Say, I have N observations stored in an array, X = [x_1,x_2,...,x_N]. What is the meaning of E[\sum_{i=1}^N{x_i}]/N? This to me appears an average operation. But not sure. Can somebody please help what is the meaning of this operator with the help of any example in Matlab?
In general, (1) what is E[x]~ wherexis a random variable,
(2)E[x^2], and
(3)E[d]whered = 1/N(sum_{i=1}^N x_i)`
Your notation is still pretty bad to the point it's hard to understand, BUT there's a limit to what you can do on this site since it does not have MathJax enabled...
The expectation value is a generalized average in the sense that it should be weighted by the distribution from which X is drawn. If X is uniformly distributed then you'll get what you would call the "average" and what I think your first formula is giving. If X is distributed by some other distribution, then you will get something else.
If the distribution is discrete, then in general
E[f(x)] = sum_{i=1}^N [f(x_i) p(x_i)],
where p(x) is the distribution for the random variable.
Related
There are many mathematical programs out there out of which some are able to solve calculus-based problems, GeoGebra, Qalculate! to name a few.
How are those programs able to solve calculus-based problems which humans need to evaluate using a long procedure?
For example, the problem:
It takes a lot of steps for humans to solve this problem as shown here on Quora.
How can those mathematical programs solve them with such a good accuracy?
The Church-Turing thesis implies that anything a human being can calculate can be calculated by any Turing-equivalent system of computation - including programs running on computers. That is to say, if we can solve the problem (or calculate an approximate answer that meets some criteria) then a computer program can be made to do the same thing. Let's consider a simpler example:
f(x) = x
a = Integral(f, 0, 1)
A human being presented with this problem has two options:
try to compute the antiderivative using some procedure, then use procedures to evaluate the definite integral over the supplied range
use some numerical method to calculate an approximate value for the definite integral which meets some criteria for closeness to the true value
In either case, human beings have a set of tools that allow them to do this:
recognize that f(x) is a polynomial in x. There are rules for constructing the antiderivatives of polynomials. Specifically, each term ax^b in the polynomial can be converted to a/(b+1)x^(b+1) and then an arbitrary constant c added to the end. We then say Sf(x)dx = (1/2)x^2 + c. Now that we have the antiderivative, we have a procedure for computing the antiderivative over a range: calculate Sf(x)dx for the high value, then subtract from that the result of calculating Sf(x)dx for the low value. This gives ((1/2)1^2) - ((1/2)0^2) = 1/2 - 0 = 1/2.
decide that for our purposes a Riemann sum with dx=1/10 is sufficient and that we'll take the midpoint value. We get 10 rectangles with base 1/10 and heights 1/20, 3/20, 5/20, 7/20, 9/20, 11/20, 13/20, 15/20, 17/20 and 19/20, respectively. The areas are 1/200, 3/200, 5/200, 7/200, 9/200, 11/200, 13/200, 15/200, 17/200 and 19/200. The sum of these is (1+3+5+7+9+11+13+15+17+19)/200 = 100/200 = 1/2. We happened to get the exact answer since we used the midpoint value and evaluated the definite integral of a linear function; in general, we'd have been close but not exact.
The only difficulty is in adequately specifying the procedure human beings use to solve these problems in various ways. Once specified, computers are perfectly capable of doing them. And make no mistake, human beings have a procedure - conscious or subconscious - for doing these problems reliably.
I am working on a code in Least Square Non Negative solution recovery context on Matlab, and I need (with no more details because it's not that important for this question) to know the number of non zero elements in my matrices and arrays.
The function NNZ on matlab does exactly what I want, but it happens that I need more information about what Matlab thinks of a "zero element", it could be 0 itself, or the numerical zero like 1e-16 or less.
Does anybody has this information about the NNZ function, cause I couldn't get the original script
Thanks.
PS : I am not an expert on Matlab, so accept my apologies if it's a really simple task.
I tried "open nnz", on Matlab but I only get a small script of commented code lines...
