(0,1)$ and $Y_i \sim \mathcal{N}(0,1)$ independent and i.i.d each. Is $V$ a Chi-Square distributed Random Variable: $V \sim \chi_n^2$?

Please, let me know what do you think since, at the end, the long term question I want to answer is: whether the following random variable is chi-squared distributed as $W \sim \chi_{n-1}^2$:

$W = \sum_{i=1}^{n-1} X_i X_{i+1}$

thanks!

kentropy
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    Notice that $V=\sum^n_{i=1}{X_iY_i}$ may take negative values with positive probability. – Mittens Dec 19 '19 at 04:32
  • Thanks for your reply. I get why you point out this, as this fact tells us that definitely, this is not chi-square. I was looking at this post https://math.stackexchange.com/questions/484300/sums-of-products-of-two-normal-variables and have the feeling that by LLN this may become a normal when n is very large. – kentropy Dec 19 '19 at 04:36

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