An upper bound of $\mathbb{P}(|S_n - \log n| \geq C \log n)$, where $S_n$ is a sum of $n$ independent Bernoulli$-\frac{1}{i}$ random variables

Question

Let $(X_i)_{i=1}^n$ be independent Bernoulli random variables with parameter $\frac{1}{i}$. Let $S_n = \sum_{i=1}^nX_i$ and $C > 0$. I need a bound of $$\mathbb{P}(|S_n - \log n| \geq C \log n)$$ and apparently I can use Chebyshev's inequality to obtain it.

My approach? Well, I first compute the expectation of $S_n$. Then $$\mathbb{E}[S_n] = \sum_{i=1}^n\mathbb{E}[X_i] = \sum_{i=1}^n\frac{1}{i}.$$ Note that $\mathbb{P}(|S_n - \log n| \geq C \log n)$ can be rewritten as $$\mathbb{P}\left(\left|S_n - \sum_{i=1}^n\frac{1}{i} + \sum_{i=1}^n\frac{1}{i} -\log n\right| \geq C \log n\right).$$ But how should I now proceed? There is a hint that $-\log n + \sum_{i=1}^n\frac{1}{i} \to \gamma$, as $n \to \infty$. Here, $\gamma$ is the Euler-Mascheroni constant.

Answer 1

Using Chebycheff's Inequality (in reality we are just using Markov's inequality)

$$P\bigg(\bigg|S_{n}-\ln(n)\bigg|\geq C\ln(n)\bigg)\leq \frac{E(|S_{n}-\ln(n)|^{2})}{C^{2}\ln^{2}(n)}.$$

Let us denote $\sum_{i=1}^{n}\frac{1}{i}$ by $H_{n}$ which means the $n-th$ Harmonic number.

$$S_{n}-\ln(n)=S_{n}-H_{n}+H_{n}-\ln(n)$$

So $(S_{n}-\ln(n))^{2}=(S_{n}-H_{n})^{2}+(H_{n}-\ln(n))^{2}+2(S_{n}-H_{n})(H_{n}-\ln(n))$

So $E((S_{n}-\ln(n))^{2})=Var(S_{n})+(H_{n}-\ln(n))^{2}$

$Var(S_{n})=\sum_{i=1}^{n}Var(X_{i})=\sum_{i=1}^{n}\frac{i-1}{i^{2}}=H_{n}-\sum_{k=1}^{n}\frac{1}{k^{2}}\leq H_{n}-M$ for some $M$ . This is because $\sum_{k=1}^{n}\frac{1}{k^{2}}\to\frac{\pi^{2}}{6}$ .

So you have $$P\bigg(\bigg|S_{n}-\ln(n)\bigg|\geq C\ln(n)\bigg)\leq \frac{1}{C^{2}\ln^{2}(n)}H_{n}-\frac{M}{C^{2}\ln^{2}(n)}+\frac{(H_{n}-\ln(n))^{2}}{C^{2}\ln^{2}(n)}$$

Now $H_{n}\sim \ln(n)+\gamma$ . So $\frac{H_{n}}{\ln(n)}\to 1$ and $\frac{H_{n}-\ln(n)}{\ln(n)}\to 0$ .

Note that this actually proves that $P(|\frac{S_{n}}{\ln(n)}-1|\geq \epsilon)\to 0$ for all $\epsilon>0$ which means $\frac{S_{n}}{\ln(n)}\xrightarrow{\text{in p}} 1$ .

A little more work can yield you almost sure convergence of $\frac{S_{n}}{\ln(n)}$ to $1$.