k c y w d k d o p e d b a i q s d f m k c n f a e...
TRANSCRIPT
A Neural Mechanism for Decision MakingK C Y W D K D O P E D B A I Q S D F M K C N F A E O I E N C V N SE N C H P D N C O E N A S H Q E N D N C K R N D N Q I O M Z C P Q
What is a decision?
• A commitment to a proposition or selectionof an action
• Based on– evidence– prior knowledge– payoff
Why study decisions?
• They are a model of higher brain function• They are experimentally tractable
– Combined behavior and physiology in rhesusmonkeys
Direction-Discrimination Task
Reward for correct choice
Direction-Discrimination TaskReaction-time version
100 ms
120
60
0
60
Spikes/sec
120
60
0
Spatially-selective, eye movement-related,persistent activity in area LIP
High motion strength
High motion stre
ngth
Low motion strength
Time~1 secStimulus
onStimulus
off
Spikes/s
Time~1 secStimulus
onStimulus
off
Spikes/s
Low motion strength
A Neural Integrator for Decisions?
MT: Sensory Evidence
Motion energy“step”
LIP: Decision Formation
Accumulation of evidence “ramp”
Threshold
Proposed by Wald, 1947 and Turing (WW II, classified); Stone, 1960; then Laming, Link, Ratcliff, Smith, . . .
Diffusion to bound modelPositive bound or Criterion to answer “1”
Negative bound or Criterion to answer “2”
Momentary evidencee.g.,
∆Spike rate:MTRight– MTLeft
Accumulated evidencefor Rightward
andagainst Leftward
Criterion to answer “Right”
Criterion to answer “Left”
Diffusion to bound model
Shadlen & Gold (2004)Palmer et al (in press)
µ = kC
C is motion strength (coherence)
µ = kC
Criterion to answer “Right”
Criterion to answer “Left”
Momentary evidencee.g.,
∆Spike rate:MTRight– MTLeft
Accumulated evidencefor Rightward
andagainst Leftward
Time (ms)
µ = kC
Criterion to answer “Right”
Criterion to answer “Left”
Momentary evidencee.g.,
∆Spike rate:MTRight– MTLeft
Accumulated evidencefor Rightward
andagainst Leftward
• LIP represents ∫ dt of momentary motion evidence
• Momentary evidence is a spike rate difference from area MT• The accumulated evidence used by the monkey is in area LIP• How and where is the integral computed?• How is the bound set?• How is a bound crossing detected?
• More RIGHT choices
• Faster RIGHT choices• Slower LEFT choices
• More RIGHT choices• Faster RIGHT choices• Slower LEFT choices
Bound for RIGHT choice
Bound for LEFT choice
Bound for RIGHT choice
Bound for LEFT choice
Stimulate RIGHTWARDMT neurons
The momentary evidence is a ∆ between opposite direction signals in area MT
The accumulated evidence used by the monkey is in area LIP
Stimulate RIGHT CHOICELIP neurons
µ = kC
Criterion to answer “Right”
Criterion to answer “Left”
Momentary evidencee.g.,
∆Spike rate:MTRight– MTLeft
Accumulated evidencefor Rightward
andagainst Leftward
• LIP represents ∫ dt of momentary motion evidence
• Momentary evidence is a spike rate difference from area MT• The accumulated evidence used by the monkey is in area LIP• How and where is the integral computed?• How is the bound set?• How is a bound crossing detected?
Fixation
1st Stim &Targets
2nd Stim
3rd Stim
4th Stim
Delay & Sacade
0 ms
500 ms
1000 ms
1500 ms
2000 ms
Time
Probabilistic categorization task:4-card stud
Tianming Yang
- -0.9 -0.7 -0.5 -0.3 0.3 0.5 0.7 0.9 +
FavoringGreen
FavoringRed
0.9 -0.9
0.7-0.9-0.2
0.61 0.39
Weight of evidence in favor of red(log10 likelihood ratio)
From sensorimotor integration tocognition and its disorders
Sensoryevidence
Motoroutput
Prior knowledgeExpected payoff
Urgency
Potentialbehavioror plan
From sensorimotor integration tocognition and its disorders
Sensoryevidence
MotoroutputArea MT Area LIP Oculomotor System
From sensorimotor integration tocognition and its disorders
Leaky integration confusion
Evanescentsensory stream
Plans forthe future
dt!
Sensoryevidence
Motorresponse
Turing’s strategy: sequential analysis
WOE =
10 log10
113( )126( )
!
" # #
$
% & &
= +3.0 db match
10 log10
1213( )2526( )
!
