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Let $$X_1,\dots,X_m$$ are i.i.d. with distribution function $F$ and $$Y_1,\dots,Y_n$$ are i.i.d. with distribution function $G$. Suppose that there exists an unknown function $\psi:\mathbb{R}\mapsto\mathbb{R}$ such that $\psi(X_i)\sim N(0,1)$ and $\psi(Y_j)\sim N(0,\sigma^2)$ for all $i=1,\dots,m$ and $j=1,\dots,n$. I'd like to estimate $\sigma^2$ in this problem.

I have obtained the following facts:

Note that $F(x)=P(X\le x)=P(\psi(X)\le \psi(x))=\Phi(\psi(x))$ where $\Phi$ is the cumulative standard normal distribution. This implies $\psi(x)=\Phi^{-1}(F(x))$.

Recall that $\psi(Y_j)\sim N(0,\sigma^2)$. So, $$\check\sigma^2=\frac1n\sum_{j=1}^n\psi^2(Y_j)$$ is an optimal estimator for $\sigma^2$. Since $F$ is unknown then I replace it with its empirical distribution function $\hat F_m$ based on $X_1,\dots,X_m$. Hence, it is natural to replace $\psi$ with $\hat\psi=\Phi^{-1}(\hat F_m)$. Therefore, I conjecture that $$\hat\sigma^2=\frac1n\sum_{j=1}^n\hat\psi^2(Y_j)$$ is an optimal estimator. I have tried to show $\check\sigma^2$ and $\hat\sigma^2$ are asymptotically equivalent, but I was failed. Could anyone help me? Or does any one have another approach?

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I think the following works:

$$\hat{\sigma}^2 = \frac{1}{n}\sum_{j=1}^n \hat{\psi}^2(Y_j) = \frac{1}{n} \sum_{j=1}^n \psi^2(Y_j) + \frac{1}{n} \sum_{j=1}^n \left[\hat{\psi}^2(Y_j) - \psi^2(Y_j)\right] \qquad (1)$$

The first term is $\breve{\sigma}^2$. Then, we just need to show that the second term is asymptotically 0 (as $m \rightarrow \infty$).

Let $g(\cdot) = \Phi^{-1}(\cdot)$. We need two things. First, the Glivenko-Cantelli Theorem says that empirical cdf $\hat{F}$ converges uniformly to $F$ i.e. $\sup_t |\hat{F}(t) - F(t)| \rightarrow 0$. Second, we use the delta method. Taylor expansion of $g(\hat{F}(x))$ at $F(x)$:

$$g(\hat{F}(x)) = g(F(x)) + g'(F(x))[\hat{F}(x) - F(x)] + o_p(1)$$

The $\hat{F}(x) - F(x)$ term goes to 0 be GC theorem for all $x \implies \hat{\psi^2}(x) \rightarrow \psi^2(x)$ as $m \rightarrow \infty \implies$ the second term in (1) is 0.

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  • $\begingroup$ First, I'd like to thank for your answer @bmciv. However, to prove that $\hat\sigma^2$ is optimal we have to show that $\sqrt n(\hat\sigma^2-\sigma^2)\to N(0,2\sigma^4)$, and it suffices to show $\sqrt n(\hat\sigma^2-\check\sigma^2)\to 0$ in probability because we will have $\hat\sigma^2$ and $\check\sigma^2$ have the same limiting distribution and we already had $\sqrt n(\check\sigma^2-\sigma^2)\to N(0,2\sigma^4)$. In your answer, we are only able to derive $(\hat\sigma^2-\check\sigma^2)\to 0$ in probability. So, how will we confirm the optimality of $\hat\sigma^2$? $\endgroup$ – Jlamprong Jun 3 '14 at 18:01
  • $\begingroup$ The Remainder of the Taylor Expansion is $o_p(1)$ when $\hat{F}(x) \rightarrow F(x)$. So we have to first write it in general as, say, $R_2(\hat{F}(x)-F(x))$, and then, invoking the GC theorem, to say that it too, goes to zero. And I think that by $m$ you meant $n$? $\endgroup$ – Alecos Papadopoulos Jun 5 '14 at 11:37
  • $\begingroup$ In case $m=n$, how would it be? $\endgroup$ – Jlamprong Jun 6 '14 at 8:34
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Let's try something, I think you need one more assumption to get optimality:

To have asymptotic equivalence you would need:

$1/\sqrt{n}\sum_{i=1}^n[\hat{\psi}^2(Y_j)-\psi^2(Y_j)]=o_p(1)$

Using the CLT and the Delta-Method you know that:

$\sqrt{m}(\hat{\psi}^2(Y_j)-\psi^2(Y_j))=O_p(1), \forall j$ or $\hat{\psi}^2(Y_j)-\psi^2(Y_j) = O_p(1/\sqrt{m})$

Let's use some intuition, for the two estimators to be equivalent, you need the estimation error from $\hat{\psi}$ to be negligible for $n \to \infty$. In the extreme case, if $m=\infty$ then the approximation error is zero and the two are equivalent. In the other extreme case, if $m$ does not go to infinity then multiplied by $\sqrt{n}$ the approximation error from using $\hat{\psi}$ becomes infinitely large.

Assume $n^2/m \to 0$ as $m,n \to \infty$ this implies that $O_p(1/\sqrt{m})=O_p(1/n)=o_p(1/\sqrt{n})$

Then you should have:

$1/\sqrt{n}\sum_{i=1}^n[\hat{\psi}^2(Y_j)-\psi^2(Y_j)]= 1/\sqrt{n}\sum_{i=1}^n[o_p(1/\sqrt{n})] =o_p(1)$

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  • $\begingroup$ Thanks @user155437 for your help. But, I'd like to ask you: 1. How did you derive $\hat{\psi}^2(Y_j)-\psi^2(Y_j) = O_p(1/\sqrt{m})$? Since using Delta-Method, I was only able so show $\hat{\psi}^2(X_j)-\psi^2(X_j) = O_p(1/\sqrt{m})$. 2. Once, (1) has been proved, how will we make sure that the sum is also $O_p(1/\sqrt{m})$? $\endgroup$ – Jlamprong Jun 9 '14 at 20:39

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