Timeline for Hypothesis test on the Euclidean length of an unknown vector
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Apr 20, 2013 at 7:35 | comment | added | M.B.M. | @whuber Is the following what you meant: focusing on the likelihood function $L(\mathbf{x})$ when $\|\mathbf{m}\|=1$. If $Y_i\sim\mathcal{N}(0,1)$ i.i.d. and $\lambda=\sqrt{\sum_{i=1}^nY_i^2}$, then $[Y_1/\lambda, \ldots, Y_n/\lambda]$ is uniformly distributed on a sphere. After algebra, $L(\mathbf{x})=\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}\exp\left[-\frac{1}{2}\sum_{i=1}^n y_i^2+\frac{\sum_{i=1}^nx_iy_i}{\sqrt{\sum_{i=1}^n y_i^2}}\right]dy_1\cdots dy_n$. However, this is a hard integral: math.stackexchange.com/questions/367129. | |
| Apr 18, 2013 at 6:47 | comment | added | M.B.M. | @whuber Thank you, this is very interesting. I think I understand that joint iid Normal yields uniform distribution on surface of a sphere (from your last comment's link, also circular symmetry), however, I think I need the "converse" for this problem (see my question on MSE: math.stackexchange.com/questions/362874/…). So, I am confused by "One way to handle that is to assume independent standard Normal distributions for all coordinates in $\mathbb{R}^{n+1}$ and ignore the radial part." Could you please elaborate? | |
| Apr 18, 2013 at 5:29 | comment | added | whuber♦ | Thanks for your replies. A uniform distribution on the sphere $S^n$ for $n\ge 2$ is not a uniform distribution on the angles. For example, on the two-sphere $S^2$ the distribution is uniform in the longitude (from $0$ to $2\pi$) but uniform in the sine of the colatitude (from $0$ to $\pi$). One way to handle that is to assume independent standard Normal distributions for all coordinates in $\mathbb{R}^{n+1}$ and ignore the radial part. See stats.stackexchange.com/questions/7977. | |
| Apr 18, 2013 at 3:06 | comment | added | M.B.M. | @whuber And for (3), yes, the posterior distribution of $l|x$ would answer my question (since the priors on $\{l_0,l_1\}$ are indeed well-defined), however, I do not know how to write it down. This is why I've been trying to figure out the likelihood function $p(x|l)$. Do you have any ideas on what $p(l|x)$ is in this problem? | |
| Apr 18, 2013 at 2:52 | comment | added | M.B.M. | @whuber Thanks for your comment. (1) was a typo, I fixed that; (2) I was basically extending the Proakis' proof on on the complex plane to an $n$-sphere. In his proof, under known magnitude of the vector, he analyzes the worst case for the detector, which is when the phase is distributed uniformly at random. I was under the impression that I could extend that to an $n$-sphere by assuming that all $n-1$ angles ("phases") are distributed uniformly at random on $[0,\pi]$ for the first $n-2$ of them and on $[0,2\pi]$ for the remaining one. Is that incorrect? | |
| Apr 18, 2013 at 1:54 | history | tweeted | twitter.com/#!/StackStats/status/324702326055120896 | ||
| Apr 17, 2013 at 17:45 | history | edited | M.B.M. | CC BY-SA 3.0 |
fixed a typo pointed out by whuber
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| Apr 17, 2013 at 13:58 | comment | added | whuber♦ | +1. Some fine points: (1) $S^2$ has the (noncentral) chi-square distribution, not $S$. (2) You seem implicitly to assume the two conditional distributions $m|l_i$ are uniform over the sphere(s). Because this does not mean uniform "on all the angles," I just want to make sure of your assumptions. (3) Why do you invoke N-P in a setting that is Bayesian? After all, you have well-determined priors on $m|l$ and $l=(l_0,l_1)$. So why doesn't the posterior distribution of $l|x$ fully answer your question? | |
| Apr 17, 2013 at 1:29 | history | edited | Glen_b | CC BY-SA 3.0 |
edited title
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| Apr 16, 2013 at 23:26 | history | asked | M.B.M. | CC BY-SA 3.0 |