Standard normal integral function in R

Question

I need to calculate the division of integrals, where the function q_alpha (z) is a quantile function of the standard normal distribution.

I have a question regarding the denominator. Since the normal standard distribution has homosedasticity, it is symmetric, continuous, etc. Is integrating the denominator word simple? Do I just need to square every quantile of this function and go to the calculation? Correctly?

This is my code in R:

library(Bolstad)

thau=1:99/100
z.standard.quantile=qnorm(thau,0,1)
z.standard.quantile.square=qnorm(thau,0,1)^2

sintegral(thau[1:50],z.standard.quantile[1:50])$value/sintegral(thau[1:50], z.standard.quantile.square[1:50])$value

Result: -0.8676396

Answer 1

qnorm, qnorm [0, 0.5] (note qnorm(0) is -Inf), .

, Bolstad::sintegral ( ); R integrate . qnorm , qnorm - , .

, , l:

## using `integrate`
f1 <- function (l) {
  a <- integrate(qnorm, lower = l, upper = 0.5)$value
  b <- integrate(function (x) qnorm(x) ^ 2, lower = l, upper = 0.5)$value
  a / b
  }

## using Trapezoidal rule, with `n` division on interval `[l, 0.5]`
f2 <- function (l, n) {
  x <- seq(l, 0.5, length = n)
  delta <- x[2] - x[1]
  y1 <- qnorm(x)
  y2 <- y1 ^ 2
  a <- sum(y1[-1] + y1[-n]) / 2 * delta
  b <- sum(y2[-1] + y2[-n]) / 2 * delta
  a / b
  }

, :

f1 (0.1)
# [1] -1.276167

f2 (0.1, 1000)
# [1] -1.276166

, , - , l -> 0 ( ).

l <- 10 ^ (- (1:16))
# [1] 1e-01 1e-02 1e-03 1e-04 1e-05 1e-06 1e-07 1e-08 1e-09 1e-10 1e-11 1e-12
# [13] 1e-13 1e-14 1e-15 1e-16

y1 <- sapply(l, f1)
# [1] -1.2761674 -0.8698411 -0.8096179 -0.7996069 -0.7981338 -0.7979341
# [7] -0.7978877 -0.7978848 -0.7978846 -0.7978846 -0.7978846 -0.7978846
# [13] -0.7978846 -0.7978846 -0.7978846 -0.7978846

## quite a dense grid; takes some time to compute
y2 <- sapply(l, f2, n = 1e+6)
# [1] -1.2761674 -0.8698411 -0.8096179 -0.7996071 -0.7981158 -0.7979137
# [7] -0.7978877 -0.7978834 -0.7978816 -0.7978799 -0.7978783 -0.7978767
# [13] -0.7978750 -0.7978734 -0.7978717 -0.7978700

, , -0.7978 l -> 0.

---

, -0.8676396 f1(0.01) f2(0.01, 1e+6).