I am trying to define a general monotonic function where linear is a special case of non-linear for certain parameter values and which can be increasing or decreasing. I think it should be possible but I cannot write it out. The non-linear curve need not have the exact shape as shown in the figure, I just need something non-linear.
Characteristics:
x between -10 and 10
y between 5 and 0.2
monotonic: increasing, decreasing, linear, non-linear
I faked a plot to show what I am looking for.
thanks
general montonic
x^α and 1 - (1 - x)^α have the desired behavior (try α < 1 and α > 1). This holds for x in [0, 1]. Rescale as needed.
For each a > 0, the function f_a defined by
f_a(x) = 4.8((x+10)/20)^a + 0.2
is a monotonic increasing function on the interval [-10,10] passing through (-10,0.2) and (10,5). f_1 is your linear function.
The class of functions g_a defined by:
g_a(x) = -4.8((x+10)/20)^a + 5
are monotonic decreasing functions on [-10,10] passing through (-10,5) and (10,0.2).
Related
Linear functions in mathematics are those polynomials whose degree is 1 and hence they are straight in nature when plotted on a graph. But the constant functions like f(x) = 3, even though their degree is 0, are straight in nature when plotted on a graph. Can’t we call them linear?
I would say they are not. There is some confusion between the equation of a straight line and the concept of linearity.
A linear function is additive, i.e. f(x+y) = f(x)+f(y), which is not true for a constant function.
The equation of a straight line through the origin y = m.x is indeed linear, but the equation of a general line y = m.x + p is not.
A linear function with an additional constant is called affine. Hence a constant function is affine.
There are both opinions, however claim of Yves Daoust is simply wrong. Consider
<i>f(x) = 2x +5 </br>
f(2) = 9</br>
f(3) = 11</br>
f(5) = 15</i>
which is definitely not 20...
I have two positive-valued vectors x,y of the same length in R. Using plot(x, y, "l",...), gives me a continuous line plot in 2 dimensions out of my finite vectors x and y. Is there a way to compute a definite integral over some range of this line plot in R?
edit1: I've looked into the integrate function in R. I'm not sure however how to make a function out of two vectors to pass to it, as my vectors are both finite.
edit2: For some more background, The length of x and y ~ 10,000. I've written a function to find periods, [xi, xj], of abnormalities in the data I'm observing. For each of these abnormalities, I've used plot to see what's going on in these snippets of my data. Now i need to compute statistics concerning the values of the integrals in these abnormal periods, so I'm trying to get as accurate as a number as possible to match with my graphs. X is a time variable, and I've taken very fine intervals of time.
You can do the integration with integrate(). To create a function out of your vectors x and y, you need to interpolate between the values. approxfun() does exactly that.
integrate takes a function and two bounds.
approxfun takes two vectors x and y just like those you have.
So my solution would be :
integrate(approxfun(x,y), range(x)[1], range(x)[2])
The approxfun function will take 2 vectors and return a function that gives the linear interpolation between the points. This can then be passed to functions like integrate. The splinefun function will also do interpolation, but based on a spline rather than piecewise linear.
In the piecewise linear case the integral will just be the sum of the trapezoids, it may be faster/simpler to just sum the areas of the trapezoids (the width, difference in x's
I landed here much later. But for future visitors,
here is some code for the suggestion from
Greg Snow's answer, for piece-wise linear functions:
line_integral <- function(x, y) {
dx <- diff(x)
end <- length(y)
my <- (y[1:(end - 1)] + y[2:end]) / 2
sum(dx *my)
}
# example
x <- c(0, 2, 3, 4, 5, 5, 6)
y <- c(0, 0, 1,-2,-1, 0, 0)
plot(x,y,"l")
line_integral(x,y)
I have a quadratic function f where, f = function (x) {2+.1*x+.23*(x*x)}. Let's say I have another quadratic fn g where g = function (x) {3+.4*x-.60*(x*x)}
Now, I want to maximize f given the constraints 1. g>0 and 2. 600<x<650
I have tried the packages optim,constrOptim and optimize. optimize does one dim. optimization, but without constraints and constrOptim I couldn't understand. I need to this using R. Please help.
