Multiple Treatments with Strategic Interaction
Sukjin Han
University of Texas at Austin
September 17, 2016
Sukjin Han (University of Texas)
Multiple Treatments with Interaction
September 17, 2016
1 / 45
Model
Semi-triangular model with multiple binary treatments
D = (D1 , ..., DS ) with Ds 2 {0, 1}
where D
Y = µ(D, X , ✏D ),
⇥
⇤
D1 = 1 ⌫ 1 (D 1 , Z1 , V1 ) 0
..
.
h
i
DS = 1 ⌫ S (D S , ZS , VS ) 0
s
= (D1 , ..., Ds
1 , Ds+1 , ..., DS )
and ✏D =
I
simultaneity in the first stage (complete info game)
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D is endogenous in Y equation
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(Z , X ) ? (✏D , V1 , ..., VS )
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P
d
1[D = d]✏d
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Parameters of Interest
Y = µ(D, X , ✏D ),
Ds = 1 [⌫ s (D
D = (D1 , ..., DS ) 2 {0, 1}S
s , Zs , V s )
0] ,
s = 1, ..., S
Parameters of interest: average structural functions (ASF)
ASF (x) ⌘ E [Yd |X = x] = E [µ(d, x, ✏d )]
and functions of them, such as average treatment e↵ects (ATE)
ATE (x) ⌘ E [Yd
Yd˜|X = x]
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e.g., d = (1, ..., 1) vs. d˜ = (0, ..., 0); or more general nonlinear e↵ects
e.g., d = (1, d s ) vs. d˜ = (0, d s ) for given d s ;
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or complementarity: E [Y11
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Sukjin Han (University of Texas)
Y01 ] > E [Y10
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Y00 ]
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Example 1
Y = µ(D, X , ✏D ),
Ds = 1 [⌫ s (D
D = (D1 , ..., DS ) 2 {0, 1}S
s , Zs , V s )
0] ,
s = 1, ..., S
“Does media a↵ect political participation or electoral
competitiveness?”
I
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Y 2 [0, 1] voter turnout, or Y 2 {0, 1} whether incumbent is re-elected
Ds : market entry decision by local newspaper s
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E [Yd
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Zs : neighborhood counties’ population size and income
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X : voter ID regulation
Yd˜]: e↵ects of newspaper entry on political outcome
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Example 2
Y = µ(D, X , ✏D ),
Ds = 1 [⌫ s (D
D = (D1 , ..., DS ) 2 {0, 1}S
s , Zs , V s )
0] ,
s = 1, ..., S
“Food deserts”
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Y 2 [0, 1]: diabetes rate
Ds : exit decision by supermarket s
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E [Yd
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Zs : change in local government’s zoning plans
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X : region’s health-related variables (num of hospitals, obesity rate)
Yd˜]: e↵ects of absence of supermarkets on health
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Example 3
Y = µ(D, X , ✏D ),
Ds = 1 [⌫ s (D
D = (D1 , ..., DS ) 2 {0, 1}S
s , Zs , V s )
0] ,
s = 1, ..., S
“Does airlines competition a↵ect local air quality or health?”
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Y : pollution or health outcome
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Ds : entry decision by airline s
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E [Yd
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Zs : cost shifters
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X : factory pollutants emission or registered cars
Yd˜]: e↵ects of competition on pollution or health
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Example 4
Y = µ(D, X , ✏D ),
Ds = 1 [⌫ s (D
D = (D1 , ..., DS ) 2 {0, 1}S
s , Zs , V s )
0] ,
s = 1, ..., S
“How incumbents respond to the threat of entry by
competitors?”
