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

normal distribution

multivariate normal distribution

multivariate normal distribution - matlab

bivariate normal distribution

mvnrnd matlab

how to find the standard deviation of a probability distribution

properties of multivariate normal distribution

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`mean10`

estimate, 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 <- as.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.

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