Partial Moments
Why is it necessary to parse the variance with partial moments? The additional information generated from partial moments permits a level of analysis simply not possible with traditional summary statistics.
Below are some basic equivalences demonstrating partial moments role as the elements of variance.
Mean
library(NNS)
set.seed(123) ; x = rnorm(100) ; y = rnorm(100)
mean(x)## [1] 0.09040591UPM(1, 0, x) - LPM(1, 0, x)## [1] 0.09040591Variance
var(x)## [1] 0.8332328# Sample Variance:
UPM(2, mean(x), x) + LPM(2, mean(x), x)## [1] 0.8249005# Population Variance:
(UPM(2, mean(x), x) + LPM(2, mean(x), x)) * (length(x) / (length(x) - 1))## [1] 0.8332328# Variance is also the co-variance of itself:
(Co.LPM(1, x, x, mean(x), mean(x)) + Co.UPM(1, x, x, mean(x), mean(x)) - D.LPM(1, 1, x, x, mean(x), mean(x)) - D.UPM(1, 1, x, x, mean(x), mean(x))) * (length(x) / (length(x) - 1))## [1] 0.8332328Standard Deviation
sd(x)## [1] 0.9128159((UPM(2, mean(x), x) + LPM(2, mean(x), x)) * (length(x) / (length(x) - 1))) ^ .5## [1] 0.9128159Covariance
cov(x, y)## [1] -0.04372107(Co.LPM(1, x, y, mean(x), mean(y)) + Co.UPM(1, x, y, mean(x), mean(y)) - D.LPM(1, 1, x, y, mean(x), mean(y)) - D.UPM(1, 1, x, y, mean(x), mean(y))) * (length(x) / (length(x) - 1))## [1] -0.04372107Covariance Elements and Covariance Matrix
PM.matrix(LPM_degree = 1, UPM_degree = 1,target = 'mean', variable = cbind(x, y), pop_adj = TRUE)## $cupm
## x y
## x 0.4299250 0.1033601
## y 0.1033601 0.5411626
##
## $dupm
## x y
## x 0.0000000 0.1469182
## y 0.1560924 0.0000000
##
## $dlpm
## x y
## x 0.0000000 0.1560924
## y 0.1469182 0.0000000
##
## $clpm
## x y
## x 0.4033078 0.1559295
## y 0.1559295 0.3939005
##
## $cov.matrix
## x y
## x 0.83323283 -0.04372107
## y -0.04372107 0.93506310# Standard Covariance Matrix
cov(cbind(x, y))## x y
## x 0.83323283 -0.04372107
## y -0.04372107 0.93506310Pearson Correlation
cor(x, y)## [1] -0.04953215cov.xy = (Co.LPM(1, x, y, mean(x), mean(y)) + Co.UPM(1, x, y, mean(x), mean(y)) - D.LPM(1, 1, x, y, mean(x), mean(y)) - D.UPM(1, 1, x, y, mean(x), mean(y))) * (length(x) / (length(x) - 1))
sd.x = ((UPM(2, mean(x), x) + LPM(2, mean(x), x)) * (length(x) / (length(x) - 1))) ^ .5
sd.y = ((UPM(2, mean(y), y) + LPM(2, mean(y) , y)) * (length(y) / (length(y) - 1))) ^ .5
cov.xy / (sd.x * sd.y)## [1] -0.04953215CDFs (Discrete and Continuous)
P = ecdf(x)
P(0) ; P(1)
LPM(0, 0, x) ; LPM(0, 1, x)
# Vectorized targets:
LPM(0, c(0, 1), x)
plot(ecdf(x))
points(sort(x), LPM(0, sort(x), x), col = "red")
legend("left", legend = c("ecdf", "LPM.CDF"), fill = c("black", "red"), border = NA, bty = "n")
# Joint CDF:
Co.LPM(0, x, y, 0, 0)
# Vectorized targets:
Co.LPM(0, x, y, c(0, 1), c(0, 1))
# Continuous CDF:
NNS.CDF(x, 1)
# CDF with target:
NNS.CDF(x, 1, target = mean(x))
# Survival Function:
NNS.CDF(x, 1, type = "survival")
PDFs
NNS.PDF(x)
Numerical Integration
Partial moments are asymptotic area approximations of \(f(x)\) akin to the familiar Trapezoidal and Simpson’s rules. More observations, more accuracy…
\[[UPM(1,0,f(x))-LPM(1,0,f(x))]\asymp\frac{[F(b)-F(a)]}{[b-a]}\] \[[UPM(1,0,f(x))-LPM(1,0,f(x))] *[b-a] \asymp[F(b)-F(a)]\]
x = seq(0, 1, .001) ; y = x ^ 2
(UPM(1, 0, y) - LPM(1, 0, y)) * (1 - 0)## [1] 0.3335\[0.3333 * [1-0] = \int_{0}^{1} x^2 dx\] For the total area, not just the definite integral, simply sum the partial moments and multiply by \([b - a]\): \[[UPM(1,0,f(x))+LPM(1,0,f(x))] *[b-a]\asymp\left\lvert{\int_{a}^{b} f(x)dx}\right\rvert\]
Bayes’ Theorem
For example, when ascertaining the probability of an increase in
\(A\) given an increase in \(B\), the
Co.UPM(degree_x, degree_y, x, y, target_x, target_y) target
parameters are set to target_x = 0 and
target_y = 0 and the
UPM(degree, target, variable) target parameter is also set
to target = 0.
\[P(A|B)=\frac{Co.UPM(0,0,A,B,0,0)}{UPM(0,0,B)}\]