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If your matrices are drawn from standard-normal iid entries, the probability of being positive-definite is approximately $p_N\approx 3^{-N^2/4}$, so for example if $N=5$, the chance is 1/1000, and goes down quite fast after that. You can find an extended discussion of this question here. You can somewhat intuit this answer by accepting that the eigenvalue ...

23

It depends which field you're in but, one of the big initial pushes for the study of random matrices came out of atomic physics, and was pioneered by Wigner. You can find a brief overview here. Specifically, it was the eigenvalues (which are energy levels in atomic physics) of random matrices that generated tons of interest because the correlations between ...

14

Since $X^TAX$ is a scalar, $$\text{Tr}(X^TAX)= X^TAX = \text{Tr}(AXX^T)$$ so that $$\text{E}(X^TAX) = \text{E}(\text{Tr}(AXX^T)) = \text{Tr}(\text{E} (A XX^T)) = \text{Tr}(A\text{E}(XX^T))$$. Here we have used that the trace of a product are invariant under cyclical permutations of the factors, and that the trace is a linear operator, so commutes with ...

13

You seem to be comfortable with applications of random vectors. For instance, I deal with this kind of random vectors every day: interest rates of different tenors. Federal Reserve Bank has H15 series, look at Treasury bills 4-week, 3-month, 6-month and 1-year. You can think of these 4 rates as a vector with 4 elements. It's quire random too, look at the ...

8

In theoretical physics random matrices play an important role to understand universal features of energy spectra of systems with particular symmetries. My background in theoretical physics may cause me to present a slightly biased point of view here, but I would even go so far to suggest that the popularity of random matrix theory (RMT) originated from its ...

6

As @cardinal said in a comment: Actually, after a little thought, I think you algorithm is exactly the Sinkhorn-Knopp algorithm with a very minor modification. Let $X$ be your original matrix and let $Y$ be a matrix of the same size such that $Y_{ij}=X^2_{ij}$. Then, your algorithm is equivalent to applying Sinkhorn-Knopp to $Y$, where at the final step ...

5

A linear map is a map between vector spaces. Suppose you have a linear map and have chosen bases for its domain and range spaces. Then you can write a matrix which encodes the linear map. If you want to consider random linear maps between those two spaces, you should come up with a theory of random matrices. Random projection is a simple example of such ...

4

Compressive sensing as an application in image processing relies on random matrices as combined measurements of a 2D signal. Specific properties of these matrices, namely coherence, are defined for these matrices and play a role in the theory. Grossly simplified, it turns out that minimizing the L1 norm of a certain product of a Gaussian matrix and an ...

4

OK, to move these efforts along (but with some diffidence) I offer this approach: generate the diagonal elements first. Make them large constants. Generate all off-diagonal elements iid according to any (non-negative) distribution you want. Normalize rows. Check the column-max condition. Repeat if violated. By making the initial constants sufficiently ...

