I am calculating a variance-covariance matrix and I see two different ways of calculating the standard errors:
sqrt(diagonal values/number of observations)
e.g. standard deviation / sqrt(number of observations)
(as is given from on how to calculate the standard error https://en.wikipedia.org/wiki/Standard_error)
or some people say it is simply
sqrt(diagonal values)
I had previously thought that the diagonal values in the variance-co-variance matrix were the variance and hence the square root would be the standard deviation (not the SE). However, the more I read the more I think I may be wrong and that it is the SE, but I am unsure why this is the case.
Can anyone help? Many thanks!!
Yes, the diagonal elements of the covariance matrix are the variances. The square root of these variances are the standard deviations. If you need the standard error you have to clarify the question "the standard error of what?" (see also the wikipedia entry of your post). If you mean the standard error of the mean then yes, "standard deviation / sqrt(number of observations)" is what you are looking for.
Related
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 am interested in extrapolating the curve from the population that I know is normally distributed. However, in my process, I am only able to get access to a section of the curve (from -3 standard deviations to -2 standard deviations). My question is what is the best way to fitting a curve to a section of a bell curve.
A normal distribution can be defined by an equation f(x) (its PDF, which is a little difficult to write in non-latex, you can check out the Wikipedia page) with two parameters: the mean and the variance (or standard deviation).
Therefore, if you want to know which variance and mean define it, you just need to solve for the mean and the variance given two known values (which you have infinitely many of, even on a short interval).
I want to extract the position of a peak from a spectrum (energy spectrum of scattered photons). To do so, I am using scipy.optimize.curve_fit to fit a Gaussian to the region of the spectrum that resembles the Gaussian.
How do I find the uncertainty of the peak value? The peak value itself will be given by the result for the mean parameter from the Gaussian regression.
There are two things that came to my mind:
I get covariance values from the minimisation routine from which I get the error on the mean parameter.
Also, I could think about using the sigma of the Gaussian to get to the
error of the mean.
My thoughts on this would be, that the error on the mean parameter cannot be the wrong way to go. And I would also wager that the standard error does not really tell us the uncertainty with which we know the peak value. It tells us about the shape of the distribution but not about the uncertainty in the peak value (which, for simplicity, we believe to have a true, sharply defined value.)
(This is a repost of a question I originally posted on stats.stackoverflow where I did not get any answers after 2 days.)
The peak value is the mean of the Gaussian distribution, so the standard error of the mean parameter gives the uncertainty of the peak. The sigma parameter describes the width of the peak and has its own uncertainty. If you are measuring a wide peak and took a good measurement, you would get a large sigma but a low peak uncertainty (or standard error).
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.
Let's say, I have two random variables,x and y, both of them have n observations. I've used a forecasting method to estimate xn+1 and yn+1, and I also got the standard error for both xn+1 and yn+1. So my question is that what the formula would be if I want to know the standard error of xn+1 + yn+1, xn+1 - yn+1, (xn+1)*(yn+1) and (xn+1)/(yn+1), so that I can calculate the prediction interval for the 4 combinations. Any thought would be much appreciated. Thanks.
Well, the general topic you need to look at is called "change of variables" in mathematical statistics.
The density function for a sum of random variables is the convolution of the individual densities (but only if the variables are independent). Likewise for the difference. In special cases, that convolution is easy to find. For example, for Gaussian variables the density of the sum is also a Gaussian.
For product and quotient, there aren't any simple results, except in special cases. For those, you might as well compute the result directly, maybe by sampling or other numerical methods.
If your variables x and y are not independent, that complicates the situation. But even then, I think sampling is straightforward.