Mean of $X_i\sim Ber(\frac{1}{\sqrt{n}})$ converges to zero

Question

Let $X_i\sim Ber\left(\frac{1}{\sqrt{n}}\right)$, that is $X_i=1$ with a probability of $\frac{1}{\sqrt{n}}$ and $X_i=0$ with a probability of $1-\frac {1}{\sqrt{n}}$. I wish to show that $P(|\frac{1}{n}\sum_{i=1}^nX_i|>\varepsilon)\rightarrow 0$ for any $\varepsilon>0$. May I use the weak law of large numbers as the expected value is $\frac {1}{\sqrt{n}}$ and variance is finite? What confuses me is that usually the expected value is fixed. If not, what other approach can I use to show the desired fact?

Answer 1

There is no assumption of independence, so Law Of Larage Numbers is not applicable.

Proof when $X_i \sim B(\frac 1{\sqrt i})$ for every $i$:

$E(\frac{1}{n}\sum\limits_{i=1}^nX_i)=\frac{1}{n}\sum_{i=1}^nEX_i=\frac{1}{n}\sum\limits_{i=1}^n \frac 1 {\sqrt i}$. Comparison of $\sum\limits_{i=1}^n \frac 1 {\sqrt i}$ with $\int_0^{n}\frac 1 {\sqrt x}$ shows that $E(\frac{1}{n}\sum\limits_{i=1}^nX_i) \to 0$. This implies that $\frac{1}{n}\sum\limits_{i=1}^nX_i \to 0$ in probability (use Markov's inequality and non-negativity of $X_i$'s) which is what we have to prove.

Proof when $X_i \sim B(\frac 1{\sqrt n})$: In this case $E(\frac{1}{n}\sum\limits_{i=1}^nX_i)=\frac 1 {\sqrt n} \to 0$ so the conlcusion holds.

Answer 2

Since $X_i\sim Ber(\frac{1}{\sqrt{n}})$, we have $$ E\left( \left| \frac 1 n \sum_{i=1}^n X_i\right| \right)= E\left(\frac 1 n \sum_{i=1}^n X_i \right)=\frac 1 n \sum_{i=1}^nE\left( X_i \right)= \frac 1 n \cdot n \cdot \frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n}} $$ and Markov's inequality yields $$ P(|\frac{1}{n}\sum_{i=1}^nX_i|>\varepsilon) \le \frac{1}{\varepsilon\sqrt{n}}. $$ As the $X_i$ on the left hand side are different random variables for each $n$, letting $n\to\infty$ would be formally clearer, if we read $X_i=X_{n,i}$. Then, this calculation yields that $$ \frac{1}{n}\sum_{i=1}^n X_{n,i} \xrightarrow{n\to\infty} 0 \quad \text{in probability.} $$