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


By the continuous mapping theorem, which states that,

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


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

| cite | improve this answer | |

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

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.