Difference between Hmisc wtd.var and SAS proc Mean generated weighted variance - r

I'm getting different results from R and SAS when I try to calculate a weighted variance. Does anyone know what might be causing this difference?
I create vectors of weights and values and I then calculate the weighted variance using the
Hmisc library wtd.var function:
library(Hmisc)
wt <- c(5, 5, 4, 1)
x <- c(3.7,3.3,3.5,2.8)
wtd.var(x,weights=wt)
I get an answer of:
[1] 0.0612381
But if I try to reproduce these results in SAS I get a quite different result:
data test;
input wt x;
cards;
5 3.7
5 3.3
4 3.5
1 2.8
;
run;
proc means data=test var;
var x;
weight wt;
run;
Results in an answer of
0.2857778

You probably have a difference in how the variance is calculated. SAS gives you an option, VARDEF, which may help here.
proc means data=test var vardef=WDF;
var x;
weight wt;
run;
That on your dataset gives a variance similar to r. Both are 'right', depending on how you choose to calculate the weighted variance. (At my shop we calculate it a third way, of course...)
Complete text from PROC MEANS documentation:
VARDEF=divisor specifies the divisor to use in the calculation of the
variance and standard deviation. The following table shows the
possible values for divisor and associated divisors.
Possible Values for VARDEF=
Value Divisor Formula for Divisor
DF degrees of freedom n - 1
N number of observations n
WDF sum of weights minus one ([Sigma]iwi) - 1
WEIGHT | WGT sum of weights [Sigma]iwi
The procedure computes the variance as CSS/Divisor, where CSS
is the corrected sums of squares and equals Sum((Xi-Xbar)^2). When you
weight the analysis variables, CSS equals sum(Wi*(Xi-Xwbar)^2), where
Xwbar is the weighted mean.
Default: DF Requirement: To compute the standard error of the mean,
confidence limits for the mean, or the Student's t-test, use the
default value of VARDEF=.
Tip: When you use the WEIGHT statement and
VARDEF=DF, the variance is an estimate of Sigma^2, where the
variance of the ith observation is Sigma^2/wi and wi is the
weight for the ith observation. This method yields an estimate of the
variance of an observation with unit weight.
Tip: When you use the
WEIGHT statement and VARDEF=WGT, the computed variance is
asymptotically (for large n) an estimate of Sigma^2/wbar, where
wbar is the average weight. This method yields an asymptotic
estimate of the variance of an observation with average weight.

Related

Generate random data based on correlation matrix for multiple timesteps in R

I would like to simulate data for some cases (e.g. nPerson=1000 obversations) at
some consecutive timesteps (e.g. ts = 3) for N intercorrelated variables (e.g. N=5).
The simulation should be based on a correlation matrix (corrMat, nrows=nPerson,.ncols = N).
corrMat should be identical for all timesteps.
I already found out that the MASS package has a function to create
random data fitting the constraints given by corrMat.
t1 <- mvrnorm(nPerson,mu=rep(0, N),Sigma=corrMat,empirical=T)
Now I would like to simulate t2 as a function of t1 and corrMat.
The data of t2 therefore should correlate according to corrMat
and they should also have same variance as the variables of t1.
One important constrained: for the intial values corrMat[i,i] = 1,
for consequtive timesteps it should be posible, that corrMat[i,i] < 1,
because each variable is depending on itsself a timestep before,
but a perfect correlation is notintended.
Maybe there is a variance decomposition of the correlation matrix,
that calculates an error variance for each of the n variables at the
next time step, so that one could calculate the
values at timestep t+1 as sum of the weighted correlations of the
variables at timestep t and then adding a random error,distributed
according to the error variance (with mean of error = 0) that replicates
the correlation matrix again at t+1.
Assuming normal errors:
getRand <- function (range) {
return (rnorm(1,mean=0, sd=range) )
}
That the (very simplified) code for the i-th variable x_i:
x_i[t+1] = 0
for (j:1..N) {
x_i[t+1] = x_i[t+1] + corrMat[i,j] * x_j[t]
}
x_i[t+1] = x_i[t+1] + getRand(sdErr)
So the question would be more specific: how to calculate sdErr?
For simplification I try to assume, that the variance for all variables
should be 1.
Thank you for any hint, how to get one step further!
I will do a mathematical formulation of the problem to stats.stackexchange.com,
as mikeck suggested to discuss details of the correlation problems more
in depth.
I still am interested in finding a geneal formula to calculate sdErr
to use it in the calculation of x_i[t+1].
But meanwhile I found a useful practical solution to the specific question "how to calculate sdErr?" without a formula for sdErr:
(1) simply calculate all variables WITHOUT errors (according to the equation above).
(2) calculate variances of the new variables
(3) calculate (for each i) differences var(x_i[t]) - var(x_i[t+1]) = sdErr ^ 2
So this sdErr can be added to each variable for each new observation.
This should lead to observations at t+1 which at least have the same variances as the observations in t.
Details concercing the question, if the model definition is adequate,
will be part of another post.

