Timeline for Is this correct ? (generating a Truncated-norm-multivariate-Gaussian)
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2015 at 15:06 | comment | added | Loves Probability | @Xi'an Thanks a lot for your patient answers. I finally understood my mistake with your stimulation, and I have also written my own detailed answer explaining the same. But sorry, hope you don't mind, I should probably accept whuber's answer for his detailed explanation that helps in actually solving the problem. | |
| Jun 26, 2015 at 14:52 | vote | accept | Loves Probability | ||
| Jun 26, 2015 at 14:47 | answer | added | Loves Probability | timeline score: 3 | |
| Jun 26, 2015 at 12:41 | comment | added | Xi'an | I am afraid you are misunderstanding the notion of a marginal distribution: If $X\sim f(x)$ and $Y|X=x\sim g(y|x)$, $f$ is the marginal density of $X$. If nothing else, because $$\int f(x)g(y|x)\text{d}y=f(x)$$ | |
| Jun 26, 2015 at 12:30 | comment | added | Loves Probability | @Xi'an Okay, my point is this. When $X_1,\ldots,X_{n-1}$ are generated as Gaussians, and later terms depend on these values, marginals of $X_1,\ldots,X_{n-1}$ won't be Gaussians. What you said is exactly the same. They might be "Conditionally Gaussian" but definitely not "marginally Gaussian". My earlier comment means that. | |
| Jun 26, 2015 at 12:22 | comment | added | Loves Probability | @Xi'an Conditional Gaussian does not mean Marginal Gaussian!! | |
| Jun 26, 2015 at 12:18 | comment | added | Xi'an | Your (incorrect) algorithm generates$$X_1,\ldots,X_{n-1}\sim \mathcal{N}(0,\sigma^2)$$first and then$$X_n\sim\mathcal{N}_T(0,\sigma^2)$$given $X_1,\ldots,X_{n-1}$. Hence, the first generation is from the marginal and the second generation is from the conditional. My proof shows the marginal is not a (n-1) dimensional Gaussian distribution. | |
| Jun 26, 2015 at 12:00 | comment | added | Loves Probability | @Xi'an Thanks for your query & interest. Here is my reasoning for your point: The algorithm in question needs RVs $X_1\ldots X_n$, that are $n-1$ Gaussians and a Truncated-Gaussian when they are seen per sample; more specifically, one of the distributions varies every sample. They are not the respective marginals. Because, each $x_i,i=1,\ldots,n-1$ appears in two terms: $x_i$ and $x_n$; and $x_n$ is clearly time varying as the truncation threshold varies for every sample. The decomposition proof that you provided has a problem in exactly the same sense. Marginals are just not available. | |
| Jun 23, 2015 at 5:11 | history | edited | Loves Probability | CC BY-SA 3.0 |
changed to a more clear picture
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| Jun 23, 2015 at 4:40 | history | edited | Loves Probability | CC BY-SA 3.0 |
fixed sentence
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| Jun 23, 2015 at 4:32 | history | edited | Loves Probability | CC BY-SA 3.0 |
picture added, fixed sentences
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| Jun 23, 2015 at 4:24 | history | edited | Loves Probability | CC BY-SA 3.0 |
picture added
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| Jun 21, 2015 at 21:35 | history | tweeted | twitter.com/#!/StackStats/status/612735692964274178 | ||
| Jun 21, 2015 at 16:32 | answer | added | Mark L. Stone | timeline score: 8 | |
| Jun 21, 2015 at 15:57 | answer | added | whuber♦ | timeline score: 15 | |
| Jun 21, 2015 at 8:41 | answer | added | Xi'an | timeline score: 5 | |
| Jun 21, 2015 at 6:48 | history | edited | Loves Probability | CC BY-SA 3.0 |
fixed grammar, added a link
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| Jun 21, 2015 at 6:38 | review | First posts | |||
| Jun 21, 2015 at 6:50 | |||||
| Jun 21, 2015 at 6:37 | history | asked | Loves Probability | CC BY-SA 3.0 |