consistent estimator

Suppose we have $X_1, X_2,X_3, ..., X_n$ independent with unknown $\theta_1, \theta_2$ with $X_i$ normally distributed with mean $\theta_1 + \theta_2 x_i$ where $x_i$ are known and variance 1. Can you suggest a simple necessary condition on a sequence of real numbers $x_1,x_2,...$ such that $(\tilde{\theta_1},\tilde{\theta_2})$ is a consistent estimator for $(\theta_1,\theta_2)$?

I guess it should be related to the MLE-estimator being consistent??

Thank you.

• Your set up is under-identified - you try to estimate n+2 unknown parameters (the two thetas and the n-real numbers that determine the mean of each random variable) with n observations. – Alecos Papadopoulos Jul 30 '13 at 22:40
• x_i are known. i have edited thank you for remark – Salih Ucan Jul 31 '13 at 22:38

I assume that $\theta_2 \neq 0$. I will denote $Y_i$ the process determining the mean of the $X_i$'s to avoid confusion. The process $\left\{X_i, i=1,...,n\right\}$ is an independent heterogeneous process (different means). Then a sufficient condition on the process for consistency in the estimation of $\theta_1,\;\theta_2$ is that the Markov condition holds, a version of which is $$E\left|X_i\right|^{1+\delta}\lt \Delta \lt \infty \;\text{for some} \;\delta \; \gt 0\; \text{and all i}$$ ...where $\Delta$ is some finite positive constant. This condition guarantees that asymptotically, $\frac1n\sum_iX_i\rightarrow_{a.s.} \mu$ i.e. it converges to some constant almost surely. Therefore if $\frac1n\sum_iX_i\rightarrow_{a.s.} \mu$ holds we have that $$\frac1n\sum_iX_i\rightarrow_{a.s.}\frac1n\sum_iE(X_i)= \frac1n\sum_i(\theta_1+\theta_2Y_i)= \theta_1+\theta_2\frac1n\sum_iY_i=\mu$$ $$\Rightarrow \frac1n\sum_iY_i=\frac{\mu -\theta_1}{\theta_2} = \mu'$$ meaning that the Markov condition must also hold for the $Y_i$'s,
$$E\left|Y_i\right|^{1+\delta}\lt \Delta' \lt \infty \;\text{for some} \;\delta \; \gt 0\; \text{and all i}$$