Calculate the values of a probability density function for multiple columns of means and a single column of sigma

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Given the sample data sampleDT below, which contains means mean1 to mean10 and the standard deviation sd2, I would appreciate any help to create a function that:

for each column of means mean1 to mean10estimate, extract and add to the data frame the values of the density function for a conditional normal distribution evaluated at the observed level of the variable dollar.wage_1 using standard deviation sd2.

Using the code below I can successfully calculate for a single column of means, but I fail to see how to specify a function to simultaneously calculate for each column of means.

#sample data

sampleDT<-structure(list(id = 1:10, N = c(10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L, 10L), A = c(62L, 96L, 17L, 41L, 212L, 143L, 143L, 
143L, 73L, 73L), B = c(3L, 1L, 0L, 2L, 170L, 21L, 0L, 33L, 62L, 
17L), C = c(0.05, 0.01, 0, 0.05, 0.8, 0.15, 0, 0.23, 0.85, 0.23
), employer = c(1L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 0L, 0L), F = c(0L, 
0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L), G = c(1.94, 1.19, 1.16, 
1.16, 1.13, 1.13, 1.13, 1.13, 1.12, 1.12), H = c(0.14, 0.24, 
0.28, 0.28, 0.21, 0.12, 0.17, 0.07, 0.14, 0.12), dollar.wage_1 = c(1.94, 
1.19, 3.16, 3.16, 1.13, 1.13, 2.13, 1.13, 1.12, 1.12), mean1 = c(1.936652081, 
3.688171386, 3.160993574, 3.768485048, 1.311370546, 0.313760016, 
-1.621000294, 1.13182676, 1.114458025, 1.119315775), mean2 = c(1.946806222, 
3.688885811, 3.15903495, 3.767778705, 1.309663497, 0.316394741, 
-1.618552806, 1.134088181, 1.117600968, 1.120688482), mean3 = c(1.893627954, 
3.689341572, 3.157622975, 3.771231512, 1.324985578, 0.318026311, 
-1.620565712, 1.13301769, 1.120760085, 1.119426932), mean4 = c(1.887509366, 
3.660243949, 3.160911994, 3.738992465, 1.331637143, 0.284716279, 
-1.655368774, 1.137338962, 1.122096234, 1.120837428), mean5 = c(7.071170501, 
3.458558276, 3.156676637, 3.160692822, 1.131841192, 1.126997224, 
1.028924299, 1.219378155, 0.118097115, 1.118108075), mean6 = c(7.010141264, 
3.434098438, 3.160978044, 3.161388054, 1.131706507, 1.131073576, 
1.044957033, 1.202376831, 0.088502176, 1.120101488), mean7 = c(6.918631396, 
3.455412441, 3.064840549, 3.158657611, 1.134281965, 1.131677907, 
1.035688483, 1.181551066, 0.542276222, 1.121549931), mean8 = c(6.980214117, 
3.513440689, 3.175191087, 3.158919334, 1.130088008, 1.131692248, 
1.12222788, 1.235102249, 0.281700405, 1.118473791), mean9 = c(6.708505027, 
3.504542699, 3.173629275, 3.158457814, 1.134560107, 1.129357587, 
1.151489857, 1.219991269, 0.364343124, 1.120228667), mean10 = c(6.883206883, 
3.467216323, 3.174805298, 3.160917024, 1.128835398, 1.128265912, 
1.084046983, 1.214981489, 0.160046133, 1.118496504), sd1 = c(2.6334129999306, 
2.6334129999306, 2.6334129999306, 2.6334129999306, 2.6334129999306, 
2.6334129999306, 2.6334129999306, 2.6334129999306, 2.6334129999306, 
2.6334129999306), sd2 = c(514.02608349227, 101.976862386691, 
8.70627514696715, 4.79710442214283, 2.45930925299156e+49, 2.01406038865916e+30, 
1.8980055884822e+34, 1.65244344266379e+28, 26.9398910547703, 
1.74978644797635)), row.names = c(NA, -10L), spec = structure(list(
    cols = list(id = structure(list(), class = c("collector_integer", 
    "collector")), N = structure(list(), class = c("collector_integer", 
    "collector")), A = structure(list(), class = c("collector_integer", 
    "collector")), B = structure(list(), class = c("collector_integer", 
    "collector")), C = structure(list(), class = c("collector_double", 
    "collector")), employer = structure(list(), class = c("collector_integer", 
    "collector")), F = structure(list(), class = c("collector_integer", 
    "collector")), G = structure(list(), class = c("collector_double", 
    "collector")), H = structure(list(), class = c("collector_double", 
    "collector")), dollar.wage_1 = structure(list(), class = c("collector_double", 
    "collector")), mean1 = structure(list(), class = c("collector_double", 
    "collector")), mean2 = structure(list(), class = c("collector_double", 
    "collector")), mean3 = structure(list(), class = c("collector_double", 
    "collector")), mean4 = structure(list(), class = c("collector_double", 
    "collector")), mean5 = structure(list(), class = c("collector_double", 
    "collector")), mean6 = structure(list(), class = c("collector_double", 
    "collector")), mean7 = structure(list(), class = c("collector_double", 
    "collector")), mean8 = structure(list(), class = c("collector_double", 
    "collector")), mean9 = structure(list(), class = c("collector_double", 
    "collector")), mean10 = structure(list(), class = c("collector_double", 
    "collector"))), default = structure(list(), class = c("collector_guess", 
    "collector"))), class = "col_spec"), class = c("tbl_df", 
"tbl", "data.frame"))