Since nnz counts everything that isn't an exact zero (i.e. 1e-100 is non-zero), you just have to apply a relational operator to your data first to find how many values exceed some tolerance around zero. For a matrix A:
n = nnz(abs(A) > 1e-16);
Also, this discussion of floating-point comparison might be of interest to you.
You can add in a tolerance by doing something like:
nnz(abs(myarray)>tol);
This will create a binary array that is 1 when abs(myarray)>tol and 0 otherwise and then count the number of non-zero entries.
I would like to partition a number into an almost equal number of values in each partition. The only criteria is that each partition must be in between 60 to 80.
For example, if I have a value = 300, this means that 75 * 4 = 300.
I would like to know a method to get this 4 and 75 in the above example. In some cases, all partitions don't need to be of equal value, but they should be in between 60 and 80. Any constraints can be used (addition, subtraction, etc..). However, the outputs must not be floating point.
Also it's not that the total must be exactly 300 as in this case, but they can be up to a maximum of +40 of the total, and so for the case of 300, the numbers can sum up to 340 if required.
Assuming only addition, you can formulate this problem into a linear programming problem. You would choose an objective function that would maximize the sum of all of the factors chosen to generate that number for you. Therefore, your objective function would be:
(source: codecogs.com)
.
In this case, n would be the number of factors you are using to try and decompose your number into. Each x_i is a particular factor in the overall sum of the value you want to decompose. I'm also going to assume that none of the factors can be floating point, and can only be integer. As such, you need to use a special case of linear programming called integer programming where the constraints and the actual solution to your problem are all in integers. In general, the integer programming problem is formulated thusly:
You are actually trying to minimize this objective function, such that you produce a parameter vector of x that are subject to all of these constraints. In our case, x would be a vector of numbers where each element forms part of the sum to the value you are trying to decompose (300 in your case).
You have inequalities, equalities and also boundaries of x that each parameter in your solution must respect. You also need to make sure that each parameter of x is an integer. As such, MATLAB has a function called intlinprog that will perform this for you. However, this function assumes that you are minimizing the objective function, and so if you want to maximize, simply minimize on the negative. f is a vector of weights to be applied to each value in your parameter vector, and with our objective function, you just need to set all of these to -1.
Therefore, to formulate your problem in an integer programming framework, you are actually doing:
(source: codecogs.com)
V would be the value you are trying to decompose (so 300 in your example).
The standard way to call intlinprog is in the following way:
x = intlinprog(f,intcon,A,b,Aeq,beq,lb,ub);
f is the vector that weights each parameter of the solution you want to solve, intcon denotes which of your parameters need to be integer. In this case, you want all of them to be integer so you would have to supply an increasing vector from 1 to n, where n is the number of factors you want to decompose the number V into (same as before). A and b are matrices and vectors that define your inequality constraints. Because you want equality, you'd set this to empty ([]). Aeq and beq are the same as A and b, but for equality. Because you only have one constraint here, you would simply create a matrix of 1 row, where each value is set to 1. beq would be a single value which denotes the number you are trying to factorize. lb and ub are the lower and upper bounds for each value in the parameter set that you are bounding with, so this would be 60 and 80 respectively, and you'd have to specify a vector to ensure that each value of the parameters are bounded between these two ranges.
Now, because you don't know how many factors will evenly decompose your value, you'll have to loop over a given set of factors (like between 1 to 10, or 1 to 20, etc.), place your results in a cell array, then you have to manually examine yourself whether or not an integer decomposition was successful.
num_factors = 20; %// Number of factors to try and decompose your value
V = 300;
results = cell(1, num_factors);
%// Try to solve the problem for a number of different factors
for n = 1 : num_factors
x = intlinprog(-ones(n,1),1:n,[],[],ones(1,n),V,60*ones(n,1),80*ones(n,1));
results{n} = x;
end
You can then go through results and see which value of n was successful in decomposing your number into that said number of factors.
One small problem here is that we also don't know how many factors we should check up to. That unfortunately I don't have an answer to, and so you'll have to play with this value until you get good results. This is also an unconstrained parameter, and I'll talk about this more later in this post.
