I'm quite new to biostatistics so I apologize if my question is too dumb.
I'm studying data transformation in biostatistics to fit my data to the normal distribution.
I started with the Poisson distribution (which is quite common in the biostatistics: daily admissions, prevalence of rare disease etc) It is recommended to use the square root to fit data to normal distribution.
I used stata and this free dataset ( https://www.kaggle.com/datasets/martj42/international-football-results-from-1872-to-2017?resource=download ) with the results of a huge amount of football matches.
I have created a new variable for this dataset, made by the whole amount of goals scored by both teams in each match. You will find that as the independent variable distributed as following:
We can see that the distribution quite approximate the Poisson's one, as confirmed by the values of mean and std deviation.
Then, I've created a new variable with the square root of this variable and the distribution is the following (blue line is how the normal distrib with the same mean and std deviation looks like):
As you can see It's quite far from a normal distribution of my data, as proven by normality tests, but also easily visible from the q-q plot:
So, my question is, why sqrt didn't work? What can I do to transform my dataset to fit the normal distribution?
Related
I have a dataset with county level data where N=3119 with 93 variables. I am trying to do a PCA, EFA and or CFA. The data has been given to me already min/max normalized, ranging from (0,1). Theory states that the data should be normally distributed for CFA/SEM, but my understanding is that min/max normalization does not change the distribution of the data, only it's scale.
It is clear to me that I do not have multivariate normality or univariate normality due to the skewness of data. I guess what's confusing me, is when people seemingly throw around the term normalization interchangeably with the meaning of normal distribution.
So can I go forward with my analysis since min/max normalization has been performed, or do I need to look more towards other log/box cox transformations to adjust the distribution prior to running my analysis? Is it okay to log transform data that has already been min/max normalized?
my understanding is that min/max normalization does not change the distribution of the data, only it's scale.
Correct. If you print a hist()ogram of original and transformed data, they should look identical. Only the x-axis scale will change.
the term normalization interchangeably with the meaning of normal distribution
Indeed, these are completely separate issues.
Is it okay to log transform data that has already been min/max normalized?
Taking the log() would affect 0--1 data differently than data further up the real-number line. But I don't see why you need to transform the data when nonnormality corrections are available for SEs (in EFA or CFA) and model-fit test statistics (relevant for CFA). Independent-components analysis might be an alternative to PCA if your data are not normal.
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.
I'm trying to find confidence intervals for the means of various variables in a database using SPSS, and I've run into a spot of trouble.
The data is weighted, because each of the people who was surveyed represents a different portion of the overall population. For example, one young man in our sample might represent 28000 young men in the general population. The problem is that SPSS seems to think that the young man's database entries each represent 28000 measurements when they actually just represent one, and this makes SPSS think we have much more data than we actually do. As a result SPSS is giving very very low standard error estimates and very very narrow confidence intervals.
I've tried fixing this by dividing every weight value by the mean weight. This gives plausible figures and an average weight of 1, but I'm not sure the resulting numbers are actually correct.
Is my approach sound? If not, what should I try?
I've been using the Explore command to find mean and standard error (among other things), in case it matters.
You do need to scale weights to the actual sample size, but only the procedures in the Complex Samples option are designed to account for sampling weights properly. The regular weight variable in Statistics is treated as a frequency weight.
I have a trading strategy on the foreign exchange market that I am attempting to improve upon.
I have a huge table (100k+ rows) that represent every possible trade in the market, the type of trade (buy or sell), the profit/loss after that trade closed, and 10 or so additional variables that represent various market measurements at the time of trade-opening.
I am trying to find out if any of these 10 variables are significantly related to the profits/losses.
For example, imagine that variable X ranges from 50 to -50.
The average value of X for a buy order is 25, and for a sell order is -25.
If most profitable buy orders have a value of X > 25, and most profitable sell orders have a value of X < -25 then I would consider the relationship of X-to-profit as significant.
I would like a good starting point for this. I have installed RapidMiner 5 in case someone can give me a specific recommendation for that.
A Decision Tree is perhaps the best place to begin.
The tree itself is a visual summary of feature importance ranking (or significant variables as phrased in the OP).
gives you a visual representation of the entire
classification/regression analysis (in the form of a binary tree),
which distinguishes it from any other analytical/statistical
technique that i am aware of;
decision tree algorithms require very little pre-processing on your data, no normalization, no rescaling, no conversion of discrete variables into integers (eg, Male/Female => 0/1); they can accept both categorical (discrete) and continuous variables, and many implementations can handle incomplete data (values missing from some of the rows in your data matrix); and
again, the tree itself is a visual summary of feature importance ranking
(ie, significant variables)--the most significant variable is the
root node, and is more significant than the two child nodes, which in
turn are more significant than their four combined children. "significance" here means the percent of variance explained (with respect to some response variable, aka 'target variable' or the thing
you are trying to predict). One proviso: from a visual inspection of
a decision tree you cannot distinguish variable significance from
among nodes of the same rank.
If you haven't used them before, here's how Decision Trees work: the algorithm will go through every variable (column) in your data and every value for each variable and split your data into two sub-sets based on each of those values. Which of these splits is actually chosen by the algorithm--i.e., what is the splitting criterion? The particular variable/value combination that "purifies" the data the most (i.e., maximizes the information gain) is chosen to split the data (that variable/value combination is usually indicated as the node's label). This simple heuristic is just performed recursively until the remaining data sub-sets are pure or further splitting doesn't increase the information gain.