I
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Y 2 {0, 1}: whether incumbents respond with price/investment
Ds : entry decision by firm s in “nearby” markets
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E [Yd
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Zs : cost shifters
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X : characteristics of incumbent’s market
F
Yd˜]: entry deterrence
e.g., in airline entry, distance btw the endpoints of incumbent’s market
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Features and Challenges
Y = µ(D, X , ✏D ),
Ds = 1 [⌫ s (D
D = (D1 , ..., DS ) 2 {0, 1}S
s , Zs , V s )
0] ,
s = 1, ..., S
1. Multiple equilibria in the first stage, thus the model as a whole is
incomplete
2. No distributional assumptions on (✏D , V1 , ..., VS )
I
arbitrary dependence: players’ correlated type and endogenous
treatments
3. No parametric functional form assumptions for µ(·) and ⌫ s (·)
4. Allow small support for excluded Zs and X
5. Can allow Zs to fail to be excluded from other players
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e.g., allow Z1 = · · · = ZS
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In This Paper
1. Partial identification (ID) of ATE with Zs (“instruments”) of small
support
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shape restrictions on µ(·) and ⌫ s (·) (e.g., monotonicity, symmetry)
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analytical characterization of equilibrium regions for S > 2
F
enables LATE and MTE frameworks
2. Tighter bounds with additional X excluded from the first-stage game
3. Sharp bounds with rectangular support for (X , Z1 , ..., ZS )
4. Point ID with player-specific Zs of large support
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Related Literature
Partial ID in triangular models with discrete endogenous variables
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Shaikh and Vytlacil (2011), Chesher (2005): single treatment
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Heckman, Urzua & Vytlacil (2006), Jun, Pinkse and Xu (2011): allow
multiple treatments, but single agent and no simultaneity
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This paper: multi-treatment, multi-agent models with strategic
interaction among agents
Partial ID of treatment e↵ects
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Manski (1990): simple bounds with excluded IV (applicable regardless
of dim of treatments)
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Manski (1997), Manski and Pepper (2000): allow multiple treatment
but no explicit simultaneity; use shape restrictions (e.g. semi-MTR)
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This paper: tighter bounds for mean by specifying “selection process”
with di↵erent shape restrictions and existence of X
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Related Literature
Partial ID of treatment response with social interaction
I
Manski (2013): multiple treatments but interaction through agents’
response variables
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This paper: interaction through treatment decisions
Discrete games with complete information
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Berry (1992), Tamer (2003), Ciliberto and Tamer (2009): partial ID in
general, point ID under large support
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Ciliberto, Murry and Tamer (2015): entry decision + pricing decision
(upon entry)
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This paper: the e↵ect of game’s equilibrium on a second-stage
outcome; di↵erent approach to solve multiplicity
Network formation and its e↵ects on outcomes
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Gilleskie and Zheng (2010), Badev (2013), Goldsmith-Pinkham and
Imbens (2013), Hsieh and Lee (2016)
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Plan of Talk
Model and parameters of interest
Point ID
Partial ID
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two-player case
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many-player case
Simulation
Conclusion
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Model and Parameters of Interest
For this talk, X and Zs scalar, Y 2 {0, 1}, and ✏d = ✏ for simplicity
Yd = 1[µ(d, X )
s
where Y =
P
Ds = 1 [⌫ (D
d
✏]
s , Zs )
Us ]
1[D = d]Yd and Us ⇠ Unif [0, 1] as normalization
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all the analyses (except sharpness) can be extended to cases of other
LDV or continuous Y (Vytlacil and Yildiz (2007))
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rank similarity can be assumed instead
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can allow Z1 = · · · = ZS instead
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when S = 2 and D = (D1 , D2 ),
Y = 1[µ(D1 , D2 , X ) ✏]
⇥
⇤
D1 = 1 ⌫ 1 (D2 , Z1 ) U1
⇥
⇤
D2 = 1 ⌫ 2 (D1 , Z2 ) U2
Parameters of interest: partial ATE, E [Yd 1 ,D 2
partition d = (d 1 , d 2 ) (more in the paper)
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Yd˜1 ,D 2 ] for some
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Point ID under Full Support
Yd = 1[µ(d, X )
s
Ds = 1 [⌫ (D
✏]
s , Zs )
Us ]
Assumption IN
(X , Z ) ? (✏d , U) 8d, where U ⌘ (U1 , ..., US ).
Under additional exclusion restriction with full support of instruments,
E [Yd |X ] is point ID’ed
“ID at infinity” solves multiple equilibria problem and endogeneity,
simultaneously
We want to avoid full support condition even if we possess such
instruments (Andrews and Schafgan (1998), Khan and Tamer (2010))
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Partial ID without Full Support
Yd = 1[µ(d, X )
s
Ds = 1 [⌫ (D
✏]
s , Zs )
Us ]
Assumption M
For given x, either µ(1, d s , x) µ(0, d
or µ(1, d s , x)  µ(0, d s , x) 8d s .
s , x)
8d
s,
M is mild monotonicity requirement
No need to know the direction
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cf. Manski (1997), Manski and Pepper (2000)
Assumption SS (Strategic Substitute)
⌫ s (d
s , zs )
is strictly decreasing in each element of d
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s.
