# Tag Info

23

You have to subclass the rv_continuous class in scipy.stats import scipy.stats as st class my_pdf(st.rv_continuous): def _pdf(self,x): return 3*x**2 # Normalized over its range, in this case [0,1] my_cv = my_pdf(a=0, b=1, name='my_pdf') now my_cv is a continuous random variable with the given PDF and range [0,1] Note that in this example ...

15

There is a straightforward solution with only two passes through the data: First compute $$K := \max_i\; a_i,$$ which tells you that, if there are $n$ terms, then $$\sum_i e^{a_i} \le n e^K.$$ Since you presumably don't have $n$ anywhere near as large as even $10^{20}$, you should have no worry about overflowing in the computation of $$\tau := \sum_i e^{... 15 You should check out sympy.stats. It provides an interface to deal with random variables. The following example provides a random variable X defined on the unit interval with density 2x In : from sympy.stats import * In : x = Symbol('x') In : X = ContinuousRV(x, 2*x, Interval(0, 1)) In : P(X>.5) Out: 0.750000000000000 In : Var(X) # ... 10 To keep precision while you add doubles together you need to use Kahan Summation, this is the software equivalent to having a carry register. This is fine for most values, but if you are getting overflow then you are hitting the limits of IEEE 754 double-precision which would be about e^{709.783}. At this point you need a new representation. You can ... 9 I would recommend the GNU Scientific Library (GSL). GSL has both gsl_ran_dirichlet and gsl_ran_multinomial for the Dirichlet and multinomial distributions respectively. See the manual for the Dirichlet distribution is here and the manual page for the multinomial distribution is here. Full documentation can be found here or online here. 9 In addition to all that Bill Barth has already said above, let me mention that people often report the fastest of several runs. The rationale is that the actual run time is the ideal run time plus any number of slow downs resulting from other processes running, OS delays, network delays, etc. Since these are all noise we are not interested in, using the ... 8 \mathbf{A} is an (n+1) \times (n+1) matrix. It can be obtained as follows: \textbf{A} = \left[ \matrix{1 & 1 & 1 & \cdots & 1 \cr x_0 & x_1 & x_2 & \cdots & x_{n} \cr \vdots & \vdots & \vdots & \cdots & \vdots \cr x_0^n & x_1^n & x_2^n & \cdots & ... 6 Since a distribution function contains more information than the finite set of numbers you start with, you clearly have to add information in the process. This information comes in the form of a model that you assume, and whose parameters you adjust to make the model fit your finite sample. Unless the nature of your problem suggests some model, the choice ... 6 Histograms are not useful for high dimensional data. The curse of dimensionality affects one quite fast. As in your case if the grid is of size 7**6, you have on average one point in one bin. Kernel density estimator are better suited as long as you keep the kernel bandwidth large enough. In my experience the top hat kernel as k-nearest neighbor yields ... 6 This doesn't randomly sample points, but instead chooses representative points deterministically. scipy.stats.norm.ppf(np.linspace(0, 1, 1000+2)[1:-1]) 6 It's a question of how you choose \lambda and \Gamma (of course). Think for a moment about what happens if you choose \Gamma=I and make \lambda large: in that case you say that it is more important to you to minimize the regularization term \|Ix\|^2=\|x\|^2 than to minimize the misfit \|Ax-b\|^2. Obviously, making the term \|x\|^2 small means ... 5 Knuth (TAOCP Volume 2, 3rd ed., pg. 232) suggests using the formula M_k = M_{k-1} + (x_k - M_{k-1})/k to calculate the mean, where x is a vector of your samples. See also: a stackoverflow question, an article on accurately computing running variance, and a paper on computing covariances and arbitrary-order statistical moments 5 You haven't specified the distribution of x(t). I'll assume that you want to use a complex normal distribution, since that choice makes it reasonably easy to solve the problem and because this assumption is quite common in signal processing. I'll also discretize the problem so that you're generating a vector X of N entries with a specified complex ... 5 It's actually not all that complicated to calculate the Tracy-Widom CDF just from its definition: see On The Numerical Evaluation Of Fredholm Determinants by Folkmar Bornemann. The Wikipedia page gives the definition as$$ F_2(s) = \det(I - A_s), $$where A_s is the integral operator on [s,\infty) with the kernel$$ \DeclareMathOperator{\Ai}{Ai} K(x,y) =...

