Probability limits and limits I'm trying to understand how to practically calculate probability limits, so I'm working on this example.  Suppose we have a sample of data $y_t$, $t=1,2,\ldots,n$ and we are trying to estimate the population mean $\mu$.  Let's take the estimator:
$$
\hat{\mu_1} = \dfrac{1}{n+1} \sum_{t=1}^n y_t
$$
This estimator is biased but we want to show that it is consistent.  We know that it is equal to:
$$
\dfrac{n}{n+1}\bar{y}_n
$$
where $\bar{y}_n$ is the sample mean of $n$ random variables.  By the WLLN, we have $\text{plim}\, \bar{y}_n = \mu$.  Looking at $\dfrac{n}{n+1}\bar{y}_n$, if this were a regular limit, the $\dfrac{n}{n+1}$ would just go to 1.  However, I know that lims are different than plims.  So, what algebraic steps would I take to show that:
$$
\text{plim} \, \dfrac{n}{n+1}\bar{y}_n = \mu
$$
 A: Let $Z_n$ be the sequence of random variables where $Z_n(\omega) = \frac{n}{n+1}$, i.e. they are constant. Clearly $Z_n \to_p 1$. Then by Slutsky's theorem $Z_n \cdot \bar Y_n \to_p 1 \cdot \bar Y$ where the $1$ is the limit of the $Z_n$ and $\bar Y = \mu$ is the limit of the $\bar Y_n$.
That's a high-powered way of showing this. But let's say you want to do a more direct proof.
Fixing some $\varepsilon > 0$ we need to show
$$
P(|\frac{n}{n+1}\bar Y_n - \mu| > \varepsilon) \to 0
$$
as $n \to \infty$. We can do this with Chebyshev's inequality. 
Note that
$$
|\frac n{n+1} \bar Y_n - \mu | = |\frac n{n+1} \bar Y_n - \frac n{n+1}\mu + \frac n{n+1}\mu - \mu |
$$
$$
\leq \frac n{n+1}| \bar Y_n - \mu| + |\mu| \cdot|\frac n{n+1} - 1 |,
$$
so for our $\varepsilon$ we know
$$
P(|\frac n{n+1} \bar Y_n - \mu | > \varepsilon) \leq P(\frac n{n+1}| \bar Y_n - \mu| + |\mu| \cdot|\frac n{n+1} - 1 | > \varepsilon)
$$
$$
= P\left[| \bar Y_n - \mu| > \frac{n+1}{n}\left(\varepsilon - |\mu| \cdot|\frac n{n+1} - 1 |\right)\right].
$$
Let $a_n = \frac{n+1}{n}\left(\varepsilon - |\mu| \cdot|\frac n{n+1} - 1 |\right)$ so that we have $P(|\bar Y_n - \mu| > a_n)$.
Now by Chebyshev's inequality we have
$$
P(| \bar Y_n - \mu| >a_n) \leq \frac{Var(\bar Y_n)}{a_n^2} = \frac{\sigma^2}{n a_n^2}
$$
under the assumption that the $Y_i$ are iid with common finite variance $\sigma^2$.
Can you finish it from here?
A: You don't need any additional moment assumptions on the $Y_i$. WLOG assume $\mu = \mathbb{E}Y_i = 0$ and let $\alpha_1,\alpha_2,\dots$ be any sequence such that $\alpha_n \to 1$. Then
\begin{align}
\mathbb{P}(|\alpha_n \bar Y_n|>\epsilon) & \le \mathbb{P}(|\alpha_n \bar Y_n - \bar Y_n|>\epsilon/2)+\mathbb{P}(|\bar Y_n|>\epsilon/2) \\
& = \mathbb{P}(|1-\alpha_n||\bar Y_n|>\epsilon/2)+ o(1) \\
&\le \mathbb{P}(|\bar Y_n|>n^2)+o(1) ,\text{ for large enough $n$} \\
&\le n\, \mathbb{P}(|Y_1|>n^2) + o(1) \\
&\le \frac{\mathbb{E}|Y_1|}{n}+o(1),\text{ by Markov's inequality} \\
&= o(1).
\end{align}
