I am a bit confused regarding what exactly is the invariance property of sufficient estimators, consistent estimators and maximum likelihood estimators. As far as I know,

  1. Invariance property of consistent estimators is : If $T$ is a consistent estimator of $\theta$, and $f$ is a continuous function then $f(T)$ is a consistent estimator of $f(\theta)$.

  2. Invariance property of sufficient estimators is : If $T$ is sufficient estimator of $\theta$ and $f$ is one-one, onto function then $f(T)$ is sufficient estimator of $f(\theta)$, also $f(T)$ is sufficient estimator of $\theta$, and $T$ is sufficient estimator of $f(\theta)$.

  3. Invariance property of maximum likelihood estimators(MLE) is : If $T$ is a MLE of $\theta$, and $f$ is a continuous/ one-one, onto function then $f(T)$ is a MLE of $f(\theta)$.

Please correct me if I am wrong somewhere and please tell me the least I need to check for it as I am appearing for a competitive exam where time really matters.


These are somewhat different properties , and throwing them together under the same name mostly is confusing. But:

  1. This is correct, but I have never before seen this named invariance.

  2. Confused. We talk about sufficient statistics not estimators. Of course an estimator is a statistic, but $T$ being sufficient statistic (in a model parametrized by $\theta$, or for $\theta$) says in itself not that $T$ is a good estimator! If $T$ is sufficient, and also a good estimator for $\theta$, then still $1000000 T$ is sufficient, but maybe not a good estimator (for $\theta$.) Of your requirements for $f$ then one-to-one is essential, onto is unnecessary. Then we can reformulate: If $T$ is sufficient for $\theta$ and $f$ is one-one, then $f(T)$ is also sufficient for $\theta$.

  3. Mostly right, but you don't need all those conditions on $f$. See Invariance property of maximum likelihood estimator?.

  • $\begingroup$ But for the sufficient Statistic, it holds both ways right? i.e, If T is sufficient statistic of theta and f is a one-one function then f(T) is a sufficient statistic of theta, and also f(T) is sufficient statistic of f(theta) , and T is also a sufficient statistic of f(theta). $\endgroup$
    – user233797
    Feb 6 '19 at 3:35
  • $\begingroup$ Yes, its right. $\endgroup$ Feb 6 '19 at 7:39

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