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Partial ID without Full Support
Assumption SY (Symmetry)
˜ ·) for any permutation d˜ of d;
(i) µ(d, ·) = µ(d,
s
(ii) ⌫ (d s , ·) = ⌫ s (d˜ s , ·) for any permutation d˜
s
of d
s.
SY is useful to make the incomplete model tractable
SY(i) can be relaxed (at the expense of having wider bounds)
SY(ii) trivially holds in two-player case
SY(ii) becomes useful with many players; related to “anonymity”
assumption in games (e.g., Berry (1992), Kalai (2004), Menzel
(2016))
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still allows heterogeneity via nonseparability in ⌫ s (d
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s , zs )
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Partial ID: Results
Theorem
Suppose Assumptions IN, M, SS, and SY hold. The upper and lower
bounds of ASF and ATE with d, d˜ 2 {0, 1}S is
Ld (x)  E [Yd |X = x]  Ud (x)
and
Ld (x)
Ud˜(x)  E [Yd
Yd˜|X = x]  Ud (x)
Ld˜(x)
where, for some d † 2 {0, 1}S ,
(
†
Ud † (x) = inf Pr[Y = 1, D = d |Z = z, X = x] +
z
(
†
Ld † (x) = sup Pr[Y = 1D = d |Z = z, X = x] +
z
Sukjin Han (University of Texas)
X
inf
X
sup
x 0 2XU (x,d † ,d 0 )
d 0 6=d †
† 0
0
d 0 6=d † x 2XL (x,d ,d )
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0
0
)
Pr[Y = 1, D = d |Z = z, X = x ]
0
0
)
Pr[Y = 1D = d |Z = z, X = x ]
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Manski’s Bound
Consider
E [Yd |X ] = E [Yd |Z , X ] = E [Yd |D = d, Z , X ] Pr[D = d|Z , X ]
X
+
E [Yd |D = d 0 , Z , X ] Pr[D = d 0 |Z , X ]
d 0 6=d
I
1st eq. by exclusion restriction of Z
E [Yd |D = d 0 , Z , X ] = Pr[Yd = 1|D = d 0 , Z , X ] not observed, can be
bounded between 0 and 1
Can narrow the bound by taking maxz over LB and minz over UB
(Manski (1990))
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Proof for Tighter Bounds
The goal is to derive bounds tighter than above for
E [Yd |D = d 0 , Z , X ]
by fully exploiting the structure of the model (esp. the “selection
process”)
Suppose S = 2 for simplicity
Suppose we know the direction of monotonicity in M. Then e.g., we
have tighter UB
Pr[Y00 = 1|D = (1, 0), Z = z, X = x]
= Pr[µ(0, 0, x)
 Pr[µ(1, 0, x)
✏|D = (1, 0), Z = z, X = x]
✏|D = (1, 0), Z = z, X = x]
= Pr[Y10 = 1|D = (1, 0), Z = z, X = x] 1
Below, we construct quantities from which we can identify the
direction of the monotonicity
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Key Lemma: Using Variation of Z
For z 6= z 0 , define
h11 (z, z 0 , x) ⌘ Pr[Y = 1, D = (1, 1)|Z = z, X = x]
Pr[Y = 1, D = (1, 1)|Z = z 0 , X = x]
h00 (z, z 0 , x) ⌘ Pr[Y = 1, D = (0, 0)|Z = z, X = x]
Pr[Y = 1, D = (0, 0)|Z = z 0 , X = x]
hM (z, z 0 , x) ⌘ Pr[Y = 1, D 2 {(1, 0), (0, 1)}|Z = z, X = x]
Pr[Y = 1, D 2 {(1, 0), (0, 1)}|Z = z 0 , X = x]
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by Assumption SS, {(1, 0), (0, 1)} are values of D predicted by possible
multiple equilibria
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h11 , h00 , and hM can be directly recovered from data
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Z changes the equilibrium behavior in the first stage without directly
a↵ecting Y
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Key Lemma: Using Variation of Z
Define
h(z, z 0 , x) ⌘ Pr[Y = 1|Z = z, X = x]
Pr[Y = 1|Z = z 0 , X = x]
which satisfies h ⌘ h11 + h00 + hM
Also define
D
h11
(z, z 0 ) ⌘ Pr[D = (1, 1)|Z = z]
Pr[D = (1, 1)|Z = z 0 ]
and
D
h00
(z, z 0 ) ⌘ Pr[D = (0, 0)|Z = z]
Pr[D = (0, 0)|Z = z 0 ]
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D
D
h11
and h00
can also be recovered from data
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can determine volume change of the equilibrium regions at Z
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Key Lemma: Using Variation of Z
Let µd1 d2 (x) ⌘ µ(d1 , d2 , x) for simplicity
Lemma
D (z, z 0 ) and
Under Assumptions of Theorem, for any (z, z 0 , x) such that h11
D (z, z 0 ) are nonzero,
h00
n
o
D
sgn h(z, z 0 , x) = sgn h11
(z, z 0 ) · sgn {µ11 (x) µ01 (x)}
n
o
D
= sgn
h00
(z, z 0 ) · sgn {µ10 (x) µ00 (x)} .
