Timeline for Hypothesis test for the presence of a Gaussian signal in i.i.d additive Gaussian noise
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 7, 2021 at 6:32 | answer | added | Ben | timeline score: 2 | |
| Mar 28, 2015 at 4:26 | history | tweeted | twitter.com/#!/StackStats/status/581673811768143873 | ||
| Jul 17, 2013 at 16:44 | comment | added | M.B.M. | @whuber Right, I was just about to post that I was confused again, as I was having trouble writing down the likelihood function for the alternate... Could you please elaborate on the "first-order approximation to the correct log-likelihood"? How does one justify an approximation of a function when the function is unknown in the first place? | |
| Jul 17, 2013 at 16:40 | comment | added | whuber♦ | I'm not convinced this is correct. The potential error I made lies in asserting that the conditional likelihood is independent of the permutation: but it does depend on it. And that, unfortunately, gets us right back to where you started. This pseudo-reasoning, though, suggests you might construct a successful test anyway by using the sum of squares as your test statistic. It can be justified as a first-order approximation to the correct log-likelihood. | |
| Jul 17, 2013 at 16:19 | history | edited | M.B.M. | CC BY-SA 3.0 |
fixed typo
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| Jul 17, 2013 at 16:18 | comment | added | M.B.M. | Aha!!! Very nice explanation. This also means that the test involving the maximum is not optimal either (which doesn't matter, as the sum of squared observations is nicer to deal with analytically). If you put your comment in an answer form, I'll accept it. | |
| Jul 17, 2013 at 16:09 | comment | added | whuber♦ | The alternate is not a mixture, that's why. You have guaranteed that precisely one of the values is different. Therefore the data are a random permutation of an experiment in which the first observation is from one distribution and all the rest are from another (and all distributional parameters are known, apparently). Because you don't label the data and they are independent, the likelihood conditional on any permutation is the product of the likelihoods. Because this conditional likelihood does not depend on the permutation, it must be the full likelihood. | |
| Jul 17, 2013 at 16:03 | comment | added | M.B.M. | @whuber Could you please explain why the sum of squares of the data is a sufficient statistic in this case, as it's not obvious to me given that the distribution of the data when the alternate hypothesis is true is a mixture? I can see the sum of the squares of data being sufficient statistic if this was a test between $n+1$ hypotheses (i.e. the current alternate hypothesis broken into $n$ separate hypotheses for the signal $y_t$ being in each of the $n$ locations, plus the current null hypothesis for absent $y_t$). However, this problem doesn't care about learning location $t$ of the signal. | |
| Jul 17, 2013 at 15:51 | comment | added | whuber♦ | I think that's an unnecessary complication. A sufficient statistic is the sum of squares of the data: this will have a distribution equal under the null to a $\chi^2(n)$ scaled by $s^2$ or, under the alternate, to the sum of a $\chi^2(n)$ scaled by $s^2$ and an independently distributed $\chi^2(1)$ scaled by $\sigma^2$. | |
| Jul 17, 2013 at 15:45 | comment | added | M.B.M. | @whuber The sum is due to the mixture distribution over the data when alternate hypothesis is true. I clarified in the question. | |
| Jul 17, 2013 at 15:44 | history | edited | M.B.M. | CC BY-SA 3.0 |
Defined likelihood functions to address whuber's comment, as well as improved notation, added asymptotics tag, and fixed a typo in the LR formula
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| Jul 17, 2013 at 14:57 | comment | added | whuber♦ | How do you obtain a sum instead of a product over the data? | |
| Jul 17, 2013 at 2:59 | history | asked | M.B.M. | CC BY-SA 3.0 |