i don't understand R & D White Paper WHP 031(page 26)
http://downloads.bbc.co.uk/rd/pubs/whp/whp-pdf-files/WHP031.pdf
why constant factor r=9.
My calcuation is factor r=7
two polynomial are belows
7*(x^2)+ 7*(x^1)+9 , 3*(x^1)+14
Would you explain how 9 is extracted ?
I try to resolve using GF multiplication table and extened Euclid algorithm.
The first polynomial, λΛ(x) = 7x^2+7x+9 is the error locator polynomial. The second polynomial, λΩ(x) = 3x+14, is used for calculating error values. Although not required, a common convention for Λ(x) is that the least significant term of Λ(x) is defined to be 1 (as opposed to the most significant term). In the example, the least significant term of λΛ(x) = 7x^2+7x+9 is the 9, so λ=9, and both polynomials are divided by λ=9. This results in (7x^2+7x+9)/9 = 14x^2+14x+1, and (3x+14)/9 = 6x+15.
A similar thing is shown in the wiki article, which uses GF(929), a non-binary field, in it's examples. In the wiki's extended Euclid example, both polynomials are divided by 544.
https://en.wikipedia.org/wiki/Reed%E2%80%93Solomon_error_correction#Euclidean_decoder
Comments about the white paper.
The Forney equation 21 on page 21 has an X^(1-b) term, which the paper notes can be ignored in the case b=1, but in the worked examples, b=0, so that X term is needed for the calculations as shown in the paper.
Some Reed Solomon implementations use a non-binary field, such as GF(929) as shown in the wiki article examples, and which is used for PDF417 barcodes. In this case, it's important to pay attention to addition versus subtraction, the signs of terms in equations, and the derivative of Λ(x) follows normal convention, the derivative of aX^b = abX^(b-1) with all math modulo 929. The wiki article examples show this.
http://www.ujamjar.com/demo/ocaml/2014/06/18/reed-solomon-demo.html
when m=6, k=2, b=1
more complicated polynomial extended euclid algorithm is done.
i just understand example in the white paper
but i test more complext example on the link.
i don't know why S4, S5
Would you explain how S4, S5 polynomial is extracted?
Related
I am learning statistics, and have some basic yet core questions on SD:
s = sample size
n = total number of observations
xi = ith observation
μ = arithmetic mean of all observations
σ = the usual definition of SD, i.e. ((1/(n-1))*sum([(xi-μ)**2 for xi in s])**(1/2) in Python lingo
f = frequency of an observation value
I do understand that (1/n)*sum([xi-μ for xi in s]) would be useless (= 0), but would not (1/n)*sum([abs(xi-μ) for xi in s]) have been a measure of variation?
Why stop at power of 1 or 2? Would ((1/(n-1))*sum([abs((xi-μ)**3) for xi in s])**(1/3) or ((1/(n-1))*sum([(xi-μ)**4 for xi in s])**(1/4) and so on have made any sense?
My notion of squaring is that it 'amplifies' the measure of variation from the arithmetic mean while the simple absolute difference is somewhat a linear scale notionally. Would it not amplify it even more if I cubed it (and made absolute value of course) or quad it?
I do agree computationally cubes and quads would have been more expensive. But with the same argument, the absolute values would have been less expensive... So why squares?
Why is the Normal Distribution like it is, i.e. f = (1/(σ*math.sqrt(2*pi)))*e**((-1/2)*((xi-μ)/σ))?
What impact would it have on the normal distribution formula above if I calculated SD as described in (1) and (2) above?
Is it only a matter of our 'getting used to the squares', it could well have been linear, cubed or quad, and we would have trained our minds likewise?
(I may not have been 100% accurate in my number of opening and closing brackets above, but you will get the idea.)
So, if you are looking for an index of dispersion, you actually don't have to use the standard deviation. You can indeed report mean absolute deviation, the summary statistic you suggested. You merely need to be aware of how each summary statistic behaves, for example the SD assigns more weight to outlying variables. You should also consider how each one can be interpreted. For example, with a normal distribution, we know how much of the distribution lies between ±2SD from the mean. For some discussion of mean absolute deviation (and other measures of average absolute deviation, such as the median average deviation) and their uses see here.
