# 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

set.seed(123) ; x = rnorm(100) ; y = rnorm(100)

mean(x)
##  0.09040591
UPM(1, 0, x) - LPM(1, 0, x)
##  0.09040591

## Variance

var(x)
##  0.8332328
# Sample Variance:
UPM(2, mean(x), x) + LPM(2, mean(x), x)
##  0.8249005
# Population Variance:
(UPM(2, mean(x), x) + LPM(2, mean(x), x)) * (length(x) / (length(x) - 1))
##  0.8332328
# Variance is also the co-variance of itself:
(Co.LPM(1, 1, x, x, mean(x), mean(x)) + Co.UPM(1, 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))
##  0.8332328

## Standard Deviation

sd(x)
##  0.9128159
((UPM(2, mean(x), x) + LPM(2, mean(x), x)) * (length(x) / (length(x) - 1))) ^ .5
##  0.9128159

## Covariance

cov(x, y)
##  -0.04372107
(Co.LPM(1, 1, x, y, mean(x), mean(y)) + Co.UPM(1, 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))
##  -0.04372107

## Covariance Elements and Covariance Matrix

PM.matrix(LPM.degree = 1, UPM.degree = 1,target = 'mean', variable = cbind(x, y), pop.adj = TRUE)
## $clpm ## x y ## x 0.4033078 0.1559295 ## y 0.1559295 0.3939005 ## ##$cupm
##           x         y
## x 0.4299250 0.1033601
## y 0.1033601 0.5411626
##
## $dlpm ## x y ## x 0.0000000 0.1469182 ## y 0.1560924 0.0000000 ## ##$dupm
##           x         y
## x 0.0000000 0.1560924
## y 0.1469182 0.0000000
##
## \$matrix
##             x           y
## x  0.83323283 -0.04372107
## y -0.04372107  0.93506310

## Pearson Correlation

cor(x, y)
##  -0.04953215
cov.xy = (Co.LPM(1, 1, x, y, mean(x), mean(y)) + Co.UPM(1, 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)
##  -0.04953215

## CDFs (Discrete and Continuous)

P = ecdf(x)
P(0) ; P(1)
##  0.48
##  0.83
LPM(0, 0, x) ; LPM(0, 1, x)
##  0.48
##  0.83
# Vectorized targets:
LPM(0, c(0, 1), x)
##  0.48 0.83
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, 0, x, y, 0, 0)
##  0.28
# Vectorized targets:
Co.LPM(0, 0, x, y, c(0, 1), c(0, 1))
##  0.28 0.73
# Continuous CDF:
plot(sort(x), LPM.ratio(1, sort(x), x), type = "l", col = "blue", lwd = 3, xlab = "x") ## PDFs

NNS.PDF(degree = 1, x) ##      Intervals          PDF
##   1: -2.309169 0.0001908930
##   2: -2.264204 0.0003890644
##   3: -2.219239 0.0004040456
##   4: -2.174274 0.0004199092
##   5: -2.129309 0.0004367259
##  ---
##  97:  2.007473 0.0014780900
##  98:  2.052438 0.0011915263
##  99:  2.097403 0.0009296535
## 100:  2.142368 0.0008033076
## 101:  2.187333 0.0003484623

## Numerical Integration

Partial moments are asymptotic area approximations of $$f(x)$$ akin to the familiar Trapezoidal and Simpson’s rules. More observations, more accuracy…

$[UPM(1,0,f(x))-LPM(1,0,f(x))]\asymp\frac{[F(b)-F(a)]}{[b-a]}$

x = seq(0, 1, .001) ; y = x ^ 2
UPM(1, 0, y) - LPM(1, 0, y)
##  0.3335

$0.3333=\frac{\int_{0}^{1} x^2 dx}{1-0}$ For the total area, not just the definite integral, simply sum the partial moments: $[UPM(1,0,f(x))+LPM(1,0,f(x))]\asymp\left\lvert{\int_{a}^{b} f(x)dx}\right\rvert$

## 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.

$P(A|B)=\frac{Co.UPM(0,0,A,B,0,0)}{UPM(0,0,B)}$