Plan ₿: UtilityFunctionExample

UtilityFunctionExample

For any stochastic process, the utility function can be found by brute force numerical calculation. However, in some cases it is easier.

For discrete time and continuous flow case, if the only outflow distribution follows exponential distribution, then a simple form is available:

$Pr[X<-m]=pe^{-am}$

The solution of $u(m)= \int {u \big(m+X\big) dF \big(X\big) }$ and $u \big(m\big)= \int_{m+X>0} {u \big(m+X\big) dF \big(X\big) }$ is of the form $1-C e^{-\gamma m}$ for some C and $\gamma$ .

By $u(m)= \int {u \big(m+X\big) dF \big(X\big) }$ and denote M as the moment generating function of positive flow, it follows

$1= \int {e^{-\gamma X}dF \big(X\big)}= (1-p)M(-\gamma)+\frac{pa}{a-\gamma}$

Note that this equation has the non-zero solution when the expectation of flow is greater than zero.

By $u \big(m\big) = \int_{m+X>0} {u \big(m+X\big) dF \big(X\big) }$ , it follows

$0=\int_{m+X \leq 0} {u \big(m+X\big) dF \big(X\big) } =\int_{m+X \leq 0} {(1-Ce^{-\gamma(m+X)}) dF \big(X\big) }=pe^{-am}-\frac{Cpa}{a-\gamma} e^{-am}$

Therefore $C=1- \frac{\gamma}{a}$

For the time preference:

$\int_{m+X>0} {u \big(m+B+X\big) dF \big(X\big) }=\int_{-m}^ \propto {(1-Ce^{-\gamma(m+B+X)}) dF \big(X\big)}$

$=Pr[X>-m]-Ce^{-\gamma(m+B)} \int_{-m}^ \propto {e^{-\gamma X}dF \big(X\big) }$

$=1-pe^{-am}-Ce^{-\gamma(m+B)} [(1-p)M(-\gamma)+p \int_{-m}^0 {e^{-\gamma X}ae^{aX}dX} ]$

$=1-pe^{-am}-Ce^{-\gamma(m+B)} [(1-p)M(-\gamma)+ \frac{pa}{a-\gamma}(1-e^{-(a-\gamma)m}) ]$

$=1-pe^{-am}-Ce^{-\gamma(m+B)}[1-\frac{pa}{a-\gamma}e^{-(a-\gamma)m}]$

$=1-Ce^{-\gamma(m+B)}-pe^{-am}[1-e^{-\gamma B}]$

Therefore the relationship between A and B is

$1-Ce^{-\gamma m}=1-Ce^{-\gamma(m-A+B)}-pe^{-a(m-A)}[1-e^{-\gamma B}]$

Rearrange then

$e^{-\gamma B}=\frac{e^{-\gamma A}-\frac{p}{C} e^{-(a-\gamma)(m-A)}}{1-\frac{p}{C} e^{-(a-\gamma)(m-A)}}$

For tiny A in comparison with m, then

$\lim_{A \rightarrow 0} \frac{B}{A} =\frac{1}{1-\frac{p}{C} e^{-(a-\gamma)m}}$

Note that B/A approaches 1 as m approaches infinite.

For continuous time and continuous flow case, if it is a Poisson process with a constant flow and a random flow whose only outflow is an exponential distribution:

$u(m,t)=(1-\lambda dt)u(m- \mu dt, t+dt)+\lambda dt \int_{-(m- \mu dt)}^ \propto {u(m- \mu dt+X,t+dt)dF(X)}$

$=(1-\lambda dt)(u(m,t)- \frac{\partial u}{\partial m} \mu dt+\frac{\partial u}{\partial t} dt)+\lambda dt \int_{-(m- \mu dt)}^ \propto {u(m- \mu dt+X,t+dt)dF(X)}$

$0=-\lambda u(m,t)- \frac{\partial u}{\partial m} \mu+\frac{\partial u}{\partial t}+\lambda \int_{-(m- \mu dt)}^ \propto {u(m- \mu dt+X,t+dt)dF(X)}$