" # #
$
% & &
= '0.17db non -match
(
)
* *
+
* *
Weight of evidencein favor of
common rotor setting
Hypothesis: Messages encrypted by Enigma devices in same stateK C Y W D K D O P E D B A I Q S D F M K C N F A E O I E N C V N S D F NE N C H P D N C O E N A S H Q E N D N C K R N D N Q I O M Z F J K C P Q
0
5
Wei
ght o
f evi
denc
e in
favo
r of
com
mon
setti
ngs
(dec
iban
s)
Turing’s strategy: sequential analysis
WOE =
10 log10
113( )126( )
!
" # #
$
% & &
= +3.0 db match
10 log10
1213( )2526( )
!
" # #
$
% & &
= '0.17db non -match
(
)
* *
+
* *
Weight of evidencein favor of
common rotor setting
Hypothesis: Messages encrypted by Enigma devices in same stateK C Y W D K D O P E D B A I Q S D F M K C N F A E O I E N C V N S D F NE N C H P D N C O E N A S H Q E N D N C K R N D N Q I O M Z F J K C P Q
0
5
Wei
ght o
f evi
denc
e in
favo
r of
com
mon
setti
ngs
(dec
iban
s)
K C Y W D K D O P E D B A I Q S D F M K C N F A E O I E N C V N S D F NE N C H P D N C O E N A S H Q E N D N C K R N D N Q I O M Z F J K C P Q
0
5
Wei
ght o
f evi
denc
e in
favo
r of
com
mon
setti
ngs
(dec
iban
s) K C Y W D K D O P E D B A I Q S D F M K C N F A E O I E N C V N S D F NE N C H P D N C O E N A S H Q E N D N C K R N D N Q I O M Z F J K C P Q
WOE =
10 log10
113( )126( )
!
" # #
$
% & &
= +3.0 db match
10 log10
1213( )2526( )
!
" # #
$
% & &
= '0.17db non -match
(
)
* *
+
* *
Weight of evidencein favor of
common rotor setting
Turing’s strategy: sequential analysis
0 10 20 30 40 500
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
Response (spikes/sec)
Pro
ba
bili
ty
Response (spikes/s)
Frequency ofoccurrence
LEFT preferringMT neurons
RIGHT preferringMT neurons
Variable response to weak RIGHTWARD motion
-40 -20 0 20 400
0.01
0.02
0.03
0.04
0.05
0.06
Frequency ofoccurrence
Response difference (spikes/s)
Difference in spike rate is proportionalto the logarithm of the likelihood ratio
Distribution ofresponseDIFFERENCES,right-left, forrightward motion
0 0.2 0.4 0.6 0.8
0
Time (s)
Accumulateddifference (R-L)
Amount of accumulatedevidence required tochoose “LEFT”
Weight ofevidence
Decibans
Belief
Log ofLikelihood
Ratio
Amount of accumulatedevidence required tochoose “RIGHT”
Yn= X
i
i=1
n
! random walk or diffusion
MX(!) " E e!X#$ %& = f (x)e!xdx
'(
(
) def. of MGF for X
MYn
(!) = MX
n(!) MGF for sums
M!Y(!) = P
+e!A
+ (1" P+)e
"!A MFG for !Y
!Y ! stopped accumulation
Random walk to bounds at ±A
E Zn[ ] = E M
X
!n(")e"Yn#$ %&!!!!!!!
= MX
!n(")E e
"Yn#$ %&
= MX
!n(")M
Yn
(")
= 1
E Zn+1
Y1,Y2,…,Y
n!" #$ = E M
X
%(n+1)(&)e&Yn+1 Y
1,Y2,…,Y
n!" #$
= E MX
%(n+1)(&)e& (Yn +Xn+1 )!" #$
= E[MX
%1(&)M
X
%n(&)e&Yn e&Xn+1 ]
= E[ZnM
X
%1(&)e&Xn+1 ]!!!
= MX
%1(&)Z
nE e
&Xn+1!" #$
= Zn
Wald’s martingale & identity
E !Z!" #$ = E Z
n[ ] = 1
E M
X
!n(")e"
!Y#$
%& = 1
If there were a value for ! such that MX(!) = 1 , it no longer matters
that n is a random number. At this special value, !1 ,
E e
!1 !Y"#
$% = 1
E.g., for the Normal distribution, with mean ! and variance !2, "
1= #
2µ
!2
M!Y(!1) = P
+e!1A + (1" P
+)e
"!1A = 1
P+=
1
1+ e!1A
Use Wald’s martingale to simplify
M!Y(!)