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Jun 6, 2023 at 23:29 comment added Iosif Pinelis In fact, if $X$ has the chi-squared distribution, then this distribution (or any distribution, which is a measure) cannot be the number $f_n(x)$ or even a function $f_n$. Also, then $(n/2)1/2(X−1)$ cannot have a normal distribution, whatever $n$ is doing. This is not just a question of having clear definitions. Without clear definitions, no proofs are possible. Accordingly, the claims you make in your post can be true only under certain conditions, not stated by you at all (if they can ever be true).
Jun 6, 2023 at 21:04 comment added Iosif Pinelis Then, what does "$(n/2)1/2(X−1)$ has a normal distribution when $n\to\infty$" mean?
Jun 6, 2023 at 19:50 comment added Carlo Beenakker if $X$ has the chi-squared distribution $f_n(x)$, then $(n/2)^{1/2}(X-1)$ has a normal distribution when $n\rightarrow\infty$
Jun 6, 2023 at 18:47 comment added Iosif Pinelis What does "$f_{n}(x)\mapsto \sqrt{\frac{n}{4\pi}}e^{-(n/4)(x-1)^2}$" mean (mathematically)?
Jun 5, 2023 at 11:26 history edited Carlo Beenakker CC BY-SA 4.0
added 16 characters in body
Jun 5, 2023 at 11:17 history answered Carlo Beenakker CC BY-SA 4.0