Formula of computing the Gini Coefficient in fastgini

I use the fastgini package for Stata (https://ideas.repec.org/c/boc/bocode/s456814.html).
I am familiar with the classical formula for the Gini coefficient reported for example in Karagiannis & Kovacevic (2000) (http://onlinelibrary.wiley.com/doi/10.1111/1468-0084.00163/abstract)
Formula I:
Here G is the Gini coefficient, µ the mean value of the distribution, N the sample size and y_i the income of the ith sample unit. Hence, the Gini coefficient computes the difference between all available income pairs in the data and calculates the total of all absolute differences.
This total is then normalized by dividing it by population squared times mean income (and multiplied by two?).
The Gini coefficient ranges between 0 and 1, where 0 means perfect equality (all individuals earn the same) and 1 refers to maximum inequality (1 person earns all the income in the country).
However the fastgini package refers to a different formula (http://fmwww.bc.edu/repec/bocode/f/fastgini.html):
Formula II:
fastgini uses formula:
i=N j=i
SUM W_i*(SUM W_j*X_j - W_i*X_i/2)
i=1 j=1
G = 1 - 2* ----------------------------------
i=N i=N
SUM W_i*X_i * SUM W_i
i=1 i=1
where observations are sorted in ascending order of X.
Here W seems to be the weight, which I don't use, therefore it should be 1 (?). I’m not sure whether formula I and formula II are the same. There are no absolute differences and the result is subtracted from 1 in formula II. I have tried to transform the equations but I don’t get any further.
Could someone give me a hint whether both ways of computing (formula I + formula II) are equivalent?

ANOVA (AOV function) in R: Misleading p_value reported on equal values

I would greatly appreciate any guidance on the following: I am running ANOVA (aov) to retrieve p_value s for a number of subsets of a larger data set. So I kind of bumped into a subset where my numeric variables/values are equally 36. Because it is a part of a loop ANOVA is still executed along with reporting an seemingly infinitely small p_value 1.2855e-134--> Correct me if I am wrong but the smaller the p_value the higher the probability that the difference between the factors is significantly different?
For simplicity this is the subset:
sUBSET_FOR_ANOVA
Here is how I calculate ANOVA and retrieve p_value, where TEMP_DF2 is just the subset you see attached:
#
anova_sweep <- aov(TEMP_DF2$GOOD_PTS~TEMP_DF2$MACH,data = TEMP_DF2)
p_value <- summary(anova_sweep)[[1]][["Pr(>F)"]]
p_value <- p_value[1]
#
Many thanks for any guidance,
I can't replicate your findings. Let's produce an example dataset with all values being 36:
df <- data.frame(gr = rep(letters[1:2], 100),
y = 36)
summary(aov(y~gr, data = df))
Gives:
Df Sum Sq Mean Sq F value Pr(>F)
gr 1 1.260e-27 1.262e-27 1 0.319
Residuals 198 2.499e-25 1.262e-27
Basically, depending on the sample size, we obtain a p-value around 0.3 or so. The F statistic is (by definition) always 1, since the between and within group variances are equal.
Are there results misleading? To some extent, yes. The estimated SS and MS should be 0, aov calculates them as very very small. Some other statistical tests in R and in some packages check for zero variance and would produce an error, but aov apparently does not.
However, more importantly, I would say your data is violating the assumptions of the ANOVA and therefore any result cannot be trusted to base conclusion on. The expectation in R when it comes to statistical tests is usually that it is upon the user to employ the tests in the correct circumstances.

