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Let $X, Y$ be iid Chi-squared random variables with parameter $k$ and consider, \begin{align*} Z = \frac{X-Y}{X+Y}. \end{align*}

I am after bounds for the tail: $\mathbb{P}[ |Z| > t ]$. I know the upper-bound is exponential, and my intuition is that there is a matching lower-bound. However, I'm not sure how to show it (or if its even true). Asymptotic results for $k$ large enough would be fine too (i.e. using approximation of Chi squared with normals). I'm a bit of a noob when it comes to this kind of analysis, so even just a general outline of how to approach this problem would be appreciated.

Im also interested in estimating $\mathbb{E}[|Z|]$ (in particular, finding a lower bound).

I found this: Distribution of Difference of Chi-squared Variables which handles the numerator, but the denominator is clearly not independent which perhaps introduces some difficulties.

Using mathematica, it seem that if I assume $X,Y\sim\mathcal{N}(k,(\sqrt{2k})^2)$, then for $k$ sufficiently large it seems that $Z$ is more or less distributed like $\mathcal{N}(0,(1/\sqrt{k})^2)$. But I have no idea how to obtain this analytically.

RobPratt
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1 Answers1

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I was talking with a professor and they gave this first order approximation.

In the large $k$ limit, $X\sim k + \sqrt{k} U$, $Y\sim k + \sqrt{k} V$ where $U,V$ are iid $\mathcal{N}(0,2)$ random variables.

Thus, \begin{align*} \frac{X-Y}{X+Y} = \frac{\sqrt{k}(U-V)}{2k + \sqrt{k}(U+V)} = \frac{1}{\sqrt{k}} \frac{U-V}{2+(U+V)/\sqrt{k}} = \frac{1}{\sqrt{k}} (U-V)\left( \frac{1}{2} + \mathcal{O}(k^{-1/2}) \right). \end{align*} Now $U-V \sim \mathcal{N}(0,4)$.

Therefore, ignoring the higher order term we have $(U-V)/(2\sqrt{k}) \sim \mathcal{N}(0,1/k)$.

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