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Table 3 Simulation study 1, scenario 1 with nX=30, nY=10 and 100 runs

From: Identification of differentially expressed genes by means of outlier detection

   

Δ=1.5

 

Nb. of

  

α

 

DE genes

 

0.1

0.05

0.01

50

\(\hat {p}_{m}\)

0.85

1

1

 

%

97.2 (2.8)

91.1 (2.1)

69.6 (6.7)

100

\(\hat {p}_{m}\)

1

1

1

 

%

93.7 (2.5)

86.0 (3.5)

60.1 (5.0)

200

\(\hat {p}_{m}\)

1

1

1

 

%

84.9 (2.6)

70.5 (3.5)

34.9 (3.4)

   

Δ=2

 

Nb. of

  

α

 

DE genes

 

0.1

0.05

0.01

50

\(\hat {p}_{m}\)

0.13

0.33

0.98

 

%

99.7 (0.7)

99.2 (1.2)

93.2 (3.6)

100

\(\hat {p}_{m}\)

0.38

0.88

1

 

%

99.6 (0.6)

98.4 (1.0)

88.4 (3.2)

200

\(\hat {p}_{m}\)

0.99

1

1

 

%

97.8 (1.1)

92.1 (1.9)

65.3 (4.0)

   

Δ=3

 

Nb. of

  

α

 

DE genes

 

0.1

0.05

0.01

50

\(\hat {p}_{m}\)

0.00

0.00

0.02

 

%

100

100

96.0 (0.3)

100

\(\hat {p}_{m}\)

0.00

0.00

0.14

 

%

100

100

99.8 (0.5)

200

\(\hat {p}_{m}\)

0.01

0.11

0.97

 

%

100.0 (0.1)

99.9 (0.2)

98.1 (1.1)

  1. Evaluation of the first step of the ORdensity method using different values of α. The Table shows the estimated probability, \(\hat {p}_{m}\), of no considering as a potential DE gene at least one gene that it really is, and the mean proportion of DE genes (row named “%”) that the procedure considered as potential DE genes. Corresponding standard deviations are in brackets