3

Here is a document about your issue: http://math.nyu.edu/faculty/avellane/LalouxPCA.pdf The idea is simple, you calculate the Marcenko-Pastur distribution with a modified variance of the elements of the matrix. The modified variance simply correspond to the variance explained by other eigenvalue than the first one. As said by john, you have to replace $\... 3 It looks like you can find it in: S. Kourouklis and P.G. Moschopoulos (1985) On the distribution of the trace of a non-central Wishart. Metron XLIII(1--2): 85--92. It looks like they cover the case of general covariance matrix$\Sigma$there. They also give pointers to in the paper to: Mathai, A.M. and Pillai, K.C.S. (1982) Further results on the trace ... 2 First, I would generate a matrix$M$such that$M_{ij}$is exponentially distributed with mean$\mu$for all$i \ne j$or$0otherwise. Then, calculate your matrix \begin{align} X_{ij} &= \frac{e^{-M_{ij}}}{\sum_je^{-M_{ij}}} \end{align} To efficiently sample from a categorical distribution (a list of probabilities totalling 1), just build a Huffman ... 2 You do not provide the whole reproduction code so I think what you are seeing is an artifact of two factors: the smoothing bandwidth used, ie. how "large of window" your smoother uses when constructing the kernel density estimate. The bell-shape-like shape you are trying to get rid off is mostly due to edge effects. As smaller bandwidth would have almost ... 2 Using the matrix computation only (although essentially, this solution is not that different from @DeltaIV's direct calculation), let me first slightly modify your definition ofS$to its centralized version$\begin{bmatrix}\hat{\theta}_1 - \theta & \cdots & \hat{\theta}_m - \theta\end{bmatrix}. We can go as follows \begin{align} & E[(\hat{\... 2 Why involving random matrices? This seems much simpler:$\mathbb{E}\left[\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)^{T}\left(\hat{\boldsymbol\theta} - \boldsymbol\theta\right)\right] = \frac{1}{m^2}\mathbb{E}\left[\sum_{i=1}^{m}\left(\hat{\boldsymbol\theta}_i-\boldsymbol\theta\right)^T\sum_{j=1}^{m}\left(\hat{\boldsymbol\theta}_j-\... 2 I am not aware of any useful distributional result here, and my suspicion is that the distribution of the eigenvalues will be so complicated that you will have to simulate it through sampling. However, it is worth noting that eigenvalues are roots of the characteristic polynomial$p_\boldsymbol{X}(t) = \det (t \boldsymbol{I} - \boldsymbol{X})$, so what you ... 1 You get eigenvalues of the correlation matrix. Marchenko-Pastur will tell you the threshold under which you can drop the eigenvalues. For instance, take a look at this dissertation p.24 where they use the theorem. What eigen decomposition does here is it lets you identivy a few important factors that represent entire correlation matrix. Instead of dealing ... 1 The rank will always be bounded by the smaller of the dimensions. For the$X$with dimensions$N\times D$, centering by the columns (i.e., find the mean of each column, and then subtract the mean from each value in that column) reduces the rank by column: now you are comparing$N$and$D-1$...$D-1$is smaller, so that is your rank. But, for the transposed$...

1

I'll put the $M$ subscript to denote matrix operators. For matrices $\mathbf{X} = [X_1\cdots X_n]$ and $\mathbf{Y} = [Y_1 \cdots Y_n]$, the law of total expectation works because we can apply the vector version (that we know to be true) to each column. \begin{align*} E_M(E_M[\mathbf{X} |\mathbf{Y} ]) &= \left[E(E_M[\mathbf{X} \vert \mathbf{Y}]_1),\...

1

Setting aside for a moment whether your sampling procedure in fact samples from the Haar measure on $SO(3)$, it samples from some distribution on $SO(3)$. To show that this measure is (proportional to) Haar measure, we need to show that it is left invariant. Now, there is a natural map $SO(3) \rightarrow S^2$ which takes an orthogonal matrix to its first ...

1

I don't think you can attack this question directly. I tested stock market data on a truncated normal, a log-normal and a truncated Cauchy density. In fact, it is none of the above, of course as it is a mixture distribution. Nonetheless, my argument was that the dominant density would by the Cauchy distribution and that I could factor out the other ...

1

Jan Magnus has his book on matrix calculus available online. Section 8.7, page 177 and on, deals with differentials in eigenproblems. See if you can apply the delta method and carry through the big $O$ and small $o$ through those differentials. Without knowing your convergence rates, though, I don't really know if your question could be answered. He ...

1

Merely doing a linear transformation is not going to help much in the classification. There two ways of linear transformation that you can do. PCA - Principal Component Analysis - Do a principal component analysis and take only few directions components that explains most of the variability in the data. Say you take 5 principal components that accounts for ...

1

This paper and R package completely solved my problem. It uses the Markov Chain Monte Carlo method, which relies on the fact that if you can find an initial solution of the constraint, through linear programming, you can find an arbitrary number of them by using a matrix that when multiplied by $E$, the constraints, gives zero. Read about it here: http://...

1

I asked a related (but not identical) question on MathOverflow and got responses that are similar to what you suggest here. However, the paper I cite in my answer there gives a distribution on O(p) with a dispersion parameter that controls how concentrated samples will be around the identity. This isn't exactly what you ask for, but it might be a useful ...

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