Timeline for Question on Maximum Likelihood Estimation
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2011 at 13:24 | vote | accept | Haining Yu | ||
| Apr 7, 2011 at 13:50 | vote | accept | Haining Yu | ||
| Apr 7, 2011 at 13:50 | |||||
| Apr 1, 2011 at 22:00 | comment | added | Michael Hardy | I don't think you've got a well-defined math problem here. $V$ could be a vector of 100 independent normally distributed random variables with uniform distributions on the interval $(0,1)$ and $X$ could be the sum of $\theta\in\mathbb{R}$ and a suitable function of the 43rd one of those uniform RVs, so that you get a normally distributed RV with expectation $\theta$. Or you could do all sorts of things like that. At best, you'd have to say more than what you've said to specify a mathematical problem. | |
| Apr 1, 2011 at 0:41 | comment | added | Haining Yu | Sorry about your confusion. Please let me clarify: 1. there exists a well-defined function $f(.)$ so that $x=f(V|\theta)$. Therefore if the pdf of V is known, the pdf of x is known. 2. The pdf of $V$ cannot be uniquely determined by the pdf of $x$ and function $f(.)$. 3. $x=f(V|\theta)$ is set up that, given enough pairs of x and V, $\theta$ can be uniquely identified Does that help? | |
| Apr 1, 2011 at 0:05 | comment | added | Michael Hardy | ....two different $\theta$'s never correspond to the same distribution of $V$. If $\theta$ is a function of $(\mu,\sigma)$---say $\theta = j(\mu,\sigma)$, then the MLE for $\theta$ would just be $j(\hat{\mu},\hat{\sigma})$, where $\hat{\mu}$ and $\hat{\sigma}$ are the respective MLEs of $\mu$ and $\simga$. So the difficulty that still remains comes only from the non-one-to-one nature of the dependence of $X$ on $V$. | |
| Apr 1, 2011 at 0:02 | comment | added | Michael Hardy | I'm not altogether sure I do understand what you mean. I think you said (among other things) that $X = f(V \mid \theta)$ does not depend in a one-to-one fashion upon $V$. If $f$ is known, and if the dependence of $X$ on $V$ were one-to-one, then the probability distribution of $V$ would essentially be known if $\mu$ and $\sigma$ were known, in the sense that there would be only one probability distribution of $V$ that would give rise to that particular probability distribution of $X$. The parameter $\theta$ would then simply be a function of $(\mu,\sigma)$, at least if...... | |
| Mar 31, 2011 at 21:54 | comment | added | Haining Yu | Thanks for the comments. My notations are to be improved, but hopefully you can understand what I mean. The gist of your point seems to be that, if I know the distribution of $V$, the distribution of $X$ can be derived. If it is indeed your point I agree with you. But in my case, the distribution of $V$ is not known. I guess I would rephase my question in the following way: if we can observe samples of $V$ but its closed-form distribution is unknown, and the distribution of $f(V|\theta)$ is known (except for $\theta$), what kind of parameter estimates can we do for $\theta$? | |
| Mar 31, 2011 at 21:36 | history | answered | Michael Hardy | CC BY-SA 2.5 |