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Table 2 Simulation study 1, scenario 1 with nX=nY=30 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

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

0.08

0.20

0.80

 

%

99.84 (0.54)

99.56 (0.92)

97.20 (2.2)

100

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

0.13

0.45

0.99

 

%

99.87 (0.34)

99.34 (0.66)

95.20 (2.3)

200

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

0.80

0.99

1.00

 

%

99.27 (0.56)

96.70 (1.5)

78.80 (4.0)

   

Δ=2

 

Nb. of

  

α

 

DE genes

 

0.1

0.05

0.01

50

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

0.00

0.00

0.00

 

%

100

100

100

100

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

0.00

0.00

0.05

 

%

100

100

99.95 (0.22)

200

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

0.01

0.03

0.95

 

%

100

99.98 (0.09)

98.50 (0.83)

   

Δ=3

 

Nb. of

  

α

 

DE genes

 

0.1

0.05

0.01

50

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

0.00

0.00

0.00

 

%

100

100

100

100

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

0.00

0.00

0.00

 

%

100

100

100

200

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

0.00

0.00

0.00

 

%

100

100

100

  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 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