#my approach

sampleDT$dens_test <- dnorm(sampleDT$dollar.wage_1, 
                      mean = sampleDT$mean1,sd = sampleDT$sd2)

Thanks in advance for any help.

It's a typical case of dplyr::mutate_at():

df %>% mutate_at(vars(matches("mean")),
                 funs(dens = dnorm(dollar.wage_1, mean = ., sd = sd2)))

The output will be a complete dataset, and you don't need to bind anything.

Data Transforms, As of SigmaPlot Version 10, the user-defined transforms can be stored as an item in a makeblock, The makeblock function converts a single column to a block of cells of defined It can be used to calculate the average across rows for worksheet columns. normden, The normal (or Gaussian) probability density function. This MATLAB function returns an n-by-1 vector y containing the probability density function (pdf) of the d-dimensional multivariate normal distribution with zero mean and identity covariance matrix, evaluated at each row of the n-by-d matrix X.

Or put into a sapply().

sapply(grep("^mean", names(sampleDT)), function(x) 
  dnorm(sampleDT$dollar.wage_1, sampleDT[[x]], sampleDT$sd2))

Mean (expected value) of a discrete random variable (video), Finding the mean (or expected value) of a discrete random variable. Constructing a Duration: 4:32 Posted: Jul 15, 2017 Parametric probability density estimation involves selecting a common distribution and estimating the parameters for the density function from a data sample. Nonparametric probability density estimation involves using a technique to fit a model to the arbitrary distribution of the data, like kernel density estimation.

We may go further with your approach:

means <- as.matrix(sampleDT[, grep("mean", names(sampleDT))])
dnorm(sampleDT$dollar.wage_1, mean = means, sd = sampleDT$sd2)

In this way we pass a matrix of means, while dollar.wage_1 and sd2 also get used correctly due to recycling.

Then you may add this result to sampleDT simply with cbind.

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this is my approach using data.table:

sampleDT <- 

for(i in c(1:10)){   
 sampleDT[, eval(paste0("dnorm",i)):=mapply(dnorm,dollar.wage_1,get(paste0("mean",i)),sd2)] }

Lesson 4: Multivariate Normal Distribution, Understand the definition of the multivariate normal distribution;; Compute eigenvalues and function is a monotone function, the normal density takes a maximum value when x is equal to μ μ . Each single variable has a univariate normal distribution. Joint Probability Density Function for Bivariate Normal Distribution� Normal distribution returns for a specified mean and standard deviation. It is a built-in function for finding mean and standard deviation for a set of values in excel. To find the mean value average function is being used. The normal distribution will calculate the normal probability density function or the cumulative normal distribution function.

Probability density function, In probability theory, a probability density function (PDF), or density of a continuous random (This definition may be extended to any probability distribution using the Densities associated with multiple variables[edit] that in order to find the expected value E(g(X)), one must first find the probability density fg(X) of the new� pdf values, evaluated at the values in x, returned as a scalar value or an array of scalar values. y is the same size as x , mu , and sigma after any necessary scalar expansion. Each element in y is the pdf value of the distribution specified by the corresponding elements in mu and sigma , evaluated at the corresponding element in x .

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