Beyond its use as a measure of spread though, SD is related to variance and this is related to some of the other reasons it's popular, because the variance has some nice mathematical properties. A mathematician or statistician would be able to provide a more informed answer here, but squared difference is a smooth function and is differentiable everywhere, allowing one to analytically identify a minimum, which helps when fitting functions to data using least squares estimation. For more detail and for a comparison with least absolute deviations see here. Another major area where variance shines is that it can be easily decomposed and summed, which is useful for example in ANOVA and regression models generally. See here for a discussion.
As to your questions about raising to higher powers, they actually do have uses in statistics! In general, the mean (which is related to average absolute mean), the variance (related to standard deviation), skewness (related to the third power) and kurtosis (related to the fourth power) are all related to the moments of a distribution. Taking differences raised to those powers and standardizing them provides useful information about the shape of a distribution. The video I linked provides some easy intuition.
For some other answers and a larger discussion of why SD is so popular, See here.
Regarding the relationship of sigma and the normal distribution, sigma is simply a parameter that stretches the standard normal distribution, just like the mean changes its location. This is simply a result of the way the standard normal distribution (a normal distribution with mean=0 and SD=variance=1) is mathematically defined, and note that all normal distributions can be derived from the standard normal distribution. This answer illustrates this. Now, you can parameterize a normal distribution in other ways as well, but I believe you do need to provide sigma, whether using the SD or precisions. I don't think you can even parametrize a normal distribution using just the mean and the mean absolute difference. Now, a deeper question is why normal distributions are so incredibly useful in representing widely different phenomena and crop up everywhere. I think this is related to the Central Limit Theorem, but I do not understand the proofs of the theorem well enough to comment further.
Hi everyone I am reading a book "Element of Statistical Learning) and came across the below paragraph which i dont I understand. (explains how the training data was generated)
We generated 10 means mk from a bivariate Gaussian distribution N((0,1)T,I) and labeled this class as blue. Similraly, 10 more were drawn from from N((0,1)T,I) and labeled class Orange. Then for each class we generated 100 observations as follows: for each observation, we picked an mk at random with probability 1/10, and then generated a N(mk, I/5), thus leading to a mixture of Gaussian cluster for each class.
I would appreciate if you could explain the above paragraph and especially N((0,1)T,I)
by the way- (0,1) to the power of T for Transpose.
Is this notation mathmatically common or related to a specific computer language.
In the paragraph N stands for the Normal distribution; more specifically, in this case it stands for the Multivariate normal distribution. It is not specific to any programming languages. It comes from statistics and probability theory, but due to numerous appealing properties and important applications of this probability distribution it is also widely used in programming, so you should be able to perform the described procedure in any language.
The part (0,1)^T is a vector of means. That is, we have in mind a random vector of length two, where the first element on average is 0, and the second one on average is 1.
"I" stands for the 2x2 identity matrix whose role is the variance-covariance matrix. That is, the variance of both random vector components is 1 (i.e., the diagonal terms), while off-diagonal points are 0 and correspond to the covariance between the two random variables.
As a passion project I'm recreating a neuronal model from XJ Wang's lab at NYU. The paper is Wei, W., & Wang, X. J. (2016). Inhibitory control in the cortico-basal ganglia-thalamocortical loop: complex regulation and interplay with memory and decision processes. Neuron, 92(5), 1093-1105.
