Building a function by defining x and y using R

Question

I have this matrix:

matrix=structure(c(0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 
0.1, 0.11, 0.12, 0.13, 0.14, 0.15, 0.16, 0.17, 0.18, 0.19, 0.2, 
0.21, 0.22, 0.23, 0.24, 0.25, 0.26, 0.27, 0.28, 0.29, 0.3, 0.31, 
0.32, 0.33, 0.34, 0.35, 0.36, 0.37, 0.38, 0.39, 0.4, 0.41, 0.42, 
0.43, 0.44, 0.45, 0.46, 0.47, 0.48, 0.49, 0.5, 0.51, 0.52, 0.53, 
0.54, 0.55, 0.56, 0.57, 0.58, 0.59, 0.6, 0.61, 0.62, 0.63, 0.64, 
0.65, 0.66, 0.67, 0.68, 0.69, 0.7, 0.71, 0.72, 0.73, 0.74, 0.75, 
0.76, 0.77, 0.78, 0.79, 0.8, 0.81, 0.82, 0.83, 0.84, 0.85, 0.86, 
0.87, 0.88, 0.89, 0.9, 0.91, 0.92, 0.93, 0.94, 0.95, 0.96, 0.97, 
0.98, 0.99, -7.38512004893287, -7.38512004893287, -6.4788834441613, 
-5.63088940915783, -4.83466644123448, -4.68738146949482, -4.28638930290018, 
-4.22411786604579, -3.59136848943044, -3.51706359680799, -3.39972014575003, 
-3.28609348968074, -3.08569873266253, -2.99764447889508, -2.89470597729108, 
-2.77488515429677, -2.67019029728821, -2.54646363628509, -2.48474483938047, 
-2.30542896070156, -2.22485510301423, -2.16689229344011, -2.10316315192181, 
-2.05135466960309, -1.90942757945567, -1.87863626704201, -1.82507998490407, 
-1.75875817642096, -1.6919717645629, -1.62396997031953, -1.56159595204983, 
-1.52152738173419, -1.46478394989911, -1.4590555309334, -1.21744398902807, 
-1.21731951113139, -1.15003007559406, -1.07321513324935, -0.993364510081357, 
-0.924402354306976, -0.885939210442384, -0.831155619244629, -0.80947326709303, 
-0.786842719842383, -0.743834513319968, -0.721194178931262, -0.593033922802471, 
-0.514780082129033, -0.50717184901095, -0.44223827942003, -0.403514759789576, 
-0.296251921664, -0.204238424399985, -0.1463212643028, -0.0982036017275267, 
-0.0705262020944892, 0.0275436976821241, 0.0601977432996216, 
0.114959963559268, 0.182222546319913, 0.236503724954577, 0.272244043950984, 
0.325188234828891, 0.347862804414816, 0.438932719815686, 0.630570414177834, 
0.805087251137292, 0.904903847087405, 0.940702374334727, 0.958351604371838, 
1.03920208406121, 1.25808734990267, 1.32634708210007, 1.34458194173569, 
1.42693337001189, 1.55016591141652, 1.5710754638668, 1.61795101580197, 
1.62472416407376, 1.70223430572367, 1.86164374636379, 1.94317125269006, 
2.03941620499986, 2.12071850455654, 2.17753890907921, 2.22227616630581, 
2.45586794615095, 2.66160802425205, 2.83084956697756, 2.94669126521054, 
3.04536994227142, 3.09217816201639, 3.42405058020625, 3.45140184734503, 
3.67343579954061, 4.64233570345934, 4.87075743677502, 5.27924539262207, 
5.56822483595709), .Dim = c(99L, 2L), .Dimnames = list(NULL, 
    c("x", "y")))

It turns out that column 1 is the domain of the function and column 2 is axis y.

plot(matrix[,1],matrix[,2])

How do I create a function that has the first column as the domain of the function and the second column as the contradomain, so that I can calculate, for example, the integral in a given interval, for example the sum of integrals between (0.08 and 0.15) and (0.40 and 0.46).

My idea is to be able, for example, to run the code of the integral:

integrando= function(x) return(minhafuncao(x)); 

integrate(integrando, lower=0.08, upper=0.15)

2nd Question: (I edited later)

This function that I have established in plot(matrix[,1],matrix[,2])

It is the quantile function: q_{theta}(x)

To get the function q_{1-theta}(x) it is sufficient to reverse the column of amounts in the decreasing format: sort(matrix[,2],decreasing=TRUE)?

The way I’m thinking is this: when alpha equals 0.15 the function q_{1-theta}(x) give me the quantile q_{0.85}(x) of function q_{theta}(x). Right? Also, the functions q_{theta}(x) and q_{1-theta}(x) would intercept in the middle?

Then when calculating an integral between 0.01 and 0.50 of the type:

Integral q_{theta}(x) - q_{1-theta}(x)

I may be proceeding this way?

sintegral(thau[1:50], (matrix[,2][1:50] - sort(matrix[,2],TRUE)[1:50])[1:50])$value



Obrigado

Answer 1

This is what you want to do Riemann sum. I would use an idea based on Simpson’s formula to obtain this result:

library(Bolstad)
integral <- sintegral(matrix[, 1], matrix[, 2])
integral$value
[1] -0.3841932

To obtain the integral of specific intervals, just choose the correct lines of the matrix with your intervals to be integrated:

integral1 <- sintegral(matrix[ 8:15, 1], matrix[ 8:15, 2])
integral2 <- sintegral(matrix[40:46, 1], matrix[40:46, 2])
integral1$value + integral2$value
[1] -0.2831704

---

Yes, just reverse it the sign (and not the order) of the second column of your matrix to have the values of qnorm(1-x). The normal distribution is symmetric in relation to the mean and median. Therefore,

qnorm(x) = -qnorm(1-x) (1)

Maybe the chart below will make it clearer:

curve(qnorm(x), from=0, to=1)
curve(qnorm(1-x), from=0, to=1, add=TRUE, col="red")

Note that the two curves are 0.5, which is equivalent to the median (and mean and fashion, in the case of normal distribution).

The value of the integral of qnorm(x)-qnorm(1-x) is the value of the 2*qnorm(x), for

qnorm(x)-qnorm(1-x) = qnorm(x) - (-qnorm(x)) (by (1))

= qnorm(x) - (-qnorm(x))

= qnorm(x) + qnorm(x)

= 2*qnorm(x)

I see no need to complicate this calculation unnecessarily. Just calculate the value of the integral already calculated and double it.