# Maximum-Likelihood estimator

Imagene we have a sequence of i.i.d random variables $(Y(t))_{1\le t\le s}$. It is possible to derive the density of $Y(t)$ and it is a function of parameters of interest $f(p,\rho, y(t))$. To have a Maximum-Likelihood estimator all i need to do is \begin{align*} \widehat{(p,\rho)}^{MLE}=\arg\max\limits_{p,\rho} \prod\limits_{t=1}^sf(p,\rho,y(t)) \end{align*} Sadly $Y(t)$ isn't directly observable. With a given dataset it is possible to construct MLE for realisation of $Y(t)$.

My question is: If $y(t)$ in above equation is replaced by $\widehat{y(t)}^{MLE}$, can the estimator above still be considered as a maximum-likelihood estimator?

I tend more to a "no" answer, but have no proof.

If it isn't MLE anymore, what about asymptotic normality? Hope at least this still usable.

• How are you producing the MLE of y(t)? Aug 17, 2016 at 12:43
• @MarkL.Stone hey, $Y(t)$ represents something like a probability of default. What i have is default dataset. I can prove that under certain assumptions MLS for y(t) will be number of defaults divided by total amount of credits given. Aug 17, 2016 at 12:57

The likelihood function you have is actually based on $f(p,\rho,\widehat{y(t)}^{MLE})$. Then, instead of having $Likelihood(p,\rho;Data)$, you have $Likelihood(p,\rho;SS)$, where $SS$ denotes a summary statistic which is the sequence $(\widehat{y(1)}^{MLE},\ldots,\widehat{y(s)}^{MLE})$. If the summary statistic $SS$ is sufficient, then the estimator is in fact the MLE. Otherwise, they are different.
• hey, thanks, but i dont think it works. Sufficient statistic means basicly that while using it, you dont lose any of important information you had before. So even if my estimator for $y(t)$ is sufficient, it is still just an estimate and not the true realisation. Maximum-Likelihood method is based on true realisations, if i got it correct. Aug 17, 2016 at 14:21