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This is concerning Eq. (3.7) of C R Rao's 1945 paper (see p.81 of this article). Can someone help me in figuring out the second equality in Eq. (3.7)?

His claim is (since $\phi(x,\theta) = \Phi(T,\theta) \psi(x_1,\dots,x_n)$ from Eq. (3.6)) can be written as $$\theta = \int t \phi \pi dx_i = \int t \Phi(T,\theta) \psi(x_1,\dots,x_n) \pi dx_i = \int f(T)\Phi(T,\theta)dT,$$ for some function $f(T)$ of $T$, independent of $\theta$.

My question is concerning the last equality. Prof Rao seems to regard $t\psi(x_1,\dots,x_n) \pi dx_i$ as $f(T)dT$. Since $\psi(x_1,\dots,x_n)$ is essentially the conditional distribution of $x_1,\dots, x_n$ given $T$ and since $T$ is a sufficient statistics, it is true that $\psi(x_1,\dots,x_n)$ is a function depending on $T$, but is independent of $\theta$. Also since the conditional distribution of $x_1,\dots, x_n$ given $T$, $t\psi(x_1,\dots,x_n)$ resembles $E[t|T]$ which is a function of $T$. However, I am not able to get these rigorously. Any help in this connection is greatly appreciated.

PS: I asked the same in math.stackexchange here, but did not receive any response, so asking here.

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$\newcommand\th\theta$$T$ is a sufficient statistic and thus a random variable. So, the integral $\int f(T)\Phi(T,\theta)dT$ cannot possibly have a meaning.

The conclusions that Rao is trying to reach here are true, though:

(i) "there exists a function $f(T)$ of $T$, independent of $\theta$ and is an unbiased estimate of $\theta$".

This is true, leaving aside the faulty grammar of this statement. Indeed, $T$ is an abbreviation of $T(X)$, where $X$ is a random sample from the distribution $P_\theta$ and $T$ is a Borel-measurable function. That $T$ is sufficient means that (some version of the conditional expectation) $E_\th(t(X)|T(X))$ does not depend on $\th$ for any Borel-measurable function $t$ such that $E_\th t(X)$ exists for all $\th$. So (in view of the Doob--Dynkin_lemma), we can write $E_\th(t(X)|T(X))=f(T(X))$ for some Borel-measurable function $f$, which is the same for all $\th$. Therefore, $$E_\th f(T(X))=E_\th E_\th(t(X)|T(X))=E_\th t(X).$$ So, if $t(X)$ is unbiased for $\th$ -- that is, if $E_\th t(X)=\th$ for all $\th$ -- then $f(T(X))$ is also unbiased for $\th$.

(ii) "the best unbiased estimate of $\th$ is an explicit function of the sufficient statistic".

This is true because $$Var_\th f(T(X))=Var_\th E_\th(t(X)|T(X))\le Var_\th t(X),$$ since, in general, $Var\,E(Y|Z)\le Var\, Y$.


As noted in the preface to this paper, "The author was just 25, and did not have a PhD degree!" This may be the reason why the paper was written in a very imprecise language, which was archaic even in 1945, when the paper was written. Later writings by Rao are much more clear and precise.

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  • $\begingroup$ Thanks Iosif. How do we prove that the $f(T(X))$ he is referring to is indeed $E_\theta(t(X)|T(X))$? $\endgroup$
    – Ashok
    Commented Jan 7, 2022 at 1:31
  • $\begingroup$ @Ashok : There is no way to prove that. Indeed, as I said, the integral $\int f(T)\Phi(T,\theta)dT$, where his $f(T)$ is introduced in his paper, cannot possibly have a meaning. Therefore, there is no rigorous way to relate his non-rigorous $f(T)$ with the rigorous notion of the conditional expectation $E_\th(t(X)|T(X))$. In general, one cannot possibly prove rigorously that something rigorous is the same as something non-rigorous. What my answer gives is a rigorous interpretation of non-rigorous results in that non-rigorous paper. $\endgroup$ Commented Jan 7, 2022 at 2:00

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