P.S. In this example, the values may be erratic as I have given two random quadratic functions, but basically I want maximization of a quadratic fn given a quadratic constraint.
If you solve g(x)=0 for x by the usual quadratic formula then that just gives you another set of bounds on x. If your x^2 coefficent is negative then g(x) > 0 between the solutions, otherwise g(x)>0 outside the solutions, so within (-Inf, x1) and (x2, Inf).
In this case, g(x)>0 for -1.927 < x < 2.59. So in this case both your constraints cannot be simultaneously achieved (g(x) is LESS THAN 0 for 600<x<650).
But supposing your second condition was 1 < x < 5, then you'd just combine the solution from g(x)>0 with that interval to get 1 < x < 2.59, and then maximise f in that interval using standard univariate optimisation.
And you don't even need to run an optimisation algorithm. Your target f is quadratic. If the coefficient of x^2 is positive the maximum is going to be at one of your limits of x, so you only have a small number of values to try. If the coefficient of x^2 is -ve then the maximum is either at a limit or at the point where f(x) peaks (solve f'(x)=0) if that is within your limits.
So you can do this precisely, there's just a few conditions to test and then some intervals to compute and then some values of f at those interval limits to calculate.
So we have a matrix like
12,32
24,12
...
with length 2xN and another
44,32
44,19
...
with length 2xN and there is some function f(x, y) that returns z[1], z[2]. That 2 matrices that we were given represent known value pairs for x,y and z[1],z[2]. What are interpolation formulas that would help in such case?
If you solve the problem for one return value, you can find two functions f_1(x,y) and f_2(x,y) by interpolation, and compose your function as f(x, y) = [f_1(x,y), f_2(x,y)]. Just pick any method for solving the interpolation function suitable for your problem.
For the actual interpolation problem in two dimensions, there are a lot of ways you can handle this. If simple is what you require, you can go with linear interpolation. If you are OK with piecewise functions, you can go for bezier curves, or splines. Or, if data is uniform, you could get away with a simple polynomial interpolation (well, not quite trivial when in 2D, but easy enough).
EDIT: More information and some links.
A piecewise solution is possible using Bilinear interpolation (wikipedia).
For polynomial interpolation, if your data is on a grid, you can use the following algorithm (I cannot find the reference for it, it is from memory).
If the data points are on a k by l grid, rewrite your polynomial as follows:
f(x,y) = cx_1(x)*y^(k-1) + cx_2(x)*y^(k-2) + ... + cx_k(x)
Here, each coefficient cx_i(x) is also a polynomial of degree l. The first step is to find k polynomials of degree l by interpolating each row or column of the grid. When this is done, you have l coefficient sets (or, in other words, l polynomials) as interpolation points for each cx_i(x) polynomials as cx_i(x0), cx_i(x1), ..., cx_i(xl) (giving you a total of l*k points). Now, you can determine these polynomials using the above constants as the interpolation points, which give you the resulting f(x,y).
The same method is used for bezier curves or splines. The only difference is that you use control points instead of polynomial coefficients. You first get a set of splines that will generate your data points, and then you interpolate the control points of these intermediate curves to get the control points of the surface curve.
Let me add an example to clarify the above algorithm. Let's have the following data points:
0,0 => 1
0,1 => 2
1,0 => 3
1,1 => 4
We start by fitting two polynomials: one for data points (0,0) and (0,1), and another for (1, 0) and (1, 1):
f_0(x) = x + 1
f_1(x) = x + 3
Now, we interpolate in the other direction to determine the coefficients.When we read these polynomial coefficients vertically, we need two polynomials. One evaluates to 1 at both 0 and 1; and another that evaluates to 1 at 0, and 3 at 1:
cy_1(y) = 1
cy_2(y) = 2*y + 1
If we combine these into f(x,y), we get:
f(x,y) = cy_1(y)*x + cy_2(y)
= 1*x + (2*y + 1)*1
= x + 2*y + 1
suppose I have the following 2 random variables :
X where mean = 6 and stdev = 3.5
Y where mean = -42 and stdev = 5
I would like to create a new random variable Z based on the first two and knowing that : X happens 90% of the time and Y happens 10% of the time.