5

Would a decomposition of the form $A = XX^T$ suffice? This would be enough, e.g., if the end goal is sampling from the Gaussian distribution with this given covariance. If so, you can use the following formula, which is quite similar to your approximation: $$X = D^{1/2} + \frac{\sqrt{u^T D^{-1} u+1}-1}{u^T D^{-1} u} u u^T D^{-1/2}$$ This follows from ...

4

One solution would be to keep your rolling window of data sorted using a self-balancing binary tree or a hash table. On every update, you pay a constant insertion cost, a $O(n)$ deletion cost, and a $O(n)$ traversal cost to update the quintiles.

4

What you are looking for is Kernel Density Estimation. Algorithms to carry this out are built into many scientific software packages and libraries. There is still a degree of arbitrariness in the choice of kernel and bandwidth, but there are heuristics for these.

4

Kernel density estimation is a good suggestion. Another option is to construct the empirical CDF and then seek the distribution that best fits. Depending on the form of your output this might be more appropriate than a KDE, which is typically used for multimodal distributions. More detail on the empirical CDF here: http://en.wikipedia.org/wiki/...

4

My experience in these types of problems is that almost any perturbation will do. I suggest you do not try to perturb $u$ or $p$, because generating a divergence-free $u$ with a perturbed component is more complicated than necessary. You can easily perturb $b$ with lots of different things including a small random field or a randomly-distributed set of ...

4

There are many reasons that a coarser simulation would give different results that a finer grained simulation. A few examples: are boundary layers being resolved differently? am I resolving new features (vortices / blockages to flow) Thinking about a grid result as purely a convolution of a fine-grid result with a Gaussian will work very well in viscous ...

4

There is no reason to use hardware random number generation for anything other than full cryptography. For everything else, including computational physics, pseudorandom generators are fine. I would suggest using the Random123 library of Salmon et al.: it's fast, trivially parallelizable, and strong (in particular stronger than Mersenne Twister). It is ...

4

Geoffrey already answered the question, but I'd like to add another perspective to it. One of the things you will have to do one way or another is to debug code. It may not necessarily be that you have to debug anything that has to do with the random numbers themselves, but, say, a bug in the function evaluation that depends on the randomly selected sample. ...

4

You can easily do this with for loops. Just use n to define your loop limits. Here's a fully functional solution for MATLAB (just define n and x first): %pre-allocate A A = zeros(n+1); %first row: for j=0:n A(1,j+1)=sum(x.^j); end %rows 2 through n for i=1:n A(i+1,1:n)=A(i,2:n+1); %copy from previous row A(i+1,n+1)=sum(x.^(n+i)); %compute last ...

3

It is imperative to compare many runs with refined meshes until you detect convergence. A single solution with no mesh refinement study shouldn't give you much confidence in your results. Comparing runs with different fluid properties tells you something different. If you think that a set of runs with different viscosities is relevant to your ultimate ...

3

I'd calculate the mean $\overline x$ and the covariance matrix $C$ of the joint data set, and then do a K/S test on the univariate quantity $V(x):=(x-\overline x)^TC^{-1}(x-\overline x)$ evaluated on the parts. If the K/S test give a significant difference between the parts, there is one. If it gives no significant difference, the test is to be regarded as ...

3

ROOT supports Kolmogorov tests on higher dimensional histograms, and the notes (for the 2D version) suggest that there is a ambiguity--which they deal with by punting: calculate it both ways. I don't know if the code contains anymore details, but the comments sometimes have references to papers and the like. There are some additional interesting comments in ...

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From the Computer Science perspective I do not think that make sense to make an general statistical model for memory access time (latency) and memory bandwidth. It does make sense to create an statistical model for a algorithm. That is because each algorithm has a specific memory access pattern, the memory access patterns are relevant to the cache hierarchy,...

3

The quantity you are measuring currently is something akin to "prominence" which is a better formed topographical quantity than numerical one. The wiki page in that link describes all sorts of strange cases that arise, even without having any image/surface borders to worry about. If you want to stick with this alignment metric I would suggest using a ...

3

You don't need to sort your data first, if you have access to additional temporary storage, in which case, you should use a selection algorithm.

3

Testing whether or not the mean is correct, or even if the histogram of your generated random variants "looks" like a certain distribution is not sufficient. Stick with much more rigorous test suites such as TestU01 or Diehard. Also, you really only have TWO random numbers in each row, because of the constraint that they sum to 1. This requires more ...

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