Timeline for Multivariate CLT with varying dimension size
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Feb 10, 2016 at 16:09 | answer | added | Guillaume Poly | timeline score: 3 | |
| Jan 14, 2016 at 17:41 | comment | added | Pig | Ah my bad, I forgot that restriction you had on X_i. | |
| Jan 14, 2016 at 11:29 | vote | accept | Simd | ||
| Jan 14, 2016 at 10:44 | comment | added | Simd | @user31814 The elements of $X_i$ are independently and uniformly chosen from $\{-1,1\}$. Therefore the mean is always $0$ and the covariance matrix is just the scaled identity matrix with $n$ on the diagonal. | |
| S Jan 14, 2016 at 4:39 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| S Jan 14, 2016 at 4:39 | history | suggested | Pig | CC BY-SA 3.0 |
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| Jan 14, 2016 at 1:47 | review | Suggested edits | |||
| S Jan 14, 2016 at 4:39 | |||||
| Jan 14, 2016 at 1:45 | comment | added | Pig | ... $n \to \infty$. The fact that the $\mu$ and covariance $\sum$ above are not indexed with $n$ is confusing. | |
| Jan 14, 2016 at 1:45 | comment | added | Pig | I still can't quite get a good interpretation of this question - following @LiviuNicolaescu's comment I can accept you are considering for each $n$ the average of $n$ random variables $X_{1n}, \cdots, X_{nn}$ being i.i.d. from one fixed probability distribution. But as you change the dimension $d_n$ for each of these $n$, you have to be picking a different probability distribution for each of these $n$ - they have different means and different covariance matrix. You would have to at least specify how you expect these probability distributions to look like/to relate to each other as | |
| Jan 14, 2016 at 0:53 | comment | added | Liviu Nicolaescu | @dorothy Thus you are talking about a triangular array $X_{in}$, $1\leq i\leq n$ where for each $n$, $X_{1n}, \dotsc, X_{nn}$ are iid of dimension $d_n$. The covariance $\Sigma_n$ is a $d_n\times d_n$ matrix. In particular it also depends on $n$. You need to formulate the question more precisely. | |
| Jan 13, 2016 at 22:16 | history | edited | Simd | CC BY-SA 3.0 |
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| Jan 13, 2016 at 22:11 | history | edited | Simd | CC BY-SA 3.0 |
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| Jan 13, 2016 at 22:07 | comment | added | Simd | @LiviuNicolaescu I apologize for any confusion. For a given value $n$ the dimension of all the variables $X_i$ is fixed to be $d$. That is each vector in the sum for a given $n$ has the same dimension. It might be easier to fix $n$ and ask for the closeness of $X_1+\dots+X_n$ to the $d$-dimensional Gaussian distribution as Ryan O'Donnell suggests. Having said that, my real interest is in a local limit types results rather than merely convergence in distribution. Overall my concern is how close $d$ can be to $n$ for us still to some sort of convergence for large $n$. | |
| Jan 13, 2016 at 20:46 | comment | added | Liviu Nicolaescu | If the dimension of $X_i$ is $d_i$,$d_i\nearrow \infty$, how do you add vectors of different dimensions? | |
| Jan 13, 2016 at 19:59 | answer | added | Ryan O'Donnell | timeline score: 8 | |
| Jan 13, 2016 at 19:29 | comment | added | Simd | @user31814 All the vectors in the sum have the same dimension $d$. My question is about whether, as $d$ increases with the number of summands $n$, we still get convergence in distribution to a multivariate Gaussian. The standard multivariate CLT talks about a fixed dimension $d$ which does not increase with the number of terms that are being summed. | |
| Jan 13, 2016 at 19:21 | comment | added | Pig | What does sum of vectors of different dimensions mean? | |
| Jan 13, 2016 at 18:57 | history | edited | Simd | CC BY-SA 3.0 |
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| Jan 13, 2016 at 17:45 | history | edited | Simd | CC BY-SA 3.0 |
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| Jan 13, 2016 at 16:51 | history | edited | Simd | CC BY-SA 3.0 |
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| Jan 13, 2016 at 16:42 | history | edited | Simd | CC BY-SA 3.0 |
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| Jan 13, 2016 at 16:36 | history | asked | Simd | CC BY-SA 3.0 |