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

\(\color{green}{\textit{Multiple reader and Multiple case}{}^{\dagger} }\)

\({}^{\dagger}\) 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 \(\sigma_{m,r}\).