Mean

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

mean(x)

## [1] 0.09040591

UPM(1, 0, x) - LPM(1, 0, x)

## [1] 0.09040591

Variance

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] 0.8332328

Standard Deviation

sd(x)

## [1] 0.9128159

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

## [1] 0.9128159

First 4 Moments

The first 4 moments are returned with the function NNS.moments. For sample statistics, set population = FALSE.

NNS.moments(x)

## $mean
## [1] 0.09040591
## 
## $variance
## [1] 0.8332328
## 
## $skewness
## [1] 0.06049948
## 
## $kurtosis
## [1] -0.161053

NNS.moments(x, population = FALSE)

## $mean
## [1] 0.09040591
## 
## $variance
## [1] 0.8249005
## 
## $skewness
## [1] 0.06235774
## 
## $kurtosis
## [1] -0.1069186

Statistical Mode of a Continuous Distribution

NNS.mode offers support for discrete valued distributions as well as recognizing multiple modes.

# Continuous
NNS.mode(x)

## [1] -0.2365625

# Discrete and multiple modes
NNS.mode(c(1, 2, 2, 3, 3, 4, 4, 5), discrete = TRUE, multi = TRUE)

## [1] 2 3 4

Covariance

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] -0.04372107

Covariance Elements and Covariance Matrix

The covariance matrix (Σ) is equal to the sum of the co-partial moments matrices less the divergent partial moments matrices. Σ = CLPM + CUPM − DLPM − DUPM

cov.mtx = PM.matrix(LPM_degree = 1, UPM_degree = 1,target = 'mean', variable = cbind(x, y), pop_adj = TRUE)
cov.mtx

## $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

# Reassembled Covariance Matrix
cov.mtx$clpm + cov.mtx$cupm - cov.mtx$dlpm - cov.mtx$dupm

##             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  0.93506310

Pearson Correlation

cor(x, y)

## [1] -0.04953215

cov.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] -0.04953215

CDFs (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))

# Copula
# Transform x and y so that they are uniform
u_x = LPM.ratio(0, x, x)
u_y = LPM.ratio(0, y, y)

# Value of copula at c(.5, .5)
Co.LPM(0, u_x, u_y, .5, .5)

# Continuous CDF:
NNS.CDF(x, 1)

# CDF with target:
NNS.CDF(x, 1, target = mean(x))

# Survival Function:
NNS.CDF(x, 1, type = "survival")

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] ≍ [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] = ∫₀¹x²dx 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] ≍ |∫_a^bf(x)dx|

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.

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