The main problem I'm having is interpreting the equation for calculating the differential of the neurons membrane voltage. They have included a bursting neuronal model for cells in the basal ganglia and subthalamic nucleus. The differential equation for membrane voltage in these regions incorporates a hyperpolarization rebound which results in bursts and tonic spiking. The equation is on page 2 of a prior paper which uses basically the exact same model. I have linked to the paper below and I have provided an image link to the exact passage as well.
http://www.cns.nyu.edu/wanglab/publications/pdf/wei.jns2015.pdf
This is the equation I'm having trouble reading, don't worry about Isyn its the input current from the synapses
The equation is taken from this paper: https://www.physiology.org/doi/pdf/10.1152/jn.2000.83.1.588
Obviously the equation will need to be discritized so I can run it with numpy but I ill ignore that for now as it will be relatively easy to do so. The middle term with all the H's is whats giving me trouble. As I understand it I should be running code which dos the following:
gt * h * H(V-Vh) * (V-Vt)
Where H(V-Vh) is the heavyside step function, V is the membrane voltage at the prior timestep Vh = -60mV and Vt = 120mV. gt is a conductance efficacy constant in nanoSiemens. I think the correct way to interpret this for python is...
gt * h * heavyside(-60, 0.5)*(V-120)
But I'm not 100% sure I'm reading the notation correctly. Could someone please confirm I've read it as it is intended?
Secondly h is the deactivation term which gives rise to bursting as described in the final paragraph on page 2 of Smith et al., 2000 (the second pdf I've linked to). I understand the differential equations that govern the evolution of h well enough but what is the value of h? In Smith et al. 2000 the authors say that h relaxes to zero with a time constant of 20ms and it relaxes to unity with a time constant of 100ms. What value is h relaxing from and what does it mean to relax to unity?
For you x1 (of the numpy.heaviside) is = V-Vh; you are comparing that difference to zero. You might try writing your own version of the Heaviside function to deepen understanding, and then move back to the numpy version if you need it for speed or compatibility. The pseudo-code wordy version would be something like,
if (V<Vh): return(0); else: return(1);
You could probably just write (V>=Vh) in your code as Python will treat the boolean as 1 if true and 0 if false.
This ignores the possibility of V==Vh in the complete version of Heaviside, but for most practical work with real values (even discretized in a computer) that is unlikely to be a case to worth concerning yourself with, but you could easily add it in.
How do I compute the generalized mean for extreme values of p (very close to 0, or very large) with reasonable computational error?
As per your link, the limit for p going to 0 is the geometric mean, for which bounds are derived.
The limit for p going to infinity is the maximum.
I have been struggling with the same problem. Here is how I handled this:
Let gmean_p(x1,...,xn) be the generalized mean where p is real but not 0, and x1, ..xn nonnegative. For M>0, we have gmean_p(x1,...,xn) = M*gmean_p(x1/M,...,xn/M) of which the latter form can be exploited to reduce the computational error. For large p, I use M=max(x1,...,xn) and for p close to 0, I use M=mean(x1,..xn). In case M=0, just add a small positive constant to it. This did the job for me.
I suspect if you're interested in very large or small values of p, it may be best to do some form of algebraic manipulation of the generalized-mean formula before putting in numerical values.
For example, in the small-p limit, one can show that the generalized mean tends to the n'th root of the product x_1*x_2*...x_n. The higher order terms in p involve sums and products of log(x_i), which should also be relatively numerically stable to compute. In fact, I believe the first-order expansion in p has a simple relationship to the variance of log(x_i):
If one applies this formula to a set of 100 random numbers drawn uniformly from the range [0.2, 2], one gets a trend like this:
which here shows the asymptotic formula becoming pretty accurate for p less than about 0.3, and the simple formula only failing when p is less than about 1e-10.
The case of large p, is dominated by that x_i which has the largest magnitude (lets call that index i_max). One can rearrange the generalized mean formula to take the following form, which has less pathological behaviour for large p:
If this is applied (using standard numpy routines including numpy.log1p) to another 100 uniformly distributed samples over [0.2, 2.0], one finds that the rearranged formula agrees essentially exactly with the simple formula, but remains valid for much larger values of p for which the simple formula overflows when computing powers of x_i.
(Note that the left-hand plot has the blue curve for the simple formula shifted up by 0.1 so that one can see where it ends due to overflows. For p less than about 1000, the two curves would otherwise be indistinguishable.)