Given the lemma, we can derive tighter bounds
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D
e.g., h 0 and h00
> 0 imply µ10 (x)
bound previously shown
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more generally, can exploit full variation of Z
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extending the idea of Shaikh and Vytlacil (2011) and Vytlacil and
Yildiz (2007) to multi-agent, incomplete model setup
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µ00 (x), thus we can use the
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Proof of Key Lemma
Let R11 (z), R00 (z) and RM (z) denote equilibrium regions
By IN and the model, can show that, for U ⌘ (U1 , U2 )
h11 (z, z 0 , x) = Pr[✏  µ11 (x), U 2 R11 (z)]
Pr[✏  µ11 (x), U 2 R11 (z 0 )]
Below,
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let ⌫ s (d
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D
suppose h11
(z, z 0 ) > 0 and
s , zs )
Sukjin Han (University of Texas)
⌘ ⌫ds s (zs ) for simplicity
D
h00
(z, z 0 ) > 0 for expositional purpose
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Proof of Key Lemma
When Z = z:
1
R10 (z)
R00 (z)
⌫11 (z1 ), ⌫02 (z2 )
U2
⌫00 (z1 ), ⌫12 (z2 )
R11 (z)
0
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0
R01 (z)
U1
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1
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Proof of Key Lemma
When Z = z:
1
⌫11 (z1 ), ⌫02 (z2 )
U2
⌫00 (z1 ), ⌫12 (z2 )
R11 (z)
0
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0
U1
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1
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Proof of Key Lemma
When Z = z 0 :
1
⌫11 (z10 ), ⌫02 (z20 )
U2
⌫01 (z10 ), ⌫12 (z20 )
R11 (z 0 )
0
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0
U1
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1
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Proof of Key Lemma
Z = z vs. Z = z 0 :
1
U2
+ R11
0
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0
U1
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1
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Proof of Key Lemma
Therefore
h11 (z, z 0 , x) = Pr[✏  µ11 (x), U 2 R11 (z)]
Pr[✏  µ11 (x), U 2 R11 (z 0 )]
= Pr[✏  µ11 (x), U 2
+ R11 ]
Recall
hM (z, z 0 , x) ⌘ Pr[Y = 1, D 2 {(1, 0), (0, 1)}|Z = z, X = x]
Pr[Y = 1, D 2 {(1, 0), (0, 1)}|Z = z 0 , X = x]
Then, using symmetry that µ10 = µ01 (SY(i))
hM (z, z 0 , x) = Pr[✏  µ10 (x), U 2 R10 (z) [ R01 (z)]
Pr[✏  µ10 (x), U 2 R10 (z 0 ) [ R01 (z 0 )]
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Proof of Key Lemma
When Z = z:
1
R10 (z)
⌫11 (z1 ), ⌫02 (z2 )
U2
⌫01 (z1 ), ⌫12 (z2 )
R01 (z)
0
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0
U1
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1
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Proof of Key Lemma
When Z = z 0 :
1
R10 (z 0 )
⌫11 (z10 ), ⌫02 (z20 )
U2
⌫01 (z10 ), ⌫12 (z20 )
R01 (z 0 )
0
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0
U1
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1
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Proof of Key Lemma
Z = z vs. Z = z 0 :
1
R00
U2
+ R11
0
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0
U1
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1
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Proof of Key Lemma
