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I'm TA'ing a course and have trouble solving an exercise.

Let $X$ be a RV defined to be the number of trials required to collect at least one of each type of coupon (of which there are $n$). Then $E[X] = nH_n = n \ln(n) + O(n)$ and $\sigma^2_X = n^2 \sum_{i=1}^n \frac{1}{i^2} - nH_n$ with $\lim_{n \to \infty} \frac{\sigma^2_X}{n^2} = \frac{\pi^2}{6}$. The exercise asks how Chebyshev's inequality can be used to upper bound the probability that $X > \beta n \ln(n)$ for some $\beta > 1$.

I tried this: \begin{align*} Pr[X>\beta n \ln(n)] &\leq Pr[|X-\mu_X| \geq (\beta - 1)n \ln n] \\ &\leq Pr\left[|X-\mu_X| \geq (\beta - 1) \left( \frac{\sqrt 6}{\pi} \sigma_X \right) \ln n\right] (1) \\ &\leq \frac{\pi}{6(\beta -1)^2 \ln^2 n} = O\left(\frac{1}{\ln^2n}\right) \end{align*}

where (1) follows because $n > \frac{\sigma_X \sqrt 6}{\pi}$. However I believe the first inequality is incorrect because I'm discarding the $- O(n)$ from $\mu_X$ in deriving an upper bound.

I also tried something else: \begin{align*} Pr[X>\beta n \ln(n)] &\leq Pr[|X-\mu_X| \geq \beta n \ln n - nH_n] \\ &\leq Pr[|X-\mu_X| \geq (\beta -1)nH_n] \\ &= Pr\left[|X-\mu_X| \geq \frac{6(\beta - 1)H_n}{\pi^2n} \left(n^2 \sum_{i=1}^\infty \frac{1}{i^2} - n H_n \right)\right] \end{align*}

but I'm not sure where to go from here. Thanks for any help!

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As you should have suspected, there is only a thin line between what you have tried and what is rigorous.

Here it is the rigorous version of your attempt. Since $\beta n\ln n\gt \mu_X$ when $n$ is large enough, the first inequality holds for $n$ large enough. \begin{align*} Pr[X>\beta n \ln(n)] &\leq Pr[|X-\mu_X| \geq \beta n \ln n - \mu_X] \\ &\le \frac{\sigma_X^2}{(\beta n \ln n - \mu_X)^2}.\\ \end{align*} Now

$$\begin{aligned} \lim_{n\to\infty}\frac{\frac{\sigma_X^2}{(\beta n \ln n - \mu_X)^2}}{\frac1{\ln^2 n}} &=\lim_{n\to\infty}{\frac{\sigma_X^2}{n^2}}\lim_{n\to\infty}\frac{(n\ln n)^2}{(\beta n \ln n - \mu_X)^2}\\ &=\frac{\pi^2}{6}\lim_{n\to\infty}\frac{1}{(\beta - \frac{\mu_X}{n\ln n})^2}\\ &=\frac{\pi^2}{6(\beta - 1)^2}, \end{aligned}$$ which implies $$\frac{\sigma_X^2}{(\beta n \ln n - \mu_X)^2} = O\left(\frac{1}{\ln^2n}\right).$$

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