Timeline for Estimating values of a sequence from observed differences
Current License: CC BY-SA 3.0
16 events
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| May 23, 2013 at 19:23 | history | edited | Danica | CC BY-SA 3.0 |
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| May 23, 2013 at 19:15 | comment | added | Danica | @ChrisTaylor Yes, you're right. I've updated the answer to reflect that. | |
| May 23, 2013 at 19:13 | history | edited | Danica | CC BY-SA 3.0 |
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| May 23, 2013 at 16:09 | comment | added | Chris Taylor | That's a good point (two comments ago). I think you could argue along the following lines - if there are two zero evalues, then there is a permutation putting $A-N$ in block form where each block has a zero eigenvalue. Hence the same applies to $N$ (since $A$ is diagonal). But then $N$ is the adjacency matrix of a graph with more than one connected component (as @whuber said) which we have ruled out. | |
| May 23, 2013 at 15:59 | comment | added | Danica | Incidentally, $\bf 1$ being a zero eigenvalue perpendicular to $b$ has a very natural interpretation: you can shift the entire sequence any amount without changing the likelihood. (It's just managed by the constraint.) Since of course it's perpendicular to the ${\bf 1}^T S = 0$ hyperplane constraint, the concern I raised about having non-unique MLEs doesn't apply. Cool. | |
| May 23, 2013 at 15:49 | comment | added | Danica | @ChrisTaylor I agree that $\mathbf 1$ is an eigenvector with value 0 (thus $A-N$ is never strictly positive definite; oh well...) and is perpendicular to $b$. But how do you know there aren't other eigenvectors with value 0? Consider the degenerate case where $A = N = 0$; the MLE is clearly nonunique but your argument still applies. | |
| May 23, 2013 at 14:24 | comment | added | Chris Taylor | I think the solution is unique. Since the row sums of $A-N$ are zero, the vector ${\bf 1}$ is an eigenvector of $A-N$ with eigenvalue zero. By definition we have ${\bf 1}^T b = 0$ as well, so the zero eigenspace is perpendicular to $b$, and the constraint ${\bf 1}^T S = 0$ fixes a unique solution. | |
| May 21, 2013 at 21:57 | comment | added | Danica | @whuber Good point about the double-counting; it should be square-rooted or changed so the products are over $j > i$ for anything that needs an actual likelihood. I just looked over everything else and didn't see anything else wrong, though. | |
| May 21, 2013 at 16:43 | comment | added | whuber♦ | (+1) This is clearly a good approach. But you ought to double-check this work. Starting at the beginning, your expression for the likelihood function causes each observation to appear twice: once for $(i,j)$ and again for $(j,i)$. Thus it seems you have written down the square of the likelihood. The solution will still be the same, but any confidence intervals (from the information matrix) will be wrong. The quadratic program is very rapidly solved. Incidentally, I believe the characterization of positive-definiteness may be equivalent to the graph having a unique connected component :-). | |
| May 21, 2013 at 11:24 | comment | added | Chris Taylor | Thanks, this was just what I needed. I'll have a crack at the confidence interval stuff myself. | |
| May 21, 2013 at 11:24 | vote | accept | Chris Taylor | ||
| May 21, 2013 at 2:51 | history | edited | Danica | CC BY-SA 3.0 |
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| May 21, 2013 at 2:45 | history | edited | Danica | CC BY-SA 3.0 |
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| May 21, 2013 at 2:31 | history | edited | Danica | CC BY-SA 3.0 |
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| May 21, 2013 at 2:20 | history | edited | Danica | CC BY-SA 3.0 |
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| May 21, 2013 at 2:10 | history | answered | Danica | CC BY-SA 3.0 |