I'd like to make an array of random samples from a Gaussian distrubution.
Mean value is 0 and variance is 1.
If I take enough samples, I would think my maximum value of a sample can be 0+1=1.
However, I find that I get values like 4.2891 ...
My code:
x = 0+sqrt(1)*randn(100000,1);
mean(x)
var(x)
max(x)
This would give me a mean like 0, a var of 0.9937 but my maximum value is 4.2891?
Can anyone help me out why it does this?
As others have mentioned, there is no bound on the possible values that x can take on in a gaussian distribution. However, the farther x is from the mean, the less likely it is to be drawn.
To give you some intuition for what the variance actually means (for any distribution, not just the gaussian case), you can look at the 68-95-99.7 rule. The rule says:
about 68% of the population will be within one sigma of the mean
about 95% of the population will be within two sigma's of the mean
about 99.7% of the population will be within three sigma's of the mean
Here sigma = sqrt(var) is the standard deviation of the distribution.
So while in theory it is possible to draw any x from a gaussian distribution, in practice it is unlikely to draw anything past 5 or 6 standard deviations away for a population of 100000.
This will yield N random numbers using the gaussian normal distribution.
N = 100;
mu = 0;
sigma = 1;
Xs = normrnd(mu, sigma, N);
EDIT:
I just realized that your code is in fact equivalent to what I've written.
As others already pointed out: variance is not the maximum distance a sample can deviate from the mean! (It is just the average of the squares of those distances)
Related
I'm new to the forum and a beginner in programming.
I have the task to program a random walk in Matlab (1D or 2D) with a variance that I can adjust. I found the code for the random walk, but I'm really confused where to put the variance. I thought that the random walk always has the same variance (= t) so maybe I'm just lost in the math.
How do I control the variance?
For a simple random walk, consider using the Normal distribution with mean 0 (also called 'drift') and a non-zero variance. Notice since the mean is zero and the distribution is symmetric, this is a symmetric random walk. On each step, the process is equally like to go up or down, left or right, etc.
One easy way:
Step 1: Generate each step
Step 2: Get the cumulative sum
This can be done for any number of dimensions.
% MATLAB R2019a
drift = 0;
std = 1; % std = sqrt(variance)
pd = makedist('Normal',drift,std);
% One Dimension
nsteps = 50;
Z = random(pd,nsteps,1);
X = [0; cumsum(Z)];
plot(0:nsteps,X) % alternatively: stairs(0:nsteps,X)
And in two dimensions:
% Two Dimensions
nsteps = 100;
Z = random(pd,nsteps,2);
X = [zeros(1,2); cumsum(Z)];
% 2D Plot
figure, hold on, box on
plot(X(1,1),X(1,1),'gd','DisplayName','Start','MarkerFaceColor','g')
plot(X(:,1),X(:,2),'k-','HandleVisibility','off')
plot(X(end,1),X(end,2),'rs','DisplayName','Stop','MarkerFaceColor','r')
legend('show')
The variance will affect the "volatility" so a higher variance means a more "jumpy" process relative to the lower variance.
Note: I've intentionally avoided the Brownian motion-type implementation (scaling, step size decreasing in the limit, etc.) since OP specifically asked for a random walk. A Brownian motion implementation can link the variance to a time-index due to Gaussian properties.
The OP writes:
the random walk has always the same variance
This is true for the steps (each step typically has the same distribution). However, the variance of the process at a time step (or point in time) should be increasing with the number of steps (or as time increases).
Related:
MATLAB: plotting a random walk
Consider the following draws for a 2x1 vector in Matlab with a probability distribution that is a mixture of two Gaussian components.
P=10^3; %number draws
v=1;
%First component
mu_a = [0,0.5];
sigma_a = [v,0;0,v];
%Second component
mu_b = [0,8.2];
sigma_b = [v,0;0,v];
%Combine
MU = [mu_a;mu_b];
SIGMA = cat(3,sigma_a,sigma_b);
w = ones(1,2)/2; %equal weight 0.5
obj = gmdistribution(MU,SIGMA,w);
%Draws
RV_temp = random(obj,P);%Px2
% Transform each component of RV_temp into a uniform in [0,1] by estimating the cdf.
RV1=ksdensity(RV_temp(:,1), RV_temp(:,1),'function', 'cdf');
RV2=ksdensity(RV_temp(:,2), RV_temp(:,2),'function', 'cdf');
Now, if we check whether RV1 and RV2 are uniformly distributed on [0,1] by doing
ecdf(RV1)
ecdf(RV2)
we can see that RV1 is uniformly distributed on [0,1] (the empirical cdf is close to the 45 degree line) while RV2 is not.
