*Bounty: 50*

*Bounty: 50*

I define the function

$$

lambda_n (theta_1,theta_2)=frac{P_{theta_1}(x)}{P_{theta_2}(x)}=frac{theta_1^x(1-theta_1)^{n-x}}{theta_2^x(1-theta_2)^{n-x}}

$$

where $0<theta_1<1$ and $0<theta_2<1$. The subscript $n$ indicates dependence of $lambda_n(cdot)$ on $n$. Furthermore, $0<lambda_n(cdot)<1$.

Let $boldsymbol{hat{theta}}=(hat{theta}*{0,n},hat{theta}_n)^T$ and $boldsymbol{theta}=(theta_0,theta)^T$ where $hat{theta}*{0,n}$ and $hat{theta}_{n}$ are maximum likelihood estimates of $theta_0$ and $theta$, respectively. I’m interested in understanding the **rate of convergence (in probability)** of $lambda_n(boldsymbol{hat{theta}})stackrel{p}{to}lambda_n(boldsymbol{theta})$. This convergence holds by consistency of the MLE estimates (assuming some regularity conditions), by continuity of $lambda_n$ and the continuous mapping theorem.

By invariance, $lambda_n(boldsymbol{hat{theta}})$ is the MLE of $lambda_n(boldsymbol{theta})$ and $(hat{theta}*{0,n},hat{theta}_n)$ converge in probability to $(theta_0,theta)$ at $sqrt{n}$-consistency, i.e. $sqrt{n}(hat{theta}*{0,n}-theta_0)=O_p(1)$ and $sqrt{n}(hat{theta}_{n}-theta)=O_p(1)$.

**Question 1: Does invariance property of MLEs allow us to retain the rate of convergence?**

If $lambda_n(cdot)$ didn’t have the dependence on $n$, then this would likely be true. To get at the rate of convergence, I started doing a first-order Taylor expansion:

$$

lambda_n(boldsymbol{hat{theta}})=lambda_n(boldsymbol{theta})+(boldsymbol{hat{theta}}-boldsymbol{theta})^Tfrac{partiallambda_n(boldsymbol{theta})}{partialboldsymbol{theta}}+o_{p,n}(|boldsymbol{hat{theta}}-boldsymbol{theta}|)

$$

Note that the remainder term has dependence on $n$. Now I’d like to show that

$$

frac{partiallambda_n(boldsymbol{theta})}{partialboldsymbol{theta}}=O_p(1)

$$

in order to get the linear term to be $o_p(1)$. Also I need to remove the dependence of the remainder term on $n$ by showing some uniform (over $n$) boundedness for the second derivative, i.e.

$$

leftvertfrac{partiallambda^2_n(boldsymbol{theta})}{partialboldsymbol{theta}^2}rightvertleq M

$$

**Question 2: Is this even possible? Have I gone about this the wrong way?**

My main concern is that the first and second derivatives depend on some factor $x-theta n$ which blows up as $ntoinfty$. Is there a way to handle this?

Thank you very much for your help!