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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.

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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 $g(x,y)=x-y$.

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$X_n=1/n$ 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 $g(y,x)=y^2-x$.

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  • $\begingroup$ 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. $\endgroup$ – JKM Jul 4 at 15:11

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