Proving consistency for an estimator. Limits and Convergence in Probability.

I need to show that $$U$$, as defined below, is a consistent estimator for $$\mu^{2}$$.

$$U=\bar{Y}^{2}-\frac{1}{n}$$

By the continuous mapping theorem, which states that,

$$X_{n} \stackrel{\mathrm{P}}{\rightarrow} X \Rightarrow g\left(X_{n}\right) \stackrel{\mathrm{P}}{\rightarrow} g(X)$$

Then,

$$\bar{Y} \stackrel{P}{\longrightarrow} \mu$$ gives me $$\bar{Y}^{2} \stackrel{P}{\longrightarrow} \mu^{2} .$$

And since $$\frac{1}{n} \rightarrow 0$$ as $$n \rightarrow \infty$$ the result for conistency seems intuitively obvious.

But I have a confusion with how to show this formally, whether using only the mapping theorem, or if I need something else. Showing how the $$\frac{1}{n} \rightarrow 0$$ part leads to consistency is the part that I'm missing, since this is a standard limit and not a convergence in probability.

Any help in completing this is greatly appreciated.

You're missing two things. First of all, saying $$1/n\to\infty$$ is a 'standard limit' means that the convergence holds a.s. and hence also in probability. The next step is then to apply the continuous mapping theorem again with the function $$g(x,y)=x-y$$.
$$X_n=1/n$$ can be thought of as a random variable with a Dirac distribution with mass at $$1/n$$. $$P(X_n\leq x)\rightarrow 1$$, for all $$x\geq 0$$, and $$0$$ for all $$x< 0$$, therefore $$X_n$$ converges in distribution to the random variable with Dirac distribution with mass at zero, which is the same to say that it converges to zero.
By Slutsky's Theorem, $$\bar{Y}^2-\frac{1}{n}\rightarrow_d \mu^2-0$$, and since convergence in distribution to a constant implies convergence in probability, you have your result.
It is also possible to show this with the Continuous Mapping Theorem, since $$(\bar{Y},X_n)$$ converges jointly in probability to $$(\mu,0)$$. Then use function $$g(y,x)=y^2-x$$.
• Thanks for both comments, really got this cleared up! So the two gaps I had were i) that a random variable can be a constant, here $X_{n} = \frac{1}{n}$ and ii) that a 'standard limit' implies a.s. convergence by definition, so $X_{n} \stackrel{as}{\rightarrow} 0$. The rest follows by the mapping theorem. – JKM Jul 4 at 15:11