However, intlinprog was only released in recent versions of MATLAB. If you want to do the same thing without it, you can use linprog, which is the floating point version of integer programming... actually, it's just the core linear programming framework itself. You would call linprog this way:
x = linprog(f,A,b,Aeq,beq,lb,ub);
All of the variables are the same, except that intcon is not used here... which makes sense as linprog may generate floating point numbers as part of its solution. Due to the fact that linprog can generate floating point solutions, what you can do is if you want to ensure that for a given value of n, you could loop over your results, take the floor of the result and subtract with the final result, and sum over the result. If you get a value of 0, this means that you had a completely integer result. Therefore, you'd have to do something like:
num_factors = 20; %// Number of factors to try and decompose your value
V = 300;
results = cell(1, num_factors);
%// Try to solve the problem for a number of different factors
for n = 1 : num_factors
x = linprog(-ones(n,1),[],[],ones(1,n),V,60*ones(n,1),80*ones(n,1));
results{n} = x;
end
%// Loop through and determine which decompositions were successful integer ones
out = cellfun(#(x) sum(abs(floor(x) - x)), results);
%// Determine which values of n were successful in the integer composition.
final_factors = find(~out);
final_factors will contain which number of factors you specified that was successful in an integer decomposition. Now, if final_factors is empty, this means that it wasn't successful in finding anything that would be able to decompose the value into integer factors. Noting your problem description, you said you can allow for tolerances, so perhaps scan through results and determine which overall sum best matches the value, then choose whatever number of factors that gave you that result as the final answer.
Now, noting from my comments, you'll see that this problem is very unconstrained. You don't know how many factors are required to get an integer decomposition of your value, which is why we had to semi-brute-force it. In fact, this is a more general case of the subset sum problem. This problem is NP-complete. Basically, what this means is that it is not known whether there is a polynomial-time algorithm that can be used to solve this kind of problem and that the only way to get a valid solution is to brute-force each possible solution and check if it works with the specified problem. Usually, brute-forcing solutions requires exponential time, which is very intractable for large problems. Another interesting fact is that modern cryptography algorithms use NP-Complete intractability as part of their ciphertext and encrypting. Basically, they're banking on the fact that the only way for you to determine the right key that was used to encrypt your plain text is to check all possible keys, which is an intractable problem... especially if you use 128-bit encryption! This means you would have to check 2^128 possibilities, and assuming a moderately fast computer, the worst-case time to find the right key will take more than the current age of the universe. Check out this cool Wikipedia post for more details in intractability with regards to key breaking in cryptography.
In fact, NP-complete problems are very popular and there have been many attempts to determine whether there is or there isn't a polynomial-time algorithm to solve such problems. An interesting property is that if you can find a polynomial-time algorithm that will solve one problem, you will have found an algorithm to solve them all.
The Clay Mathematics Institute has what are known as Millennium Problems where if you solve any problem listed on their website, you get a million dollars.
Also, that's for each problem, so one problem solved == 1 million dollars!
(source: quickmeme.com)
The NP problem is amongst one of the seven problems up for solving. If I recall correctly, only one problem has been solved so far, and these problems were first released to the public in the year 2000 (hence millennium...). So... it has been about 14 years and only one problem has been solved. Don't let that discourage you though! If you want to invest some time and try to solve one of the problems, please do!
Hopefully this will be enough to get you started. Good luck!
I have three values X,Y and Z. These values have a range of values between 0 and 1 (0 and 1 included).
When I call a function f(X,Y,Z) it returns a value V (value between 0 and 1). My Goal is to choose X,Y,Z so that the returned value V is as close as possible to 1.
The selection Process should be automated and the right values for X,Y,Z are unknown.
Due to my Use Case it is possible to set Y and Z to 1 (the value 1 hasn't any influence on the output) and search for the best value of X.
After that I can replace X by that value and do the same for Y. Same procedure for Z.