It is easy to calculate the mean for Z : 0.9 * 6 + 0.1 * -42 = 1.2
But is it possible to generate random values for Z in a single function?
Of course, I could do something along those lines :
if (randIntBetween(1,10) > 1)
GenerateRandomNormalValue(6, 3.5);
else
GenerateRandomNormalValue(-42, 5);
But I would really like to have a single function that would act as a probability density function for such a random variable (Z) that is not necessary normal.
sorry for the crappy pseudo-code
Thanks for your help!
Edit : here would be one concrete interrogation :
Let's say we add the result of 5 consecutives values from Z. What would be the probability of ending with a number higher than 10?
But I would really like to have a
single function that would act as a
probability density function for such
a random variable (Z) that is not
necessary normal.
Okay, if you want the density, here it is:
rho = 0.9 * density_of_x + 0.1 * density_of_y
But you cannot sample from this density if you don't 1) compute its CDF (cumbersome, but not infeasible) 2) invert it (you will need a numerical solver for this). Or you can do rejection sampling (or variants, eg. importance sampling). This is costly, and cumbersome to get right.
So you should go for the "if" statement (ie. call the generator 3 times), except if you have a very strong reason not to (using quasi-random sequences for instance).
If a random variable is denoted x=(mean,stdev) then the following algebra applies
number * x = ( number*mean, number*stdev )
x1 + x2 = ( mean1+mean2, sqrt(stdev1^2+stdev2^2) )
so for the case of X = (mx,sx), Y= (my,sy) the linear combination is
Z = w1*X + w2*Y = (w1*mx,w1*sx) + (w2*my,w2*sy) =
( w1*mx+w2*my, sqrt( (w1*sx)^2+(w2*sy)^2 ) ) =
( 1.2, 3.19 )
link: Normal Distribution look for Miscellaneous section, item 1.
PS. Sorry for the wierd notation. The new standard deviation is calculated by something similar to the pythagorian theorem. It is the square root of the sum of squares.
This is the form of the distribution:
ListPlot[BinCounts[Table[If[RandomReal[] < .9,
RandomReal[NormalDistribution[6, 3.5]],
RandomReal[NormalDistribution[-42, 5]]], {1000000}], {-60, 20, .1}],
PlotRange -> Full, DataRange -> {-60, 20}]
It is NOT Normal, as you are not adding Normal variables, but just choosing one or the other with certain probability.
Edit
This is the curve for adding five vars with this distribution:
The upper and lower peaks represent taking one of the distributions alone, and the middle peak accounts for the mixing.
The most straightforward and generically applicable solution is to simulate the problem:
Run the piecewise function you have 1,000,000 (just a high number) of times, generate a histogram of the results (by splitting them into bins, and divide the count for each bin by your N (1,000,000 in my example). This will leave you with an approximation for the PDF of Z at every given bin.
Lots of unknowns here, but essentially you just wish to add the two (or more) probability functions to one another.
For any given probability function you could calculate a random number with that density by calculating the area under the probability curve (the integral) and then generating a random number between 0 and that area. Then move along the curve until the area is equal to your random number and use that as your value.
This process can then be generalized to any function (or sum of two or more functions).
Elaboration:
If you have a distribution function f(x) which ranges from 0 to 1. You could calculate a random number based on the distribution by calculating the integral of f(x) from 0 to 1, giving you the area under the curve, lets call it A.
Now, you generate a random number between 0 and A, let's call that number, r. Now you need to find a value t, such that the integral of f(x) from 0 to t is equal to r. t is your random number.
This process can be used for any probability density function f(x). Including the sum of two (or more) probability density functions.
I'm not sure what your functions look like, so not sure if you are able to calculate analytic solutions for all this, but worse case scenario, you could use numeric techniques to approximate the effect.