Necessary and sufficient conditions in the problem of optimal investment with intermediate consumption Oleksii Mostovyi Department of Mathematical Sciences Carnegie Mellon University
Ann Arbor, May 19, 2011
Market d + 1 traded assets: 1. Price process S = {S i }i=1,...,d - (locally bounded) semimartingale on (Ω, F, {Ft }0≤t≤T , P). 2.
Savings account with zero interest rate.
M - family of martingale measures M = {Q ∼ P : S is a Q−local martingale} Assumption (No Arbitrage) M= 6 ∅
Agent Portfolio (x, H, c) : initial wealth x > 0; number of shares H predictable, S−integrable; consumption rate c ≥ 0, optional. Wealth process Rt Rt Xt = x + 0 Hu dSu − 0 cu du. Admissibility Assumption X ≥ 0. Utility Stochastic Fields U1 : [0, T ] × Ω × R++ → R,
U2 : Ω × R++ → R :
1. strictly increasing, strictly concave, differentiable. 2. Inada conditions hold: U10 (t, ω, 0) = +∞, U20 (ω, 0) = +∞, U10 (t, ω, ∞) = 0, U20 (ω, ∞) = 0.
Primal Problem "Z u(x) =
sup c,XT : X0 =x,X ≥0
#
T
U1 (t, ω, ct )dt + U2 (ω, XT ) .
E 0
Is u a utility function? ˆT ) exist? Does the optimal solution (cˆ, X
Dual Problem Conjugate Fields V1 (t, ω, y ) = sup (U1 (t, ω, x) − xy ) , x>0
V2 (ω, y ) = sup (U2 (ω, x) − xy ) . x>0
Y(y ) the set of nonnegative supermartingales Y : 1. Y0 = y , 2. {Xt Yt }t∈[0,T ] is a supermartingale for all X ≥ 0.
"Z v (y ) =
inf
Y ∈Y(y )
T
# V1 (t, ω, Yt )dt + V2 (ω, YT ) .
E 0
Karatzas, I. and Žitkovi´c, G. (2003). ! lim sup ess sup x→∞
(t,ω)
xU10 (t,ω,x) U1 (t,ω,x)
<1
+ several other conditions
Kramkov, D. and Schachermayer, W. (2003). v (y ) < ∞ for all y > 0
Our conditions: v (y ) < ∞ for all y > 0 and u(x) > −∞ for all x > 0
Theorem Let v (y ) < ∞ for all y > 0, u(x) > −∞ for all x > 0. Then 1
u(x) < ∞ for all x > 0. u and v are convex conjugate u(x) = inf (v (y ) + xy ) ⇐⇒ y >0
v (y ) = sup (u(x) − xy ). x>0
u, −v are utility functions 2
ˆT (x)), the optimizer to the Primal Problem exists (cˆ(x), X and unique. ˆ (y ), the optimizer of the Dual Problem exists and unique. Y
How to check these conditions? To check that v (y ) < ∞ for all y > 0 it suffices to find 1 Y ∈ Y(y ): "Z # T
V1 (t, ω, Yt )dt + V2 (ω, YT ) < ∞,
E 0
t Candidate Yt = y dQ dPt
To check that u(x) > −∞ for all x > 0 it suffices to find 1 admissible pair (c, XT ): "Z # T
U1 (t, ω, ct )dt + U2 (ω, XT ) > −∞,
E 0
Candidate c ≡
x T +1 , XT
=
x T +1
