---
title: "Probability of response vectors in conquestr"
author: "Dan Cloney"
date: "`r Sys.Date()`"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{Probability of response vectors in conquestr}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

```{r setup, include = FALSE}
knitr::opts_chunk$set(
  collapse = TRUE,
  comment = "#>"
)
```

# Introduction

In some cases users might like to return the probability of a response on a given item.
For example, given a fixed set of item parameters, return the probabilities at varying levels of theta to produce custom probability plots. 

# Example

## dichotomous case

The probability of responding correctly to a dichotomous item under Rasch-like models (e.g., 1PL models) is often expressed as:

\begin{equation} 
  p(x_{ni} = 1)=\frac{exp(\theta_{n} - \delta_{i})}{1 + (\theta_{n} - \delta_{i})}
  (\#eq:slm)
\end{equation} 

Imagine the item parameters of a single item represented as:

```{r slmItem}
library(conquestr)
myItem <- matrix(c(0, 0, 0, 1, 1, 0), ncol =3, byrow=TRUE)
colnames(myItem)<- c("k", "d", "t")
print(myItem)
```

Then the probability of scoring 0 and 1 on this item, at \theta = 0.5:

```{r slmProbs}
myProbs <- simplep(0.5, myItem)
print(myProbs)
```

A simple ICC can be drawn:

```{r slmICC}
myProbsList <- list()
myThetaRange <- seq(-4, 4, by = 0.1)
for (i in seq(myThetaRange)) {
  myProbsList[[i]] <- pX(x = 1, probs = simplep(myThetaRange[i], myItem))
}
plot(unlist(myProbsList))
```

## polytomous case

In the case of polytomously scored items, the probability model can be generalised:

\begin{equation} 
  p(X_{ni} = x)=\frac{exp\sum\limits_{k=0}^{x}(\theta_{n} - (\delta_{i} + \tau_{ik}))}{\sum\limits_{j=0}^{m}exp(\sum\limits_{k=0}^{j} (\theta_{n} - (\delta_{i} + \tau_{ik})))}
  (\#eq:pcm)
\end{equation} 

An item can them be represented such that:

```{r polyItem}
library(conquestr)
myItem <- matrix(c(0, 0, 0, 1, 1, -0.2, 2, 1, 0.2), ncol =3, byrow=TRUE)
colnames(myItem)<- c("k", "d", "t")
print(myItem)
```

Then the probability of scoring 0, 1 and 2 on this item, at \theta = 0.5:

```{r polyProbs}
myProbs <- simplep(0.5, myItem)
print(myProbs)
```

A simple ICC can be drawn:

```{r polyICC}
myProbsList <- list()
myThetaRange <- seq(-4, 4, by = 0.1)
for (i in seq(myThetaRange)) {
  myProbsList[[i]] <- simplep(myThetaRange[i], myItem)
}
myProbs <- (matrix(unlist(myProbsList), ncol = 3, byrow = TRUE))
plot(myThetaRange, myProbs[,1])
points(myThetaRange, myProbs[,2])
points(myThetaRange, myProbs[,3])
abline(v = c(myItem[2, 2], sum(myItem[2, 2:3]), sum(myItem[3, 2:3])))
```

## Expected scores 

The expected score for the an item can be calculated at a given value of theta. 
Taking an aribitary set of items, it is possible therefor to calculate the test expected score.

```{r Expected}
library(conquestr)
myItems <- list()
myItems[[1]] <- matrix(c(0, 0, 0, 1, 1, -0.2, 2, 1, 0.2), ncol =3, byrow=TRUE)
myItems[[2]] <- matrix(c(0, 0, 0, 1, -1, -0.4, 2, -1, 0.4), ncol =3, byrow=TRUE)
myItems[[3]] <- matrix(c(0, 0, 0, 1, 1.25, -0.6, 2, 1.25, 0.6), ncol =3, byrow=TRUE)
myItems[[4]] <- matrix(c(0, 0, 0, 1, 2, 0.2, 2, 2, -0.2), ncol =3, byrow=TRUE)
myItems[[5]] <- matrix(c(0, 0, 0, 1, -2.5, -0.2, 2, -2.5, 0.2), ncol =3, byrow=TRUE)
for (i in seq(myItems)) {
  colnames(myItems[[i]])<- c("k", "d", "t")
}
print(myItems)

expectedRes <- list()
for (i in seq(myThetaRange)) {
  tmpExp <- 0
  for (j in seq(myItems)) {
    tmpE <- simplef(myThetaRange[i], myItems[[j]])
    tmpExp <- tmpExp + tmpE
  }
  expectedRes[[i]] <- tmpExp
}

plot(myThetaRange, unlist(expectedRes))

```






```{r working, include=FALSE}

simplep(0.5, myItem)

myProbs <- simplep(0.5, myItem)

pX(2, simplep(0.5, myItem))

tTheta <- 0.5
p0tmp <- exp((0*tTheta) - (0)) # by def. this = 1
p1tmp <- exp((1*tTheta) - (0 + 0.8))
p2tmp <- exp((2*tTheta) - (0 + 0.8 + 1.2))
p_denom <- sum(p0tmp, p1tmp, p2tmp)
p_of_0 <- p0tmp/p_denom
p_of_1 <- p1tmp/p_denom
p_of_2 <- p2tmp/p_denom


myItem1 <- matrix(c(0, 0, 0, 1, 1, 0), ncol =3, byrow=TRUE)
simplep(0.5, myItem1)

exp(0.5-1)/(1+exp(0.5-1))
sum(exp(2*0.5-(0.8+1.2)))/sum(1, exp(0.5-0.8), exp(2*0.5-(0.8+1.2)))

```
    
