## UtilityFunctionExample

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 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: The solution of and is of the form for some C and . By and denote M as the moment generating function of positive flow, it follows Note that this equation has the non-zero solution when the expectation of flow is greater than zero. By , it follows Therefore For the time preference: Therefore the relationship between A and B is Rearrange then For tiny A in comparison with m, then 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: Let dt approach zero Again, the solution to as well is the form when u is eventually irrelevant to t: Therefore, Therefore For time preference, the lending formula is: Denote then $\dpi{0} \bg_white =(1-\lambda dt)u(m- \mu dt +A \delta dt, t+dt)+\lambda dt \int_{-(m-A- \mu dt)}^ \propto {u(m- \mu dt+A \delta dt+X,t+dt)dF(X)} =(1-\lambda dt)(u + \frac{\partial u}{\partial m} (A \delta-\mu)dt +\frac{\partial u}{\partial t}dt)+\lambda dt \int_{-(m-A- \mu dt)}^ \propto {u(m- \mu dt+A \delta dt+X,t+dt)dF(X)}$ Therefore the equation about the time preference is: After simplification and dt approaching zero, it is But by the equation for u previous, then Therefore Also note the typical phenomenon that the wealthier a person is, the less time preference it is. When A is small, it becomes: 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.