Therefore,
hM (z, z 0 , x) = Pr[✏  µ10 (x), U 2 R10 (z) [ R01 (z)]
Pr[✏  µ10 (x), U 2 R10 (z 0 ) [ R01 (z 0 )]
= Pr[✏  µ10 (x), U 2
Pr[✏  µ10 (x), U 2
R00 ]
+ R11 ]
Also can show
h00 (z, z 0 , x) = Pr[✏  µ00 (x), 2
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R00 ]
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Proof of Key Lemma
Therefore,
h(x) =h11 (x) + h00 (x) + hM (x)
= Pr[✏  µ11 (x), U 2
+ R11 ]
+ Pr[✏  µ10 (x), U 2
R00 ]
Pr[✏  µ10 (x), U 2
+ R11 ]
Pr[✏  µ00 (x), U 2
Now, µ11 (x) µ10 (x) = µ01 (x) and µ10 (x)
directions by M) i↵
R00 ]
µ00 (x) (same
h(x) = Pr[µ10 (x)  ✏  µ11 (x), U 2
+ Pr[µ00 (x)  ✏  µ10 (x), U 2
+ R11 ]
R00 ]
which is non-negative (sum of two probs)
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conversely, µ11 (x) < µ10 (x) and µ10 (x) < µ00 (x) i↵ h < 0
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Discussions
1. Exploiting additional variation from X : e.g., can show if, for x 6= x 0
h(x 0 ) ⌘ h11 (x 0 ) + hM (x 0 ) + h00 (x 0 )
0
0
0
0
0
h(x, x , x ) ⌘ h11 (x ) + hM (x ) + h00 (x) < 0
then
µ00 (x)
µ10 (x 0 )
2. Sharp bounds with supp(X , Z ) = supp(X ) ⇥
binary Y
3. SY(i) can be relaxed given knowledge on
QS
s=1 supp(Zs )
for
µ10 (·) ? µ01 (·)
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if µ10 (·) = µ01 (·) is implausible, then it can be because the sign of the
ineq. is known
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the bounds on ATE are not as tight as the ones under symmetry, but
tighter than Manski (1990)’s bounds
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The General Case of Many Players (S > 2)
As seen in S = 2 case, need to fully characterize equilibrium regions
For j = 0, ..., S
1, define
e j ⌘ (e1j , ..., eSj ) ⌘ (1, ..., 1, 0, ..., 0)
| {z } | {z }
j
S j
Given d j = ( (e1j ), ..., (eSj )) for some permutation function (·),
Rd j (z) ⌘
(
U : (U
(1) , ..., U (S) )
2
j ⇣
Y
s=1
⇥
S ⇣
Y
s=j+1
Can similarly define Rd 0 and Rd S
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(s)
0, ⌫ẽ j 1 (z
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(s) )
i
9
⌘=
(s)
⌫ẽ j (z (s) ), 1
;
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The General Case of Many Players (S > 2)
The region of all equilibria with j entrants is denoted as
R̄j (z) ⌘ [d2Mj Rd (z)
where Mj is a set of all possible permutations of e j
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with S = 3, e.g., R̄2 = R110 [ R101 [ R011
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Visual Illustration with S = 3
(1, 1, 1)
R000
1 , ⌫2 )
(⌫00
00
3
⌫00
U3
3
⌫11
R111
(0, 0, 0)
1 , ⌫2 )
(⌫11
11
U2
U1
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Visual Illustration with S = 3
R010
1 , ⌫2 )
(⌫00
00
3
⌫00
U3
3
⌫11
1 , ⌫2 )
(⌫11
11
3 = ⌫3
⌫10
01
U2
U1
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Visual Illustration with S = 3
R̄1 = R010 [ R100 [ R001
1
2
00 , ⌫00 )
1 , ⌫ 2 (⌫
(⌫10
)
10
3
⌫00
U3
3
⌫11
1
2
01 , ⌫01 )
1 , ⌫ 2 (⌫
(⌫11
)
11
3 = ⌫3
⌫10
01
U2
U1
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Characterization of Equilibrium Regions
Proposition
Under Assumptions SS and SY(ii), the following holds:
(i) R̄j \ R̄j 0 = ; for j, j 0 = 0, ..., S with j 6= j 0 ;
(ii) R̄j and R̄j 1 are neighboring sets for j = 1, ..., S;
(iii) R̄j and R̄j t are not neighboring sets for t 2 and j = t, ..., S;
(iv) [Sj=0 R̄j = (0, 1)S .