I don't understand why. It seems that the more distant are mu_a(2)and mu_b(2), the worse the job done by ksdensity with a reasonable number of draws. Why?
When you have a mixture of N(0.5,v) and N(8.2,v) then the range of the generated data is larger than if you had expectation which were closer, like N(0,v) and N(0,v), as you have in the other dimension. Then you ask ksdensity to approximate a function using P points inside this range.
Like in standard linear interpolation, the denser the points the better approximation of the function (inside the range), this is the same case here. Thus in the N(0.5,v) and N(8.2,v) where the points are "sparse" (or sparser, is that a word?) the approximation is worse than in the N(0,v) and N(0,v) where the points are denser.
As a small side note, are there any reason that you do not apply ksdensity directly on the bivariate data? Also I cannot reproduce your comment where you say that 5e2points are also good. Final comment, 1e3 is typically prefered over 10^3.
I think this is simply about the number of samples you're using. For the first example, the means of the two Gaussians are relatively close, hence a thousand samples are enough to obtain a cdf really close the the U[0,1] cdf. On the second vector though, you have a higher difference, and need more samples. With 100000 samples, I obtained the following result:
With 1000 I obtained this:
Which is clearly farther from the Uniform cdf function. Try to increase the number of samples to a million and check if the result is again getting closer.
I've got an arbitrary probability density function discretized as a matrix in Matlab, that means that for every pair x,y the probability is stored in the matrix:
A(x,y) = probability
This is a 100x100 matrix, and I would like to be able to generate random samples of two dimensions (x,y) out of this matrix and also, if possible, to be able to calculate the mean and other moments of the PDF. I want to do this because after resampling, I want to fit the samples to an approximated Gaussian Mixture Model.
I've been looking everywhere but I haven't found anything as specific as this. I hope you may be able to help me.
Thank you.
If you really have a discrete probably density function defined by A (as opposed to a continuous probability density function that is merely described by A), you can "cheat" by turning your 2D problem into a 1D problem.
%define the possible values for the (x,y) pair
row_vals = [1:size(A,1)]'*ones(1,size(A,2)); %all x values
col_vals = ones(size(A,1),1)*[1:size(A,2)]; %all y values
%convert your 2D problem into a 1D problem
A = A(:);
row_vals = row_vals(:);
col_vals = col_vals(:);
%calculate your fake 1D CDF, assumes sum(A(:))==1
CDF = cumsum(A); %remember, first term out of of cumsum is not zero
%because of the operation we're doing below (interp1 followed by ceil)
%we need the CDF to start at zero
CDF = [0; CDF(:)];
%generate random values
N_vals = 1000; %give me 1000 values
rand_vals = rand(N_vals,1); %spans zero to one
%look into CDF to see which index the rand val corresponds to
out_val = interp1(CDF,[0:1/(length(CDF)-1):1],rand_vals); %spans zero to one
ind = ceil(out_val*length(A));
%using the inds, you can lookup each pair of values
xy_values = [row_vals(ind) col_vals(ind)];
I hope that this helps!
Chip
I don't believe matlab has built-in functionality for generating multivariate random variables with arbitrary distribution. As a matter of fact, the same is true for univariate random numbers. But while the latter can be easily generated based on the cumulative distribution function, the CDF does not exist for multivariate distributions, so generating such numbers is much more messy (the main problem is the fact that 2 or more variables have correlation). So this part of your question is far beyond the scope of this site.
Since half an answer is better than no answer, here's how you can compute the mean and higher moments numerically using matlab:
%generate some dummy input
xv=linspace(-50,50,101);
yv=linspace(-30,30,100);
[x y]=meshgrid(xv,yv);
%define a discretized two-hump Gaussian distribution
A=floor(15*exp(-((x-10).^2+y.^2)/100)+15*exp(-((x+25).^2+y.^2)/100));
A=A/sum(A(:)); %normalized to sum to 1
%plot it if you like
%figure;
%surf(x,y,A)
%actual half-answer starts here
%get normalized pdf
weight=trapz(xv,trapz(yv,A));
A=A/weight; %A normalized to 1 according to trapz^2
%mean
mean_x=trapz(xv,trapz(yv,A.*x));
mean_y=trapz(xv,trapz(yv,A.*y));
So, the point is that you can perform a double integral on a rectangular mesh using two consecutive calls to trapz. This allows you to compute the integral of any quantity that has the same shape as your mesh, but a drawback is that vector components have to be computed independently. If you only wish to compute things which can be parametrized with x and y (which are naturally the same size as you mesh), then you can get along without having to do any additional thinking.