What does this tell you about the "importance" of the variables in your data set? Well importance is indicated by proximity to the root node--i.e., hierarchical level or rank.
One suggestion: decision trees handle both categorical and discrete data usually without problem; however, in my experience, decision tree algorithms always perform better if the response variable (the variable you are trying to predict using all other variables) is discrete/categorical rather than continuous. It looks like yours is probably continuous, in which case in would consider discretizing it (unless doing so just causes the entire analysis to be meaningless). To do this, just bin your response variable values using parameters (bin size, bin number, and bin edges) meaningful w/r/t your problem domain--e.g., if your r/v is comprised of 'continuous values' from 1 to 100, you might sensibly bin them into 5 bins, 0-20, 21-40, 41-60, and so on.
For instance, from your Question, suppose one variable in your data is X and it has 5 values (10, 20, 25, 50, 100); suppose also that splitting your data on this variable with the third value (25) results in two nearly pure subsets--one low-value and one high-value. As long as this purity were higher than for the sub-sets obtained from splitting on the other values, the data would be split on that variable/value pair.
RapidMiner does indeed have a decision tree implementation, and it seems there are quite a few tutorials available on the Web (e.g., from YouTube, here and here). (Note, I have not used the decision tree module in R/M, nor have i used RapidMiner at all.)
The other set of techniques i would consider is usually grouped under the rubric Dimension Reduction. Feature Extraction and Feature Selection are two perhaps the most common terms after D/R. The most widely used is PCA, or principal-component analysis, which is based on an eigen-vector decomposition of the covariance matrix (derived from to your data matrix).
One direct result from this eigen-vector decomp is the fraction of variability in the data accounted for by each eigenvector. Just from this result, you can determine how many dimensions are required to explain, e.g., 95% of the variability in your data
If RapidMiner has PCA or another functionally similar dimension reduction technique, it's not obvious where to find it. I do know that RapidMiner has an R Extension, which of course let's you access R inside RapidMiner.R has plenty of PCA libraries (Packages). The ones i mention below are all available on CRAN, which means any of the PCA Packages there satisfy the minimum Package requirements for documentation and vignettes (code examples). I can recommend pcaPP (Robust PCA by Projection Pursuit).
In addition, i can recommend two excellent step-by-step tutorials on PCA. The first is from the NIST Engineering Statistics Handbook. The second is a tutorial for Independent Component Analysis (ICA) rather than PCA, but i mentioned it here because it's an excellent tutorial and the two techniques are used for the similar purposes.
I am using Octave and I would like to use the anderson_darling_test from the Octave forge Statistics package to test if two vectors of data are drawn from the same statistical distribution. Furthermore, the reference distribution is unlikely to be "normal". This reference distribution will be the known distribution and taken from the help for the above function " 'If you are selecting from a known distribution, convert your values into CDF values for the distribution and use "uniform'. "
My question therefore is: how would I convert my data values into CDF values for the reference distribution?
Some background information for the problem: I have a vector of raw data values from which I extract the cyclic component (this will be the reference distribution); I then wish to compare this cyclic component with the raw data itself to see if the raw data is essentially cyclic in nature. If the the null hypothesis that the two are the same can be rejected I will then know that most of the movement in the raw data is not due to cyclic influences but is due to either trend or just noise.
If your data has a specific distribution, for instance beta(3,3) then
p = betacdf(x, 3, 3)
will be uniform by the definition of a CDF. If you want to transform it to a normal, you can just call the inverse CDF function
x=norminv(p,0,1)
on the uniform p. Once transformed, use your favorite test. I'm not sure I understand your data, but you might consider using a Kolmogorov-Smirnov test instead, which is a nonparametric test of distributional equality.
Your approach is misguided in multiple ways. Several points:
The Anderson-Darling test implemented in Octave forge is a one-sample test: it requires one vector of data and a reference distribution. The distribution should be known - not come from data. While you quote the help-file correctly about using a CDF and the "uniform" option for a distribution that is not built in, you are ignoring the next sentence of the same help file:
Do not use "uniform" if the distribution parameters are estimated from the data itself, as this sharply biases the A^2 statistic toward smaller values.
So, don't do it.
Even if you found or wrote a function implementing a proper two-sample Anderson-Darling or Kolmogorov-Smirnov test, you would still be left with a couple of problems:
Your samples (the data and the cyclic part estimated from the data) are not independent, and these tests assume independence.
Given your description, I assume there is some sort of time predictor involved. So even if the distributions would coincide, that does not mean they coincide at the same time-points, because comparing distributions collapses over the time.
The distribution of cyclic trend + error would not expected to be the same as the distribution of the cyclic trend alone. Suppose the trend is sin(t). Then it never will go above 1. Now add a normally distributed random error term with standard deviation 0.1 (small, so that the trend is dominant). Obviously you could get values well above 1.
We do not have enough information to figure out the proper thing to do, and it is not really a programming question anyway. Look up time series theory - separating cyclic components is a major topic there. But many reasonable analyses will probably be based on the residuals: (observed value - predicted from cyclic component). You will still have to be careful about auto-correlation and other complexities, but at least it will be a move in the right direction.