By (i), unique equilibrium in terms of number of entrants
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similar result as in Berry (1992) but under weaker assumptions
(ii) and (iii) are important in equating “inflow” and “outflow”
(i) and (iv) imply that R̄j for j = 1, ..., S partition the entire space
(0, 1)S of U
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Visual Illustration with S = 3
R010 [ R100
1
2
00 , ⌫00 )
1 , ⌫ 2 (⌫
(⌫10
)
10
3
⌫00
U3
3
⌫11
1 , ⌫2 )
(⌫11
11
3 = ⌫3
⌫10
01
U2
U1
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Visual Illustration with S = 3
R010 [ R100
R011 [ R101
1
2
00 , ⌫00 )
1 , ⌫ 2 (⌫
(⌫10
)
10
3
⌫00
U3
3
⌫11
1
2
01 , ⌫01 )
1 , ⌫ 2 (⌫
(⌫11
)
11
3 = ⌫3
⌫10
01
U2
U1
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Visual Illustration with S = 3
[3j=0 R̄j = (0, 1)3
R̄0
R̄1
R̄2
R̄3
= R000
= R010 [ R100 [ R001
= R011 [ R101 [ R110
= R111
1
2
00 , ⌫00 )
1 , ⌫ 2 (⌫
(⌫10
)
10
3
⌫00
U3
3
⌫11
1
2
01 , ⌫01 )
1 , ⌫ 2 (⌫
(⌫11
)
11
3 = ⌫3
⌫10
01
U2
U1
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Visual Illustration with S = 3
Multiple equil. w/ 1 entrant
Multiple equil. w/ 2 entrant
1
2
00 , ⌫00 )
1 , ⌫ 2 (⌫
(⌫10
)
10
U3
1
2
01 , ⌫01 )
1 , ⌫ 2 (⌫
(⌫11
)
11
3 = ⌫3
⌫10
01
U2
U1
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Key Lemma: Using Variation of Z
Based on the previous results, define, for z 6= z 0 ,
hj (z, z 0 , x) = Pr[Y = 1, D 2 Mj |Z = z, X = x]
Pr[Y = 1, D 2 Mj |Z = z 0 , X = x].
Recall,
h(z, z 0 , x) = Pr[Y = 1|Z = z, X = x]
Pr[Y = 1|Z = z 0 , X = x]
I
since Mj are disjoint, h(z, z 0 , x) =
Also define
PS
j=0
hj (z, z 0 , x)
hjD (z, z 0 ) = Pr[D 2 Mj |Z = z]
Pr[D 2 Mj |Z = z 0 ]
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Key Lemma: Using Variation of Z
Lemma
Under
of Theorem, for any (z, z 0 , x) such that
PS Assumptions
D
0
k=j hk (z, z ) 6= 0 and j = 1, ..., S,
sgn h(z, z 0 , x) = sgn
8
S
<X
:
k=j
hkD (z, z 0 )
9
=
;
· sgn {µe j (x)
µe j 1 (x)} .