You could also define a function for the integration:
function res=trapz2(xv,yv,A,arg)
if ~isscalar(arg) && any(size(arg)~=size(A))
error('Size of A and var must be the same!')
end
res=trapz(xv,trapz(yv,A.*arg));
end
This way you can compute stuff like
weight=trapz2(xv,yv,A,1);
mean_x=trapz2(xv,yv,A,x);
NOTE: the reason I used a 101x100 mesh in the example is that the double call to trapz should be performed in the proper order. If you interchange xv and yv in the calls, you get the wrong answer due to inconsistency with the definition of A, but this will not be evident if A is square. I suggest avoiding symmetric quantities during the development stage.
I am working in image classification. I am using an information that called prior probability (in Bayesian rule). It has range in [0,1]. And it requires computing in logarithm. However, as you know, logarithm of zero number is Inf.
For example, given an pixel x in image I (size 3 by 3) with an cost function such as
Cost(x)=30+log(prior(x))
where prior is an matrix 3 by 3
prior=[ 0 0 0.5;
1 1 0.2;
0.4 0 0]
I =[ 1 2 3;
4 5 6;
7 8 9]
I want to compute cost of x=1 then
cost(x=1)=30+log(0)
Now, log(0) is Inf. Then result cost(x=1) also Inf. Based on my assumption that prior=0 that mean the given pixel belongs to background, and prior=1 that mean the given pixel belongs to foreground.
My question is that how to compute log(prior) satisfy my assumption.
I am using Matlab to do it. I think that log(0) becomes very small negative value. And I just set it is -9 as my code
%% Handle with log(0)
prior(prior==0.0) = NaN;
%% Compute log
log_prior=log(prior);
%% Assume that e^-9 very near 0.
log_prior(isnan(log_prior)) = -9;
UPDATE: To make clearly what I am doing. Let see the Bayesian rule. My task is that how to assign an given pixel x belongs to Background (BG) or Foreground (FG). It will depends on the probability
P(x∈BG|x)=P(x|x∈BG)P(x∈BG)/P(x)
In which P(x|x∈BG) is likelihood function and assume that it is approximated by Gaussian distribution, P(x∈BG) is prior term and P(x) can be ignore due to it is const
Using Maximum-a-Posteriori (MAP) Estimation we can map the above equation in to log space (to resolve exponential in Gaussian function)
Cost(x)=log(P(x∈BG|x))=log(P(x|x∈BG))+log(P(x∈BG))
To make simple, let assume log(P(x|x∈BG))=30, log(P(x∈BG)) is log(prior) then my cost function can rewritten as
Cost(x)=30+log(prior(x))
Now problem is that prior is within [0,1] then it logarithm is -Inf. As the chepner said, we can add eps value as
log(prior+eps)
However, log(eps) is very a lager negative number. It will be affected my cost function (also becomes very large negative number). Then the first term in my cost function (30) becomes not necessary. Based on my assumption that log(x)=1 then the pixel x will be BG and prior(x)=1 will be FG. How to make handle with my log(prior) when I compute my cost function?
The correct thing to do, before fiddling with Matlab, is to try to understand your problem. Ask yourself "what does it mean for the prior probability to vanish?". The answer is given by Bayes theorem, one form of which is:
posterior = likelihood * prior / normalization
So places where the prior is nil are, by definition, places where you are certain that your events (the things whose probabilities you are computing) cannot happen, regardless of their apparent likelihood (i.e. "cost"). So they are not interesting for you. You just recognize that and skip them.
Consider a Normal distribution with mean 0 and standard deviation 1. I would like to divide this distribution into 9 regions of equal probability and take a random sample from each region.
It sounds like you want to find the values that divide the area under the probability distribution function into segments of equal probability. This can be done in matlab by applying the norminv function.
In your particular case:
segmentBounds = norminv(linspace(0,1,10),0,1)
Any two adjacent values of segmentBounds now describe the boundaries of segments of the Normal probability distribution function such that each segment contains one ninth of the total probability.
I'm not sure exactly what you mean by taking random numbers from each sample. One approach is to sample from each region by performing rejection sampling. In short, for each region bounded by x0 and x1, draw a sample from y = normrnd(0,1). If x0 < y < x1, keep it. Else discard it and repeat.
It's also possible that you intend to sample from these regions uniformly. To do this you can try rand(1)*(x1-x0) + x0. This will produce problems for the extreme quantiles, however, since the regions extend to +/- infinity.