I think the answer here should be to use a recursive solution. In the same way that mean(1,2,3,4)=mean(mean(1,2),mean(3,4)), you can do this kind of recursion for generalized means. What this buys you is that you won't need to do as many sums of really large numbers and you decrease the likelihood of creating an overflow. Also, the other danger when working with floating point numbers is when adding numbers of very different magnitudes (or subtracting numbers of very similar magnitudes). So to avoid these kinds of rounding errors it might help to sort your data before you try and calculate the generalized mean.
Here's a hunch:
First convert all your numbers into a representation in base p. Now to raise to a power of 1/p or p, you just have to shift them --- so you can very easily do all powers without losing precision.
Work out your mean in base p, then convert the result back to base two.
If that doesn't work, an even less practical hunch:
Try working out the discrete Fourier transform, and relating that to the discrete Fourier transform of the input vector.
I've been trying to find an answer to this for months (to be used in a machine learning application), it doesn't seem like it should be a terribly hard problem, but I'm a software engineer, and math was never one of my strengths.
Here is the scenario:
I have a (possibly) unevenly weighted coin and I want to figure out the probability of it coming up heads. I know that coins from the same box that this one came from have an average probability of p, and I also know the standard deviation of these probabilities (call it s).
(If other summary properties of the probabilities of other coins aside from their mean and stddev would be useful, I can probably get them too.)
I toss the coin n times, and it comes up heads h times.
The naive approach is that the probability is just h/n - but if n is small this is unlikely to be accurate.
Is there a computationally efficient way (ie. doesn't involve very very large or very very small numbers) to take p and s into consideration to come up with a more accurate probability estimate, even when n is small?
I'd appreciate it if any answers could use pseudocode rather than mathematical notation since I find most mathematical notation to be impenetrable ;-)
Other answers:
There are some other answers on SO that are similar, but the answers provided are unsatisfactory. For example this is not computationally efficient because it quickly involves numbers way smaller than can be represented even in double-precision floats. And this one turned out to be incorrect.
Unfortunately you can't do machine learning without knowing some basic math---it's like asking somebody for help in programming but not wanting to know about "variables" , "subroutines" and all that if-then stuff.
The better way to do this is called a Bayesian integration, but there is a simpler approximation called "maximum a postieri" (MAP). It's pretty much like the usual thinking except you can put in the prior distribution.
Fancy words, but you may ask, well where did the h/(h+t) formula come from? Of course it's obvious, but it turns out that it is answer that you get when you have "no prior". And the method below is the next level of sophistication up when you add a prior. Going to Bayesian integration would be the next one but that's harder and perhaps unnecessary.
As I understand it the problem is two fold: first you draw a coin from the bag of coins. This coin has a "headsiness" called theta, so that it gives a head theta fraction of the flips. But the theta for this coin comes from the master distribution which I guess I assume is Gaussian with mean P and standard deviation S.
What you do next is to write down the total unnormalized probability (called likelihood) of seeing the whole shebang, all the data: (h heads, t tails)
L = (theta)^h * (1-theta)^t * Gaussian(theta; P, S).
Gaussian(theta; P, S) = exp( -(theta-P)^2/(2*S^2) ) / sqrt(2*Pi*S^2)
This is the meaning of "first draw 1 value of theta from the Gaussian" and then draw h heads and t tails from a coin using that theta.
The MAP principle says, if you don't know theta, find the value which maximizes L given the data that you do know. You do that with calculus. The trick to make it easy is that you take logarithms first. Define LL = log(L). Wherever L is maximized, then LL will be too.
so
LL = hlog(theta) + tlog(1-theta) + -(theta-P)^2 / (2*S^2)) - 1/2 * log(2*pi*S^2)
By calculus to look for extrema you find the value of theta such that dLL/dtheta = 0.
Since the last term with the log has no theta in it you can ignore it.
dLL/dtheta = 0 = (h/theta) + (P-theta)/S^2 - (t/(1-theta)) = 0.
