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Chi-squared distribution
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=== Concentration === The chi-squared distribution exhibits strong concentration around its mean. The standard Laurent-Massart<ref>{{Cite journal |last1=Laurent |first1=B. |last2=Massart |first2=P. |date=2000-10-01 |title=Adaptive estimation of a quadratic functional by model selection |journal=The Annals of Statistics |volume=28 |issue=5 |doi=10.1214/aos/1015957395 |s2cid=116945590 |issn=0090-5364|doi-access=free }}</ref> bounds are: : <math>\operatorname{P}(X - k \ge 2 \sqrt{k x} + 2x) \le \exp(-x)</math> : <math>\operatorname{P}(k - X \ge 2 \sqrt{k x}) \le \exp(-x)</math> One consequence is that, if <math>Z \sim N(0, 1)^k</math> is a gaussian random vector in <math>\R^k</math>, then as the dimension <math>k</math> grows, the squared length of the vector is concentrated tightly around <math>k</math> with a width <math>k^{1/2 + \alpha}</math>:<math display="block">Pr(\|Z\|^2 \in [k - 2k^{1/2+\alpha}, k + 2k^{1/2+\alpha} + 2k^{\alpha}]) \geq 1-e^{-k^\alpha}</math>where the exponent <math>\alpha</math> can be chosen as any value in <math>\R</math>. Since the cumulant generating function for <math>\chi^2(k)</math> is <math>K(t) = -\frac k2 \ln(1-2t) </math>, and its [[Convex conjugate|convex dual]] is <math>K^*(q) = \frac 12 (q-k + k\ln\frac kq) </math>, the standard [[Chernoff bound]] yields<math display="block">\begin{aligned} \ln Pr(X \geq (1 + \epsilon) k) &\leq -\frac k2 ( \epsilon - \ln(1+\epsilon)) \\ \ln Pr(X \leq (1 - \epsilon) k) &\leq -\frac k2 ( -\epsilon - \ln(1-\epsilon)) \end{aligned}</math>where <math>0< \epsilon < 1</math>. By the union bound,<math display="block">Pr(X \in (1\pm \epsilon ) k ) \geq 1 - 2e^{-\frac k2 (\frac 12 \epsilon^2 - \frac 13 \epsilon^3)} </math>This result is used in proving the [[Johnson–Lindenstrauss lemma]].<ref>[https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/f9261308512f6b90e284599f94055bb4_MIT18_S096F15_Ses15_16.pdf MIT 18.S096 (Fall 2015): Topics in Mathematics of Data Science, Lecture 5, Johnson-Lindenstrauss Lemma and Gordons Theorem]</ref>
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