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If I have two statistic say $T_1$ and $T_2$ both of which are consistent for $\theta_1$ and $\theta_2$, then will the ratio $\frac{T_1}{T_2}$ be also consistent for $\frac{\theta_1}{\theta_2}$ in general.

We have an invariance property for a continuous function of a consistent estimator. Can that property be extended for the function of two different estimator.

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    $\begingroup$ This looks like a one-step demonstration: provided $\theta_2\ne 0,$ isn't the function $(\theta_1,\theta_2)\to\theta_1/\theta_2$ continuous? $\endgroup$
    – whuber
    Commented Feb 17, 2021 at 14:14
  • $\begingroup$ Yes, it is. I was just wondering if we can extend the invariance property for the two estimator case too. $\endgroup$
    – userNoOne
    Commented Feb 17, 2021 at 14:18
  • $\begingroup$ Like if I have a normal distribution and I know $\bar{X}$ is consistent for $\mu$ and $S^2$ is consistent for $\sigma^2$, then can i say that $\frac{\bar{x}}{S^2}$ will be consistent for $\frac{\mu}{\sigma^2}$ $\endgroup$
    – userNoOne
    Commented Feb 17, 2021 at 14:20
  • $\begingroup$ My point is that this is a one estimator case: the random variable $(T_1,T_2)$ estimates $(\theta_1,\theta_2).$ $\endgroup$
    – whuber
    Commented Feb 17, 2021 at 14:20
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    $\begingroup$ Ahh. I see what you are saying, we can look at this as a single estimator. I get it. $\endgroup$
    – userNoOne
    Commented Feb 17, 2021 at 14:21

1 Answer 1

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Answered in comments, summarizing here:

What you need is the theorem that if $T_n$ is consistent for some parameter $\theta$, and $g$ is a continuous function, then $g(T_n)$ is consistent for $g(\theta)$. This is the continuous mapping theorem, see Two Different Proofs of Continuous Mapping Theorem.

You have $g(\theta_1, \theta_2)=\theta_1/\theta_2$ which is continuous on $\theta_2 \not= 0$. So, assuming that, $g(T_1, T_2)=T_1/T_2$ is consistent for $\theta_1 / \theta_2$.

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