If you can solve this equation for theta you will get an answer, the MAP estimate for theta given the number of heads h and the number of tails t.
If you want a fast approximation, try doing one step of Newton's method, where you start with your proposed theta at the obvious (called maximum likelihood) estimate of theta = h/(h+t).
And where does that 'obvious' estimate come from? If you do the stuff above but don't put in the Gaussian prior: h/theta - t/(1-theta) = 0 you'll come up with theta = h/(h+t).
If your prior probabilities are really small, as is often the case, instead of near 0.5, then a Gaussian prior on theta is probably inappropriate, as it predicts some weight with negative probabilities, clearly wrong. More appropriate is a Gaussian prior on log theta ('lognormal distribution'). Plug it in the same way and work through the calculus.
You can use p as a prior on your estimated probability. This is basically the same as doing pseudocount smoothing. I.e., use
(h + c * p) / (n + c)
as your estimate. When h and n are large, then this just becomes h / n. When h and n are small, this is just c * p / c = p. The choice of c is up to you. You can base it on s but in the end you have to decide how small is too small.
You don't have nearly enough info in this question.
How many coins are in the box? If it's two, then in some scenarios (for example one coin is always heads, the other always tails) knowing p and s would be useful. If it's more than a few, and especially if only some of the coins are only slightly weighted then it is not useful.
What is a small n? 2? 5? 10? 100? What is the probability of a weighted coin coming up heads/tail? 100/0, 60/40, 50.00001/49.99999? How is the weighting distributed? Is every coin one of 2 possible weightings? Do they follow a bell curve? etc.
It boils down to this: the differences between a weighted/unweighted coin, the distribution of weighted coins, and the number coins in your box will all decide what n has to be for you to solve this with a high confidence.
The name for what you're trying to do is a Bernoulli trial. Knowing the name should be helpful in finding better resources.
Response to comment:
If you have differences in p that small, you are going to have to do a lot of trials and there's no getting around it.
Assuming a uniform distribution of bias, p will still be 0.5 and all standard deviation will tell you is that at least some of the coins have a minor bias.
How many tosses, again, will be determined under these circumstances by the weighting of the coins. Even with 500 tosses, you won't get a strong confidence (about 2/3) detecting a .51/.49 split.
In general, what you are looking for is Maximum Likelihood Estimation. Wolfram Demonstration Project has an illustration of estimating the probability of a coin landing head, given a sample of tosses.
Well I'm no math man, but I think the simple Bayesian approach is intuitive and broadly applicable enough to put a little though into it. Others above have already suggested this, but perhaps if your like me you would prefer more verbosity.
In this lingo, you have a set of mutually-exclusive hypotheses, H, and some data D, and you want to find the (posterior) probabilities that each hypothesis Hi is correct given the data. Presumably you would choose the hypothesis that had the largest posterior probability (the MAP as noted above), if you had to choose one. As Matt notes above, what distinguishes the Bayesian approach from only maximum likelihood (finding the H that maximizes Pr(D|H)) is that you also have some PRIOR info regarding which hypotheses are most likely, and you want to incorporate these priors.
So you have from basic probability Pr(H|D) = Pr(D|H)*Pr(H)/Pr(D). You can estimate these Pr(H|D) numerically by creating a series of discrete probabilities Hi for each hypothesis you wish to test, eg [0.0,0.05, 0.1 ... 0.95, 1.0], and then determining your prior Pr(H) for each Hi -- above it is assumed you have a normal distribution of priors, and if that is acceptable you could use the mean and stdev to get each Pr(Hi) -- or use another distribution if you prefer. With coin tosses the Pr(D|H) is of course determined by the binomial using the observed number of successes with n trials and the particular Hi being tested. The denominator Pr(D) may seem daunting but we assume that we have covered all the bases with our hypotheses, so that Pr(D) is the summation of Pr(D|Hi)Pr(H) over all H.
Very simple if you think about it a bit, and maybe not so if you think about it a bit more.