Sum of N independent standard normal variables

I wanted to simulate sum of N independent standard normal variables.
sums <- c(1:5000)
for (i in 1:5000) {
sums[i] <- sum(rnorm(5000,0,1))
}
I tried to draw N=5000 standard normal and sum them. Repeat for 5000 simulation paths.
I would expect the expectation of sums be 0, and variance of sums be 5000.
> mean(sums)
[1] 0.4260789
> var(sums)
[1] 5032.494
The simulated expectation is too big. When I tried it again, I got 1.309206 for the mean.
#ilir is correct, the value you get is essentially zero.
If you look at the plot, you get values between -200 and 200. 0.42 is for all intents and purposes 0.
You can test this with t.test.
> t.test(sums, mu = 0)
One Sample t-test
data: sums
t = -1.1869, df = 4999, p-value = 0.2353
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
-3.167856 0.778563
sample estimates:
mean of x
-1.194646
There is no evidence that your mean values differs from zero (given the null hypothesis is true).
This is just plain normal that the mean does not fall exactly on 0, because it is an empirical mean computed from "only" 5000 realizations of the random variable.
However, the distribution of your realizations contained in the sumsvector should "look" Gaussian.
For example, when I try to plot the histogram and the qqplot obtained of 10000 realizations of the sum of 5000 gaussian laws (created in this way: sums <- replicate(1e4,sum(rnorm(5000,0,1)))), it looks normal, as you can see on the following figures:
hist(sums)
qqnorm(sums)
Sum of the independent normals is again normal, with mean the sum of the means and the variance the sum of variance. So sum(rnorm(5000,0,1)) is equivalent to rnorm(1,0,sqrt(5000)). The sample average of normals is again the normal variable. In your case you take a sample average of 5000 independent normal variables with zero mean and variance 5000. This is a normal variable with zero mean and unit variance, i.e. the standard normal.
So in your case mean(sums) is identical to rnorm(1). So any value from interval (-1.96,1.96) will come up 95% of the time.