Let dt approach zero

$0=-\lambda u(m,t)- \frac{\partial u}{\partial m} \mu+\frac{\partial u}{\partial t}+\lambda \int_{-m}^ \propto {u(m+X,t)dF(X)}$

Again, the solution to $0=-\lambda u(m,t)- \frac{\partial u}{\partial m} \mu+\frac{\partial u}{\partial t}+\lambda \int {u(m+X,t)dF(X)}$ as well is the form $u(m)=1-Ce^{-\gamma m}$ when u is eventually irrelevant to t:

$0=-\lambda (1-Ce^{-\gamma m})-\mu C \gamma e^{-\gamma m}+\lambda \int {(1-Ce^{-\gamma (m+X)})dF(X)}$

Therefore,

$1 -\frac{\mu \gamma}{\lambda}=\int {e^{-\gamma X}dF(X)}=(1-p)M(-\gamma)+\frac{pa}{a-\gamma}$

$0= \int_{- \propto }^{-m} {(1-Ce^{-\gamma (m+X)})dF(X)}=pe^{-am}-Cpae^{-\gamma m} \int_{- \propto }^{-m} {e^{(a-\gamma) X}dX}$

$=(1- \frac{Ca}{a-\gamma})pe^{-am}$

Therefore $C=1- \frac{\gamma}{a}$

For time preference, the lending formula is:

$(1-\lambda dt)u(m-A- \mu dt +B, t+dt)+\lambda dt \int_{-(m-A- \mu dt)}^ \propto {u(m-A- \mu dt+B+X,t+dt)dF(X)}$

Denote $\delta dt = \frac{B}{A}-1$ then

$=(1-\lambda dt)u(m- \mu dt +A \delta dt, t+dt)+\lambda dt \int_{-(m-A- \mu dt)}^ \propto {u(m- \mu dt+A \delta dt+X,t+dt)dF(X)} =(1-\lambda dt)(u + \frac{\partial u}{\partial m} (A \delta-\mu)dt +\frac{\partial u}{\partial t}dt)+\lambda dt \int_{-(m-A- \mu dt)}^ \propto {u(m- \mu dt+A \delta dt+X,t+dt)dF(X)}$

Therefore the equation about the time preference is:

$u=(1-\lambda dt)(u + \frac{\partial u}{\partial m} (A \delta-\mu)dt +\frac{\partial u}{\partial t}dt)+\lambda dt \int_{-(m-A- \mu dt)}^ \propto {u(m- \mu dt+A \delta dt+X,t+dt)dF(X)}$

After simplification and dt approaching zero, it is

$0=-\lambda u+\frac{\partial u}{\partial m} (A \delta-\mu) +\frac{\partial u}{\partial t}+\lambda \int_{-(m-A)}^ \propto {u(m+X,t)dF(X)}$

But by the equation for u previous, then

$0=\frac{\partial u}{\partial m}A \delta-\lambda \int_{-m}^{-(m-A)} {u(m+X,t)dF(X)}$

$=C \gamma \delta Ae^{-\gamma m}-\lambda[pe^{-a(m-A)}-pe^{-am}-pe^{-\gamma m} (e^{-(a-\gamma)(m-A)}-e^{-(a-\gamma)m}) ]$

Therefore

$\delta= \frac{p\lambda}{C\gamma A} (e^{aA}-e^{(a-\gamma)A})e^{-(a-\gamma)m}$

Also note the typical phenomenon that the wealthier a person is, the less time preference it is. When A is small, it becomes:

$\delta= \frac{p\lambda}{C} e^{-(a-\gamma)m}$

Note that B/A approaches 1 as m approaches infinite.

The backward induction reasoning also suggests a dynamic of the survival probability (till some date) function, therefore a solution can be obtained numerically this way. It also distinguishes some unstable solution from the stable solution. For example, in the above Poisson process example, p=0 and C=1 is a solution which is u(m)=1, but then some disturbance in the process will cause the unstable solution slide down to the indicated stable solution, if any.