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I want to know how $\mathbb{E}[|\chi_n^2 - n|]$ behaves as $n\to \infty$. Simulating this in R suggest that it grows at a rate of $\sqrt{n}$, but I am unable to prove it. Setting $\chi_n^2 =\sum_{i=1}^n Z_i^2$, where $Z_i$ are iid standard normal random variables, there is a trivial upper bound $$ \mathbb{E}[|\chi_n^2 - n]|] = \mathbb{E}[|\sum_{i=1}^n (Z_i^2-1)|]\\ \leq \sum_{i=1}^n \mathbb{E}[|Z_i^2 - 1|], $$ which grows at a rate of $n$ as $n\to\infty$. However, this bound is clearly too high. What can we say about the limiting behaviour of $\mathbb{E}[|\chi_n^2 - n|]$, or is there even an explicit expression for it?

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Your simulation is correct. Let us observe that $$ \frac1{\sqrt n}\sum_{i=1}^n(Z_i^2-1)\to\mathcal N(0,2) $$ in distribution as $n\to\infty$ by the central limit theorem. We can show that $$ \operatorname E\biggl|\frac1{\sqrt n}\sum_{i=1}^n(Z_i^2-1)\biggr|\to\operatorname E|Y| $$ as $n\to\infty$, where $Z\sim\mathcal N(0,2)$ (see Theorem 25.12 and its Corollary on p. 338 of Billingsley's textbook). Hence, $$ \operatorname E|\chi_n^2-n|=\sqrt n\operatorname E\biggl|\frac1{\sqrt n}\sum_{i=1}^n(Z_i-1)^2\biggr|\sim\sqrt n\operatorname E|Y| $$ as $n\to\infty$.

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By the central limit theorem we have $\frac{1}{\sqrt{n}}(\chi_n^2 - n) \xrightarrow{n \to \infty} \mathcal{N}(0, 2)$, which suggests that the scaling speed of $\sqrt{n}$ is correct.

If you just need an upper bound, we can use Jensen's inequality. If $X_n$ is chi-square distributed with $n$ degrees of freedom and $Z_i$ are independent standard Gaussians, then $$E[|X_n - n|]^2 \le E[(X_n - n)^2] = E\left[\left(\sum \limits_{i = 1}^n (Z_i^2 - 1)\right)^2\right] = \sum \limits_{i = 1}^n E[(Z_i^2 - 1)^2] + \sum \limits_{i < j} \operatorname{cov}(Z_i^2, Z_j^2) = 2n$$

This shows that actually $E[|X_n - n|] \le \sqrt{2n}$.

More precisely, if we actually had $E[|X_n - n|] \le \sqrt{n} a_n$ for some null sequence $a_n$, we would get $E\left[\frac{1}{\sqrt{n}}|X_n - n|\right] \le a_n$, which would imply that $\frac{1}{\sqrt{n}} |X_n - n|$ converges to zero in distribution (but this would contradict the central limit theorem).

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