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Assume we have some money. At every point in time $0\le t \le T$, we can take either action 1 that is to keep our money until $T$ say in a bank and have an expected return of $f(t)$ or take action 2 that is to buy a lottery ticket, have an expected return of $g(t)$ and the system disappears then!

The problem is that we don't have a closed-form expression for $f(t)$ but we know that $$\frac{df(t)}{dt}-f(t)+\alpha=0.$$ Similarly, we don't know what $g(t)$ is but we know that $$g(t)=h(t)-\gamma,$$ where $$\frac{dh(t)}{dt}-h(t)+\beta=0.$$

$\alpha, \beta, \gamma>0$ and we also know that no matter what action we take, the system disappears at $t=T$ so $f(T)=h(T)=0$, so we would like to take an action that maximizes our return.

If we know that there is a time interval, or even a point in time, say $\tau$ where it's best to buy the lottery ticket, how can we compute the return corresponding to that time interval?

We can solve the two ODEs and we see that both $f(t)$ and $h(t)$ and consequently $g(t)$ are decreasing and concave in $t$, if we use $f(T)=h(T)=0$ as the boundary condition. And that's all I could do!

Can anyone give me a hint?

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  • $\begingroup$ If I understand correctly, you are looking for some time $t<T$ for which $f(t)>g(t)$, no? If that’s the case, I’d solve for both f and g and then equate the two, then solving for some $t^*$. Does that make sense? $\endgroup$
    – Kermittfrog
    Commented Jan 24, 2022 at 19:23
  • $\begingroup$ @Kermittfrog thank you. That's what I thought but the boundary condition confuses me! Should I solve both ODEs with $f(T)=h(T)=0$ as B.C. then see for what range of $t$ $f(t)>g(t)$ or should I solve the first ODE for $f(t)$ with $f(t^\ast)=h(t^\ast)-\gamma$ as the B.C. since after $t^\ast$ there is no system to study? $\endgroup$
    – user57062
    Commented Jan 24, 2022 at 19:38
  • $\begingroup$ Quick question: How does the accrual on the risk free asset work? Is $f(t)$ a short rate, i.e. the valid rate from $t\to t + dt$, or is it a yield / total return, i.e. valid from $t\to T$? What is the total payoff from the risk-free asset at any point in time? $\endgroup$
    – Kermittfrog
    Commented Jan 25, 2022 at 7:34
  • $\begingroup$ @Kermittfrog If I understood the questions correctly, the former so the rate from $t\rightarrow t+dt$. $f$ is the expected return from risk-free asset. Does this change the solution though? $\endgroup$
    – user57062
    Commented Jan 25, 2022 at 10:16

1 Answer 1

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I may be completely off, but let me give it a try.

We assume some kind of risk-neutral agent who decides on expected returns, only. As you have specified in the comments, the function $f(t)$ represents the short rate process, i.e. the interest paid in the interval $(t,t+dt)$ is $f(t)dt$. At any point in time $t$, the if we switch from investing at the risk free rate towards buying the lottery ticket with expected (total) return from that point on forward equal to $g(t)$. Both $f(T)$ and $g(T)$ are zero.

The decision maker chooses some optimal time $t^*$ to optimize their expected total future wealth, which we call $W_T$ not as to clutter notation:

$$ W_T=\left( \int_0^{t^*}f(s)ds \right)g(t^*) $$ I.e. they earn at the risk free short rate $f$ until decision time and then use up all their money, invest in the lottery ticket and expect $g(t^*)$ times whatever they invested.

Solving for $f$, we get $f(t)=\alpha+c_1e^{t}$, and the condition $f(T)=0$ fixes the constant $c_1=-\alpha e^{-T}$ so that

$$ f(t)=\alpha\left(1-e^{-(T-t)}\right) $$

Likewise,

$$ h(t)=\beta\left(1-e^{-(T-t)}\right) $$

or

$$ g(t)=h(t)-\gamma = \beta\left(1-e^{-(T-t)}\right) -\gamma $$

Effectively, $W_T$ is now a function of $t^*$. Given the parameters, we could trace out $t^*$ between $0$ and $T$ and select the value that maximizes $W_T$, or we calculate the first derivative of $W_T$ w.r.t. $t^*$, set it zero and solve for $t^*$ (checking that it's a maximum, of course), thus finding the optimal level.

Does that make sense?

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  • $\begingroup$ Thank you. Could you please explain how we can find $t^\ast$? I'm not sure if I understood "You could now solve the integral, write out the product of $\int f ds \times g$ and apply the 'standard' methods to find an optimum." Is this a problem of optimal stopping? $\endgroup$
    – user57062
    Commented Jan 25, 2022 at 12:26
  • $\begingroup$ I've updated the last paragraph. Does that help? $\endgroup$
    – Kermittfrog
    Commented Jan 25, 2022 at 12:54
  • $\begingroup$ Yes, I see it now thank you. $\endgroup$
    – user57062
    Commented Jan 25, 2022 at 13:42
  • $\begingroup$ Quick question. Should we have $W(t)=\left( \int_{t}^{t^*}f(s)ds \right)g(t^*)$ and then maximize $W(t)$ which is the expected wealth at time $t$? $\endgroup$
    – user57062
    Commented Jan 26, 2022 at 10:30
  • $\begingroup$ That depends on your problem definition,of course. I’d optimize expected final wealth $E(W_T)$ $\endgroup$
    – Kermittfrog
    Commented Jan 26, 2022 at 11:34

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