How can I find the "maximum of the function"? Is there somekind of "gradient descend" or hill climbing algorithm or something like that?
The whole modul is written in perl so maybe there is an package for perl that can solve that problem?
You can use Simulated Annealing. Its a multi-variable optimization technique. It is also used to get a partial solution for the Travelling Salesperson problem. Its one of the search algorithms mentioned in Peter Norvig's Intro to AI book as well.
Its a hill climbing algorithm which depends on random variables. Also it won't necessarily give you the 'optimal' answer. You can also vary the iterations required by it as per your computational/time needs.
http://en.wikipedia.org/wiki/Simulated_annealing
http://www1bpt.bridgeport.edu/sed/projects/449/Fall_2000/fangmin/chapter2.htm
I suggest you take a look at Math::Amoeba which implements the Nelder–Mead method for finding stationary points on functions.
I have 500,000 values for a variable derived from financial markets. Specifically, this variable represents distance from the mean (in standard deviations). This variable has a arbitrary distribution. I need a formula that will allow me to select a range around any value of this variable such that an equal (or close to it) amount of data points fall within that range.
This will allow me to then analyze all of the data points within a specific range and to treat them as "similar situations to the input."
From what I understand, this means that I need to convert it from arbitrary distribution to uniform distribution. I have read (but barely understood) that what I am looking for is called "probability integral transform."
Can anyone assist me with some code (Matlab preferred, but it doesn't really matter) to help me accomplish this?
Here's something I put together quickly. It's not polished and not perfect, but it does what you want to do.
clear
randList=[randn(1e4,1);2*randn(1e4,1)+5];
[xCdf,xList]=ksdensity(randList,'npoints',5e3,'function','cdf');
xRange=getInterval(5,xList,xCdf,0.1);
and the function getInterval is
function out=getInterval(yPoint,xList,xCdf,areaFraction)
yCdf=interp1(xList,xCdf,yPoint);
yCdfRange=[-areaFraction/2, areaFraction/2]+yCdf;
out=interp1(xCdf,xList,yCdfRange);
Explanation:
The CDF of the random distribution is shown below by the line in blue. You provide a point (here 5 in the input to getInterval) about which you want a range that gives you 10% of the area (input 0.1 to getInterval). The chosen point is marked by the red cross and the
interval is marked by the lines in green. You can get the corresponding points from the original list that lie within this interval as
newList=randList(randList>=xRange(1) & randList<=xRange(2));
You'll find that on an average, the number of points in this example is ~2000, which is 10% of numel(randList)
numel(newList)
ans =
2045
NOTE:
Please note that this was done quickly and I haven't made any checks to see if the chosen point is outside the range or if yCdfRange falls outside [0 1], in which case interp1 will return a NaN. This is fairly straightforward to implement, and I'll leave that to you.
Also, ksdensity is very CPU intensive. I wouldn't recommend increasing npoints to more than 1e4. I assume you're only working with a fixed list (i.e., you have a list of 5e5 points that you've obtained somehow and now you're just running tests/analyzing it). In that case, you can run ksdensity once and save the result.
I do not speak Matlab, but you need to find quantiles in your data. This is Mathematica code which would do this:
In[88]:= data = RandomVariate[SkewNormalDistribution[0, 1, 2], 10^4];
Compute quantile points:
In[91]:= q10 = Quantile[data, Range[0, 10]/10];
Now form pairs of consecutive quantiles:
In[92]:= intervals = Partition[q10, 2, 1];
In[93]:= intervals
Out[93]= {{-1.397, -0.136989}, {-0.136989, 0.123689}, {0.123689,
0.312232}, {0.312232, 0.478551}, {0.478551, 0.652482}, {0.652482,
0.829642}, {0.829642, 1.02801}, {1.02801, 1.27609}, {1.27609,
1.6237}, {1.6237, 4.04219}}
Verify that the splitting points separate data nearly evenly:
In[94]:= Table[Count[data, x_ /; i[[1]] <= x < i[[2]]], {i, intervals}]
Out[94]= {999, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000, 1000}