# Brief description

Nowadays many people have no time, so, it is for such busy person.

The author entrust complement of the logic to reader.

# $$\color{green}{\textit{Single reader and Single modality }}$$

##### Data
Confidence Level No. of Hits No. of False alarms
5 = definitely present $$H_{5}$$ $$F_{5}$$
4 = probably present $$H_{4}$$ $$F_{4}$$
3 = equivocal $$H_{3}$$ $$F_{3}$$
2 = probably absent $$H_{2}$$ $$F_{2}$$
1 = questionable $$H_{1}$$ $$F_{1}$$

Note that $$H_{c},F_c \in \mathbb{N} \cup\{0\}$$ for $$c=1,2,...,5$$.

##### Model

$\begin{eqnarray*} H_{c } & \sim &\text{Binomial} ( p_{c}, N_{L} ), \text{ for c=1,2,...,C.}\\ F_{c } & \sim &\text{Poisson}( (\lambda _{c} -\lambda _{c+1} )\times N_{I} ), \text{ for c=1,2,...,C-1.}\\ \lambda _{c}& =& - \log \Phi ( z_{c } ),\text{ for c=1,2,...,C.}\\ p_{c} &=&\Phi (\frac{z_{c +1}-\mu}{\sigma})-\Phi (\frac{z_{c}-\mu}{\sigma}), \text{ for c=1,2,...,C-1.}\\ p_C & =& 1-\Phi (\frac{z_{C}-\mu}{\sigma}),\\ F_{C} & \sim & \text{Poisson}( (\lambda _{C} - 0)N_I),\\ dz_c=z_{c+1}-z_{c} &\sim& \text{Uniform}(0,\infty), \text{ for c=1,2,...,C-1.}\\ \mu &\sim& \text{Uniform}(-\infty,\infty),\\ \sigma &\sim& \text{Uniform}(0,\infty),\\ \end{eqnarray*}$ Our model has parameters $$z_{1}, dz_1,dz_2,\cdots, dz_{C-1}$$, $$\mu$$, and $$\sigma$$. Notation $$\text{Uniform}( -\infty,100000)$$ means the improper uniform distribution of its support is the unbounded interval $$( -\infty,100000)$$.

Or equivalently, if $$C=5$$

$\begin{eqnarray*} H_{1 } & \sim &\text{Binomial} ( p_{1}, N_{L} ) \\ H_{2 } & \sim &\text{Binomial} ( p_{2}, N_{L} ) \\ H_{3 } & \sim &\text{Binomial} ( p_{3}, N_{L} ) \\ H_{4 } & \sim &\text{Binomial} ( p_{4}, N_{L} ) \\ H_{5 } & \sim &\text{Binomial} ( p_{4}, N_{L} ) \\ F_{1 } & \sim &\text{Poisson}( (\lambda _{1} -\lambda _{2} )\times N_{I} ), \\ F_{2 } & \sim &\text{Poisson}( (\lambda _{2} -\lambda _{3} )\times N_{I} ), \\ F_{3 } & \sim &\text{Poisson}( (\lambda _{3} -\lambda _{4} )\times N_{I} ), \\ F_{4 } & \sim &\text{Poisson}( (\lambda _{4} -\lambda _{5} )\times N_{I} ), \\ F_{5 } & \sim &\text{Poisson}( (\lambda _{5} -0 )\times N_{I} ), \text{Be careful !!:'-D}\\ \lambda _{1}& =& - \log \Phi ( z_{1 } ),\\ \lambda _{2}& =& - \log \Phi ( z_{2 } ),\\ \lambda _{3}& =& - \log \Phi ( z_{3 } ),\\ \lambda _{4}& =& - \log \Phi ( z_{4 } ),\\ \lambda _{5}& =& - \log \Phi ( z_{5 } ),\\ p_{1} &:=&\Phi (\frac{z_{2}-\mu}{\sigma})-\Phi (\frac{z_{1}-\mu}{\sigma}), \\ p_{2} &:=&\Phi (\frac{z_{3}-\mu}{\sigma})-\Phi (\frac{z_{2}-\mu}{\sigma}), \\ p_{3} &:=&\Phi (\frac{z_{c4}-\mu}{\sigma})-\Phi (\frac{z_{3}-\mu}{\sigma}), \\ p_{4} &:=&\Phi (\frac{z_{5}-\mu}{\sigma})-\Phi (\frac{z_{4}-\mu}{\sigma}), \\ p_5 &:=& 1-\Phi (\frac{z_{5}-\mu}{\sigma}),\text{Be careful !!:'-D}\\ dz_c=z_{2}-z_{1} &\sim& \text{Uniform}(0,\infty), \\ dz_c=z_{3}-z_{2} &\sim& \text{Uniform}(0,\infty), \\ dz_c=z_{4}-z_{3} &\sim& \text{Uniform}(0,\infty), \\ dz_c=z_{5}-z_{4} &\sim& \text{Uniform}(0,\infty), \\ \mu &\sim& \text{Uniform}(-\infty,\infty),\\ \sigma &\sim& \text{Uniform}(0,\infty),\\ \end{eqnarray*}$ Our model has parameters $$z_{1}, dz_1,dz_2,\cdots, dz_{C-1}$$, $$\mu$$, and $$\sigma$$. Notation $$\text{Uniform}( -\infty,100000)$$ means the improper uniform distribution of its support is the unbounded interval $$( -\infty,100000)$$.

