Probabilistic modeling of sensor artifacts in critical care Norm - PowerPoint PPT Presentation

Probabilistic modeling of sensor artifacts in critical care Norm Aleks and Stuart J. Russell U.C. Berkeley Computer Science Division Michael G. Madden N.U.I. Galway Dept. of Computer Science Kristan Staudenmayer, Mitchell Cohen, Diane Morabito, and Geoffrey Manley U.C. San Francisco Dept. of Neurological Surgery and San Francisco General Hospital

Raw data

Remaining in this talk • Arterial blood pressure (ABP) and its measurement artifacts • Modeling sub-interval events • Our ABP/monitoring artifact model • Experimental results

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ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10

ABP measurements, artifacts, and the effects of averaging systolic (maximum) 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10

ABP measurements, artifacts, and the effects of averaging systolic (maximum) mean 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10

ABP measurements, artifacts, and the effects of averaging systolic (maximum) mean diastolic (minimum) 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10

ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10

ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 3 4 5 6

ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 blood draw 3 4 5 6

ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 blood draw line flush 3 4 5 6

ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 blood draw line flush 3 4 5 6 3 4 5 6

ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 blood draw line flush zeroing 3 4 5 6 3 4 5 6

ABP measurements, artifacts, and the effects of averaging 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 blood draw line flush zeroing 3 4 5 6 3 4 5 6

Physiologic ABP changes Physiologic hypertension (high blood pressure)

Physiologic ABP changes Physiologic hypertension (high blood pressure) Physiologic hypotension (low blood pressure)

How can we detect and correct for these artifacts?

How can we detect and correct for these artifacts? • By hand?

How can we detect and correct for these artifacts? • By hand? • No: too costly, doesn’t scale

How can we detect and correct for these artifacts? • By hand? • No: too costly, doesn’t scale • Low-pass filter?

How can we detect and correct for these artifacts? • By hand? • No: too costly, doesn’t scale • Low-pass filter? • No: frequencies of physiologic events overlap with those of artifactual events

How can we detect and correct for these artifacts? • By hand? • No: too costly, doesn’t scale • Low-pass filter? • No: frequencies of physiologic events overlap with those of artifactual events • Careful system modeling?

How can we detect and correct for these artifacts? • By hand? • No: too costly, doesn’t scale • Low-pass filter? • No: frequencies of physiologic events overlap with those of artifactual events • Careful system modeling? • Of course!

What should the model’s timestep be?

What should the model’s timestep be? • One second? (“fast model”)

What should the model’s timestep be? • One second? (“fast model”) ✓ artifact occurrence modeling is natural

What should the model’s timestep be? • One second? (“fast model”) ✓ artifact occurrence modeling is natural - inference runs 59x between evidence

What should the model’s timestep be? • One second? (“fast model”) ✓ artifact occurrence modeling is natural - inference runs 59x between evidence • One minute? (“slow model”)

What should the model’s timestep be? • One second? (“fast model”) ✓ artifact occurrence modeling is natural - inference runs 59x between evidence • One minute? (“slow model”) ✓ this is the frequency of the sensor data

What should the model’s timestep be? • One second? (“fast model”) ✓ artifact occurrence modeling is natural - inference runs 59x between evidence • One minute? (“slow model”) ✓ this is the frequency of the sensor data - it’s difficult to model event occurrence

Event occurrence: fast model f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8

Event occurrence: fast model f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 f i : 1 if event is in process at timestep i , 0 if it is not.

Event occurrence: fast model f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 f i : 1 if event is in process at timestep i , 0 if it is not. p = P( f i =1 | f i–1 =1) ... “P(event continues)” q = P( f i =1 | f i–1 =0) ... “P(new event starts)”

Relate fast model to slow f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8

Relate fast model to slow f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 ∑ ∑ G 0 G 4 G 8

Relate fast model to slow f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 ∑ ∑ G 0 G 4 G 8 E 0 E 4 E 8

Sum out extra fast nodes f 0 f 4 f 8 G 0 G 4 G 8 E 0 E 4 E 8

f 0 f 1 f 2 G 0 G 1 G 2 E 0 E 1 E 2

f 0 f 1 G 0 G 1 E 0 E 1

Computational cost • Using a dynamic program, computing new CPTs has time complexity O( j 2 N j +1 ), where the fast model has ... • j mutually exclusive events to count • N timesteps per slow-model timestep • Markovian transitions • Space complexity is O( j 2 N j )

Parameterizing the fast model for “bag” events time t+1 normal bag Valve position normal 1 - q q time t bag 1 - p p

Parameterizing the fast model for “bag” events time t+1 normal bag Valve position normal 1 - q q time t bag 1 - p p q = event count / time = 53 / 557,893 = 0.000129

Parameterizing the fast model for “bag” events time t+1 normal bag Valve position normal 1 - q q time t bag 1 - p p q = event count / time = 53 / 557,893 = 0.000129 p = 1 - (1 / average event length) = 1-1/26.38 = 0.962

Event durations fit a geometric model well )!" !"#$%&"'&()*&+,+$%-&./"-+&0#1)2"$&3&4%/+&56)57-&,)8#+& (!" '!" &!" %!" $!" #!" !" #" %" '" )" *" ##"#%"#'"#)"#*"$#"$%"$'"$)"$*"%#"%%"%'"%)"%*"&#"&%"&'"&)"&*"'#"'%"''"')"'*"(#"(%"('"()"(*")#")%")'"))")*"+#" Bag event durations

P(G | f 0 =0) P(f 1 | G, f 0 =0) f 1 =0 f 1 =1 P(G | f 0 =1) P(f 1 | G, f 0 =1) f 1 =0 f 1 =1

f 0 f 1 G 0 G 1 E 0 E 1

valve- valve- pos 0 pos 1

valve- valve- pos 0 pos 1 bag- time 1 zero- time 1

valve- valve- pos 0 pos 1 bag- true- time 1 BP 1 bag- zero- press 1 time 1 zero- press 1 appa- rent 1 Apparent BP = (1/60) • ( (60 – bag-time – zero-time) • true-BP + bag-time • bag-press + zero-time • zero-press )

valve- valve- pos 0 pos 1 bag- true- time 1 BP 1 bag- zero- press 1 time 1 zero- press 1 appa- rent 1 meas- ured 1