My question is about estimating a normal distribution for a row of points of an interval that are extracted from the data, just knowing the maximum = 200 and minimum = 100, to distribute them randomly by an average and an associated standard deviation that we will estimate. Is there a method in Matlab, or the literature to do it, such that when you draw the associated law; we must not exceed the min and max values I used this trick there:
x = xmin + rand (1, n) * (xmax-xmin);
but I do not know how to extract the parameters of the normal distribution (the mean, the standard deviation).
I'm looking for the mean and the std.
Related
I'm building up on my preivous question because there is a further issue.
I have fitted in Matlab a normal distribution to my data vector: PD = fitdist(data,'normal'). Now I have a new data point coming in (e.g. x = 0.5) and I would like to calculate its probability.
Using cdf(PD,x) will not work because it gives the probability that the point is smaller or equal to x (but not exactly x). Using pdf(PD,x) gives just the densitiy but not the probability and so it can be greater than one.
How can I calculate the probability?
If the distribution is continuous then the probability of any point x is 0, almost by definition of continuous distribution. If the distribution is discrete and, furthermore, the support of the distribution is a subset of the set of integers, then for any integer x its probability is
cdf(PD,x) - cdf(PD,x-1)
More generally, for any random variable X which takes on integer values, the probability mass function f(x) and the cumulative distribution F(x) are related by
f(x) = F(x) - F(x-1)
The right hand side can be interpreted as a discrete derivative, so this is a direct analog of the fact that in the continuous case the pdf is the derivative of the cdf.
I'm not sure if matlab has a more direct way to get at the probability mass function in your situation than going through the cdf like that.
In the continuous case, your question doesn't make a lot of sense since, as I said above, the probability is 0. Non-zero probability in this case is something that attaches to intervals rather than individual points. You still might want to ask for the probability of getting a value near x -- but then you have to decide on what you mean by "near". For example, if x is an integer then you might want to know the probability of getting a value that rounds to x. That would be:
cdf(PD, x + 0.5) - cdf(PD, x - 0.5)
Let's say you have a random variable X that follows the normal distribution with mean mu and standard deviation s.
Let F be the cumulative distribution function for the normal distribution with mean mu and standard deviation s. The probability the random variableX falls between a and b, that is P(a < X <= b) = F(b) - F(a).
In Matlab code:
P_a_b = normcdf(b, mu, s) - normcdf(a, mu, s);
Note: observe that the probability X is exactly equal to 0.5 (or any specific value) is zero! A range of outcomes will have positive probability, but an insufficient sum of individual outcomes will have probability zero.
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.
What test can I do in MATLAB to test the spread of a histogram? For example, in the given set of histograms, I am only interested in 1,2,3,5 and 7 (going from left to right, top to bottom) because they are less spread out. How can I obtain a value that will tell me if a histogram is positively skewed?
It may be possible using Chi-Squared tests but I am not sure what the MATLAB code will be for that.
You can use the standard definition of skewness. In other words, you can use:
You compute the mean of your data and you use the above equation to calculate skewness. Positive and negative skewness are like so:
Source: Wikipedia
As such, the larger the value, the more positively skewed it is. The more negative the value, the more negatively skewed it is.
Now to compute the mean of your histogram data, it's quite simple. You simply do a weighted sum of the histogram entries and divide by the total number of entries. Given that your histogram is stored in h, the bin centres of your histogram are stored in x, you would do the following. What I will do here is assume that you have bins from 0 up to N-1 where N is the total number of bins in the histogram... judging from your picture:
x = 0:numel(h)-1; %// Judging from your pictures
num_entries = sum(h(:));
mu = sum(h.*x) / num_entries;
skew = ((1/num_entries)*(sum((h.*x - mu).^3))) / ...
((1/(num_entries-1))*(sum((h.*x - mu).^2)))^(3/2);
skew would contain the numerical measure of skewness for a histogram that follows that formula. Therefore, with your problem statement, you will want to look for skewness numbers that are positive and large. I can't really comment on what threshold you should look at, but look for positive numbers that are much larger than most of the histograms that you have.
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)
I'm trying to create confidence interval for a set of data not randomly distributed and very skewed at right. Surfing, I discovered a pretty rude method that consists in using the 97.5% percentile (of my data) for the upperbound CL and 2.5% percentile for your lower CL.
Unfortunately, I need a more sophisticated way!
Then I discovered the bootstrap, precisley the MATLAB bootci function, but I'm having hard time to undestand how to used it properly.
Let's say that M is my matrix containing my data (19x100), and let's say that:
Mean = mean(M,2);
StdDev = sqrt(var(M'))';
How can I compute the asymmetrical CI for every row of the Mean vector using bootci?
Note: earlier, I was computing the CI in this very wrong way: Mean +/- 2 * StdDev, shame on me!
Let's say you have a 100x19 data set. Each column has a different distribution. We'll choose the log normal distribution, so that they skew to the right.
means = repmat(log(1:19), 100, 1);
stdevs = ones(100, 19);
X = lognrnd(means, stdevs);
Notice that each column is from the same distribution, and the rows are separate observations. Most functions in MATLAB operate on the rows by default, so it's always preferable to keep your data this way around.
You can compute bootstrap confidence intervals for the mean using the bootci function.
ci = bootci(1000, #mean, X);
This does 1000 resamplings of your data, calculates the mean for each resampling and then takes the 2.5% and 97.5% quantiles. To show that it's an asymmetric confidence interval about the mean, we can plot the mean and the confidence intervals for each column
plot(mean(X), 'r')
hold on
plot(ci')