##### R script for the above model
d   <- BayesianFROC::d
fit <- fit_Bayesian_FROC(  dataList = d )

# $$\color{green}{\textit{Multiple reader and Multiple case}{}^{\dagger} }$$

$${}^{\dagger}$$ traditionally, case means modality in this context.

#### Example Data format

Two readers and two modalities and three kind of confidence levels.

Confidence Level Modality ID Reader ID Number of Hits Number of False alarms
3 = definitely present 1 1 $$H_{3,1,1}$$ $$F_{3,1,1}$$
2 = equivocal 1 1 $$H_{2,1,1}$$ $$F_{2,1,1}$$
1 = questionable 1 1 $$H_{1,1,1}$$ $$F_{1,1,1}$$
3 = definitely present 1 2 $$H_{3,1,2}$$ $$F_{3,1,2}$$
2 = equivocal 1 2 $$H_{2,1,2}$$ $$F_{2,1,2}$$
1 = questionable 1 2 $$H_{1,1,2}$$ $$F_{1,1,2}$$
3 = definitely present 2 1 $$H_{3,2,1}$$ $$F_{3,2,1}$$
2 = equivocal 2 1 $$H_{2,2,1}$$ $$F_{2,2,1}$$
1 = questionable 2 1 $$H_{1,2,1}$$ $$F_{1,2,1}$$
3 = definitely present 2 2 $$H_{3,2,2}$$ $$F_{3,2,2}$$
2 = equivocal 2 2 $$H_{2,2,2}$$ $$F_{2,2,2}$$
1 = questionable 2 2 $$H_{1,2,2}$$ $$F_{1,2,2}$$

#### Model

$\begin{eqnarray*} H_{c,m,r} & \sim &\text{Binomial }( p_{c,m,r}, N_L ),\\ F_{c,m,r} &\sim& \text{Poisson }( ( \lambda _{c} - \lambda _{c+1})N_L ),\\ \lambda _{c}& =& - \log \Phi (z_{c }),\\ p_{c,m,r} &:=&\Phi (\frac{z_{c +1}-\mu_{m,r}}{\sigma_{m,r}})-\Phi (\frac{z_{c}-\mu_{m,r}}{\sigma_{m,r}}), \\ p_C & =& 1-\Phi (\frac{z_{C}-\mu_{m,r}}{\sigma_{m,r}}),\\ F_{C,m,r} & \sim &\text{Poisson } ( (\lambda _{C} - 0)N_I),\\ A_{m,r}&:=&\Phi (\frac{\mu_{m,r}/\sigma_{m,r}}{\sqrt{(1/\sigma_{m,r})^2+1}}), \\ A_{m,r}&\sim&\text{Normal} (A_{m},\sigma_{r}^2), \\ dz_c&:=&z_{c+1}-z_{c},\\ dz_c, \sigma_{m,r} &\sim& \text{Uniform}(0,\infty),\\ z_{c} &\sim& \text{Uniform}( -\infty,100000),\\ A_{m} &\sim& \text{Uniform}(0,1).\\ \end{eqnarray*}$ Our new model has parameters $$z_{1}, dz_1,dz_2,\cdots, dz_{C}$$, $$A_{m}$$, $$\sigma_{r}$$, $$\mu_{m,r}$$, and $$\sigma_{m,r}$$.

##### R script for the above model
dd  < -BayesianFROC::dd
fit <- fit_Bayesian_FROC(  dataList = dd )