Convergence in probability of an estimator

I'm asked to verify if the following estimator of $\mu$: $$\hat{\mu}=\frac{1}{n+1}\sum_{t=1}^{n}y_t$$ converges in probability by computing its $p \lim$ (note that $E[y_t]=\mu$). I've made the following computations:

\begin{align} p \lim(\hat{\mu})&=p\lim \left ( \frac{1}{n+1}\sum_{t=1}^{n}y_t \right) \\ &=p\lim \left ( \frac{1}{n+1} \frac{n}{n}\sum_{t=1}^{n}y_t \right ) \\ &=p\lim \left ( \frac{n}{n+1} \frac{1}{n}\sum_{t=1}^{n}y_t \right ) \\ &=\frac{n}{n+1}\ p\lim \left ( \frac{1}{n}\sum_{t=1}^{n}y_t \right ) \\ &=\frac{n}{n+1}\ p\lim \left ( E[y_t] \right ) \\ &=\frac{n}{n+1}\ \mu \end{align}

Can I say that for $n \to +\infty$ the fraction $n/(n+1)$ is equal to 1 and thus $p \lim(\hat{\mu})=\mu$ or must I leave the ratio in front of $\mu$?

• In the first three lines of your calculations, "$n$" was a bound variable: you could have replaced it by any variable name you please without changing the meaning. At the fourth line, where $n/(n+1)$ appears outside the limit, "$n$" suddenly appears unbound--and thereby is meaningless. That line is invalid.
– whuber
Oct 19, 2015 at 16:03

You can say that the limit of the fraction $\frac{n}{n+1}$ is $1$ and use the Law of Large Numbers for the sample mean, yes. It's a mistake however to take the limits sequentially. They have to be taken at the same time. Here for instance, you can use the fact that if $A_n \xrightarrow{P} A$ and $B_n \xrightarrow{P} B$ then $A_n B_n \xrightarrow{P} AB$. And you are done.