Timeline for How to prove the equation holds in asymptotic sense

Current License: CC BY-SA 4.0

9 events

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May 26, 2023 at 17:51	history	edited	Christophe Leuridan	CC BY-SA 4.0	added 59 characters in body
May 26, 2023 at 10:38	comment	added	Thomas Lehéricy		Thanks for correcting the typos. Your lower bound on $\sum_i X_i \ln \frac{X_i}{np_i}$ is still $0$ though, which is strictly weaker than OP’s. Do you think there is a way to improve your method to get the correct lower bound?
May 26, 2023 at 7:01	comment	added	Christophe Leuridan		Sorry, I wrote that too quickly. I hope it is fine now. Pearson Theorem follows from central limit Theorem by the application of a suitable quadratic form.
May 26, 2023 at 6:58	history	edited	Christophe Leuridan	CC BY-SA 4.0	I corrected the answer.
May 25, 2023 at 21:03	comment	added	Thomas Lehéricy		I like the double use of convexity to obtain inequalities that sandwich the quantity! But the bounds don’t match, do they? You get $0\leq \sum \dots \leq \sum \frac{(X_i^{(n)}-np_i)^2}{np_i} + O_P(1)$. – In the first one, why is the first occurence of $f$ outside of the sum? – In the second one, you forgot the $\ln$ in the first sum. – Second paragraph, the first sentence, "Note that saying that a sequence..." is unfinished.
May 25, 2023 at 20:54	history	edited	Christophe Leuridan	CC BY-SA 4.0	I corrected the answer.
May 25, 2023 at 20:54	comment	added	Christophe Leuridan		You are right. I corrected my answer. Pearson theorem can be found in statistic books, since it is the theoretical foundation of the $\chi^2$ test. Yet, I did not find a reference on internet.
May 25, 2023 at 11:29	comment	added	Thomas Lehéricy		Could you add a link to a statement (or reference) of the convergence towards the $\chi^2_{k-1}$? It is interesting though not related to OP’s original question. Also, I cannot find what LHS converges in probability to 0.
May 25, 2023 at 10:41	history	answered	Christophe Leuridan	CC BY-SA 4.0