Question on Maximum Likelihood Estimation My question concerns using Maximum Likelihood to estimate unknown parameters. I will sincerely appreciate if anyone can help me find out if my approach is flawed.
Assume we have a random vector $V$, and we can observe $M$ samples of it denoted by $V_1,V_2,\ldots,V_M$. Define a scalar random variable $X = f(V\mid\theta)$ where $f(\cdot)$ is function whose closed form is known but parameter $\theta$ is unknown. I cannot observe $X$.
We do not know the distribution of $V$, but we do know the distribution of $X$. In fact, $X$ follows a Normal distribution $N(\mu,\sigma^2)$ where $\mu$ and $\sigma$ are unknown. 
My purpose is to JOINTLY estimate $\theta$, $\mu$, and 
$\sigma$. I take the following approach: let us use $g(.|\mu,\sigma^2)$ to denote the pdf for $N(\mu,\sigma)$. The likelihood function $L$ is then $g(f(V\mid\theta)\mid\mu,\sigma^2)$. Based on MLE, we then maximize $\sum_{m=1}^M \log(g(f(V_m\mid\theta)\mid\mu,\sigma^2))$
of all $M$ samples. Since $f(\cdot)$ and $g(\cdot)$ are both known, this maximization problem can be solved by nonlinear programming (at least in theory). The optimal solution 
would provide the best estimates for $\theta$, $\mu$, and $\sigma$ simultaneously.
I have multiple questions regarding the approach described above:


*

*Deviation from traditional MLE definition: in textbook examples, we should formulate the joint distribution of $V$ (joint because $V$ is a vector) and maximize it. In my problem, however, the joint distribution of $V$ is unknown. The only thing we know is that it has to satisfy $X = f(V)$. For example, let us assume $V$ has two dimensions, and $ X = V_{(1)} + \theta V_{(2)}$. There are many joint distributions that satisfy this equation 
above. 

*Identification: as pointed by the example above, There is no one-to-one mapping between $V$ and $X$. Do I have to run into identification issues when I do the estimation?

*Do consistency, asymptotic normality and other decent properties for MLE still apply?

*Is there any alternative besides MLE I can take?
Many thanks! Any comments are more than welcome.
 A: You definitely would run into identification issues. Suppose $f(x|\mid\theta) = x \theta$. Then 
multiplying $\theta$ and $\sigma$ by any common factor would result in an identical situation. In this case the MLE would not really be defined.
Generally, as $\theta$ changes, $X$ is revealing a different facet of $V$. For example, if $f$ is a projection map and $\theta$ determines the angle of projection, then $f(V\mid\theta)$ is the shadow of $V$ in the direction of $\theta$. Many of the shadows could be normally distributed, and could be essentially unrelated to each other.
A: The correct usage is "a sample of size $M$", rather than "$M$ samples".
The function $f$ and the probability distribution of $V$ would completely determine the distribution of $X = f(V\mid\theta)$, and therefore would completely determine $\mu$ and $\sigma$.  Hence form some function $h$ we have $h(\theta) = (\mu,\sigma)$.  Since the parameter $\theta$ then determines the probability distribution of $X$, if there is such a thing as the MLE $\hat{\theta}$ for $\theta$, then $h(\hat{\theta})$ would be the MLE for the pair $(\mu,\sigma)$.  So you can't treat $\theta$ and the pair $(\mu,\sigma)$ as three independent parameters to be estimated.
Your assertion about what the likelihood function is doesn't make sense.  You could say $L_1(\mu,\sigma) = g(X\mid\mu,\sigma)$ is a likelihood function.  How to find the values of $\mu$ and $\sigma$ that maximize that is well-known.  But in order to find a likelihood function $L_2(\theta)$ you would need to know a way in which the probability distribution of the data depends on $\theta$.  You seem to be saying you don't know that.