In practice, for efficiency, we determine the sign of µe j (x)
by the sign of
2
3
S
X
Hj (x) ⌘ E 4h(Z , Z 0 , x)
hkD (z, z 0 ) > 0 5
µe j 1 (x)
k=j
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Discussions
4. Again, symmetry in µd (·) can be relaxed if µd (·) can be ordered
within d 2 Mj
5. Based on Proposition, versions of LATE and MTE can be considered
I
cf. Angrist & Imbens (1995), Heckman, Urzua & Vytlacil (2006)
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e.g., IV estimand has LATE interpretation
PS
h(z, z 0 )
k=j
hkD (z, z 0 )
⌘
Pr[Y = 1|Z = z]
Pr[D 2 M j |Z = z]
= E [YM
j
Pr[Y = 1|Z = z 0 ]
Pr[D 2 M j |Z = z 0 ]
YM j 1 |D(z) 2 M
j
, D(z 0 ) 2 M j
1
]
e.g., when S = 2,
E [Y11
Y{(1,0),(0,1),(0,0)} |D(z) = (1, 1), D(z 0 ) 2 {(1, 0), (0, 1), (0, 0)}]
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LATE subgroup
Z = z vs. Z = z 0 :
1
U2
+ R11
0
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U1
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Monte Carlo Simulation
DGP:
Yd = 1{µ̃d + X
✏}
D1 = 1{ 2 D2 +
1 Z1
D2 = 1{ 1 D1 +
2 Z2
V1 }
V2 }
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(✏, V1 , V2 ) drawn from joint normal, mean zero, indep. of (X , Z )
I
Z 2 { 1, 1} and X 2 { 1, 0, 1} or X 2 { 1,
drawn from multinomial
I
1
< 0 and
2
6
7,
5
5 6
7 , ..., 7 , 7 , 1}
< 0 (SS)
Design 1: µ̃11 > µ̃10 = µ̃01 > µ̃00 (consistent with M and SY)
Design 2: µ̃11 > µ̃01 , µ̃11 > µ̃10 , µ̃01 > µ̃00 , and µ̃10 > µ̃00 , but
µ̃10 6= µ̃01 (consistent with M)
Parameter
ATE (0) = E [Y11
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Design 1: Variation of Z
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Design 1: Additional X with |supp(X )| = 3
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Design 1: Additional X with |supp(X )| = 15
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Design 2: Additional X with |supp(X )| = 15
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Conclusions
Partial ID of ATE in a model for heterogeneous e↵ects of multiple
treatments, where the treatments are equilibrium of a complete info
game
I
no distributional assumptions nor parametric functional forms
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shape restrictions and excluded variables with small supports
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symmetry can be relaxed with some knowledge on e↵ectiveness of each
treatment
Inference
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modification of Andrews and Shi (2013), Chernozhukov, Lee and Rosen
(2013), Armstrong and Chan (2015)
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possible first step: Armstrong and Shen (2015)
Empirical works: 1. pollution with airline entry; 2. entry deterrence
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data is collected from Dept. of Transportation and US Environmental
Protection Agency
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Exploiting Additional Variation from X
Analogously define
X ⌘ x 0 : h(x 0 ) < 0
X2+ (x) ⌘ x 0 : h(x, x 0 , x 0 )
0
0
X1+ (x) ⌘ x : h(x, x, x )
0
0
Then, similarly as above, the UB’s are
Pr[Y00 = 1|D = (1, 0), Z , X = x] 
Pr[Y00 = 1|D = (0, 1), Z , X = x] 
Pr[Y00 = 1|D = (1, 1), Z , X = x] 
inf
Pr[Y10 = 1|D = (1, 0), Z , X = x 0 ]
inf
Pr[Y01 = 1|D = (0, 1), Z , X = x 0 ]
inf
Pr[Y11 = 1|D = (1, 1), Z , X = x 0 ]
x 0 2X \X2+ (x)
x 0 2X \X2+ (x)
x 0 2X1+ (x 00 )
x 00 2X
\X2+ (x)
Can analogously derive LB’s and UB’s for Pr[Y11 = 1|D = d 0 , Z , X ]
for d 0 2 {(0, 0), (1, 0), (0, 1)}
Lastly, take supz and inf z
Return
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Relaxing Symmetry
Since the equilibrium selection rule is unknown,
⇤
⇤
h10 + h01 = Pr[✏  µ10 , U 2 R10
(z)] + Pr[✏  µ01 , U 2 R01
(z)]
⇤
Pr[✏  µ10 , U 2 R10
(z 0 )]
I
⇤
Pr[✏  µ01 , U 2 R01
(z 0 )]
which, following Tamer (2003), has a LB of
min
min
Pr[✏  µ10 , U 2 R10
(z)] + Pr[✏  µ01 , U 2 R01
(z)]
max 0
Pr[✏  µ10 , U 2 R10
(z )]
I
max 0
Pr[✏  µ01 , U 2 R01
(z )]
but, following our approach, has a LB of
⇤
⇤
Pr[✏  µ10 ^ µ01 , U 2 R10
(z)] + Pr[✏  µ10 ^ µ01 , U 2 R01
(z)]
⇤
Pr[✏  µ10 _ µ01 , U 2 R10
(z 0 )]
⇤
Pr[✏  µ10 _ µ01 , U 2 R01
(z 0 )]
⇤
⇤
and R10
[ R01
= R10 [ R01
Return
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