Spearman correlation and ties

I'm computing Spearman's rho on small sets of paired rankings.
Spearman is well known for not handling ties properly. For example, taking 2 sets of 8 rankings, even if 6 are ties in one of the two sets, the correlation is still very high:
> cor.test(c(1,2,3,4,5,6,7,8), c(0,0,0,0,0,0,7,8), method="spearman")
Spearman's rank correlation rho
S = 19.8439, p-value = 0.0274
sample estimates:
rho
0.7637626
Warning message:
Cannot compute exact p-values with ties
The p-value <.05 seems like a pretty high statistical significance for this data.
Is there a ties-corrected version of Spearman in R?
What is the best formula to date to compute it with a lot of ties?
Well, Kendall tau rank correlation is also a non-parametric test for statistical dependence between two ordinal (or rank-transformed) variables--like Spearman's, but unlike Spearman's, can handle ties.
More specifically, there are three Kendall tau statistics--tau-a, tau-b, and tau-c. tau-b is specifically adapted to handle ties.
The tau-b statistic handles ties (i.e., both members of the pair have the same ordinal value) by a divisor term, which represents the geometric mean between the number of pairs not tied on x and the number not tied on y.
Kendall's tau is not Spearman's--they are not the same, but they are also quite similar. You'll have to decide, based on context, whether the two are similar enough such one can be substituted for the other.
For instance, tau-b:
Kendall_tau_b = (P - Q) / ( (P + Q + Y0)*(P + Q + X0) )^0.5
P: number of concordant pairs ('concordant' means the ranks of each member of the pair of data points agree)
Q: number of discordant pairs
X0: number of pairs not tied on x
Y0: number of pairs not tied on y
There is in fact a variant of Spearman's rho that explicitly accounts for ties. In situations in which i needed a non-parametric rank correlation statistic, i have always chosen tau over rho. The reason is that rho sums the squared errors, whereas tau sums the absolute
discrepancies. Given that both tau and rho are competent statistics and we are left to choose, a linear penalty on discrepancies (tau) has always seemed to me, a more natural way to express rank correlation. That's not a recommendation, your context might be quite different and dictate otherwise.
I think exact=FALSE does the trick.
cor.test(c(1,2,3,4,5,6,7,8), c(0,0,0,0,0,0,7,8), method="spearman", exact=FALSE)
Spearman's rank correlation rho
data: c(1, 2, 3, 4, 5, 6, 7, 8) and c(0, 0, 0, 0, 0, 0, 7, 8)
S = 19.8439, p-value = 0.0274
alternative hypothesis: true rho is not equal to 0
sample estimates:
rho
0.7637626
cor.test with method="spearman" actually calculates Spearman coefficient corrected for ties.
I've checked it by "manually" calculating tie-corrected and tie-uncorrected Spearman coefficients from equations in Zar 1984, Biostatistical Analysis. Here's the code - just substitute your own variable names to check for yourself:
ym <- data.frame(lousy, dors) ## my data
## ranking variables
ym$l <- rank(ym$lousy)
ym$d <- rank(ym$dors)
## calculating squared differences between ranks
ym$d2d <- (ym$l-ym$d)^2
## calculating variables for equations 19.35 and 19.37 in Zar 1984
lice <- as.data.frame(table(ym$lousy))
lice$t <- lice$Freq^3-lice$Freq
dorsal <- as.data.frame(table(ym$dors))
dorsal$t <- dorsal$Freq^3-dorsal$Freq
n <- nrow(ym)
sum.d2 <- sum(ym$d2d)
Tx <- sum(lice$t)/12
Ty <-sum(dorsal$t)/12
## calculating the coefficients
rs1 <- 1 - (6*sum.d2/(n^3-n)) ## "standard" Spearman cor. coeff. (uncorrected for ties) - eq. 19.35
rs2 <- ((n^3-n)/6 - sum.d2 - Tx - Ty)/sqrt(((n^3-n)/6 - 2*Tx)*((n^3-n)/6 - 2*Ty)) ## Spearman cor.coeff. corrected for ties - eq.19.37
##comparing with cor.test function
cor.test(ym$lousy,ym$dors, method="spearman") ## cor.test gives tie-corrected coefficient!
Ties-corrected Spearman
Using method="spearman" gives you the ties-corrected Spearman. Spearman's rho, according to the definition, is simply the Pearson's sample correlation coefficient computed for ranks of sample data. So it works both in presence and in absence of ties.
You can see that after replacing your original data with their ranks (midranks for ties) and using method="pearson", you will get the same result:
> cor.test(rank(c(1,2,3,4,5,6,7,8)), rank(c(0,0,0,0,0,0,7,8)), method="pearson")
Pearson's product-moment correlation
data: rank(c(1, 2, 3, 4, 5, 6, 7, 8)) and rank(c(0, 0, 0, 0, 0, 0, 7, 8))
t = 2.8983, df = 6, p-value = 0.0274
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.1279559 0.9546436
sample estimates:
cor
0.7637626
Notice, there exists a simplified no-ties Spearman version, that is in fact used in cor.test() implementation in absence of ties, but it is equivalent to the definition above.
P-value
In case of ties in data, exact p-values are not computed neither for Spearman nor for Kendall measures (within cor.test() implementation), hence the warning. As mentioned in Eduardo's post, for not to get a warning you should set exact=FALSE,
The paper "A new rank correlation coefficient with application to the consensus ranking problem" is aimed to solve the ranking with tie problem. It also mentions that Tau-b should not be used as a ranking correlation measure for measuring agreement between weak orderings.
Emond, E. J. and Mason, D. W. (2002), A new rank correlation coefficient with application to the consensus ranking problem. J. Multi‐Crit. Decis. Anal., 11: 17-28. doi:10.1002/mcda.313
I was having a similar problem and by reading the answers here and the help file on R I saw that, when you have ties, you have to add the parameter exact = FALSE) to the cor.test() function. By adding this, it does not try to calculate an exact P value, but instead "the test statistic is the estimate scaled to zero mean and unit variance, and is approximately normally distributed".
The result, in my case, was exactly the same, but without the warning about ties.
cor.test(x, y, method = "spearm", exact = FALSE)
The R package ConsRank contains an implementation of Edmon and Mason's Tau_X. This appears to be the (mathematically) best currently known method for handling ties.
See the docs, which give the usage as
Tau_X(X, Y=NULL)
where X can be a matrix.
As pointed out by #wibeasley, Emond and Mason (2002) proposed Tau_X, a new rank correlation coefficient which appears to be superior to Kendal's Tau-b. NelsonGon was concerned that the paper is from 2002, predating the question by a few years, but seems to have overlooked that Spearman's correlation dates from 1904, and Kendall's Tau from 1938.

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