Timeline for Standard deviation of a particular dimension in a multivariate Gaussian distribution
Current License: CC BY-SA 3.0
26 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Feb 12, 2013 at 15:12 | comment | added | user603 | @Aly strictly speaking here but these are implemented in many statistical packages (for example R in the function qF) | |
| Feb 12, 2013 at 14:38 | comment | added | Aly | @user603 thanks, I have accepted. Can you tell me how/where I can look up values of $F_0.95(p,n-p)$ ? | |
| Feb 12, 2013 at 14:37 | vote | accept | Aly | ||
| Feb 12, 2013 at 14:00 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 12, 2013 at 13:55 | comment | added | user603 | @Aly: the $n$ refers to the sample size used to estimate the parameters ('x_bar' and 'sigma').... | |
| Feb 12, 2013 at 13:42 | comment | added | Aly | Also, in my case I am just checking one sample so n=1? If so then the distance metric used above n/(n-1)^2 * d will give a divide by zero | |
| Feb 12, 2013 at 13:37 | comment | added | Aly | @user603 Looking at the link you provided, and the fact that my observation is independent of the samples used to estimate the distribution, it would appear I should be using Fishers F-ratio distribution. If this is correct, please update your answer and I will accept | |
| Feb 11, 2013 at 20:50 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 11, 2013 at 19:53 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 11, 2013 at 19:51 | comment | added | whuber♦ | Thank you! That was exactly it--I missed the fact that the denominator was squared but the numerator not. I feel much better about the situation now :-). | |
| Feb 11, 2013 at 18:53 | comment | added | user603 | @Aly: i've provided a link for the beta distribution, my $n$ is your $M$ (and your $n$ is my $p$) $X$ is not a covariance matrix (it is your dataset: you have n obserations each a p variate vector). $\hat{\mu}_x$ ($\hat{\mu}_X$) is the mean of the entries of $x$ ($X$). In other words, for example in the univariate case, $x$ is the dataset you have used to compute "x_bar" (my $\hat{\mu}_x$) and you "sigma" (my $\hat{\sigma}_x$). | |
| Feb 11, 2013 at 18:48 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 11, 2013 at 18:15 | comment | added | whuber♦ | Where do these formulas come from? For large $n$, and because clearly the right hand sides cannot exceed $1$, the formulas would indicate that any $x_i$ for which $|x_i - \hat{\mu}_x| \gt \sigma_x$ is (or perhaps is not?) a member of the cluster. I cannot find any way to interpret the question that makes this correct. What am I missing? And why does this answer deal with a vector of observations rather than a single observation as posed in the question? | |
| Feb 11, 2013 at 18:02 | comment | added | Aly | Additionally, why are we using the beta distribution and not Z scores with the phi distribution? | |
| Feb 11, 2013 at 17:41 | comment | added | Aly | If my M samples were of dimension n, then I construct an n-dimensional mean and a nxn dimension covariance matrix. If I then have a new n-dimension sample and I want to figure out if its distance from the mean is less than two standard deviations I should use the second formula? If so, I think I am getting confused by the notion of the nxp dimension covariance matrix as I thought it had to be square. Also, the subscripts on mu and sigma. Thanks | |
| Feb 11, 2013 at 17:39 | comment | added | Aly | Sorry, in the first formula. xi refers to a scalar,so what is n? For the univariate case, let's say for example I have had M samples and calculated the mean and variance, given another sample x, if I want to test that its distance from the mean is less than two standard deviations I should use the first formula? If so, what does the subscript x mean on mu and sigma and also what is n. | |
| Feb 11, 2013 at 17:37 | comment | added | user603 | If your $x_i$ is a scalar, use the first inequality. If your $X_i$ is a vector (of p measurments) use the second inequality. Can you explain where my formulation is confusing? I can edit the answer | |
| Feb 11, 2013 at 17:35 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 11, 2013 at 17:30 | comment | added | Aly | But can you use this formula for a multi-dimension vector? for my purpose I have a a cluster of n-dimension vectors and I wish to compute a mean and std deviation so that given another n-dimension vector I can calculate the likelihood that it belongs to this cluster. Should I be using a multivariate gaussian or have I misunderstood something and can just use the univariate? | |
| Feb 11, 2013 at 17:27 | comment | added | user603 | it's a notation convention thing. I've made the dimensions explicit in all cases. | |
| Feb 11, 2013 at 17:26 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 11, 2013 at 17:21 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 11, 2013 at 17:17 | comment | added | Aly | I thought in the univariate case the vectors are of length 1? | |
| Feb 11, 2013 at 17:15 | history | edited | user603 | CC BY-SA 3.0 |
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| Feb 11, 2013 at 17:10 | history | answered | user603 | CC BY-SA 3.0 |