Utility-optimal leverage with costs

Question

Say I have a portfolio, $X_t$, using a leverage of $f$, such that the dynamics are given by \begin{equation} dX_t = \mu f X_t dt + \sigma f X_t dW_t \end{equation} I want to optimize the expected utility after some time $T$, $E[U(V_T)]$, and find the optimal leverage $f$. With the utility function $U(x)=\frac{x^\gamma}{\gamma}$ this is fairly easy. The SDE can be solved and the expected utility is maximized with $f^* = \frac{\mu}{\sigma^2 (1-\gamma)}$. With $\gamma=0$ and log-utility this is just the Kelly criterion.

But what if I also have to pay a constant cost $C$ such that the dynamics are \begin{equation} dX_t = (\mu f X_t - C) dt + \sigma f X_t dW_t, \quad X_t > 0 \end{equation} and $dX_t = 0$ when $X_t=0$ (i.e., I go bankrupt). The utility function would need to be altered to account for the non-zero probability of bankruptcy, so $U(x)=\frac{(x + b)^\gamma}{\gamma}$ with some $b>0$ so that the utility is bounded at bankruptcy.

Is there any way I can formulate the problem such that I can get an expression for the optimal $f^*$ that maximizes the expected utility $E[U(V_T)]$ when costs are included?

Answer 1

I hope my computations are correct.

Let $u(t,x)=\max_{(f_s)_{s\geq t}}\mathbb{E}[(b+X^{f_.}_T)^\gamma]$.

Using HJB (you have to prove that it is ok to use it).

$$0=\max_{f}\partial_t u(t,x)+(\mu f x - C)\partial_x u(t,x)+\frac{\sigma^2}{2}f^2x^2\partial_{xx}u(t,x)$$ Since $\partial_{xx}u(t,x)<0$ (prove it), maximum is hit at $f=\frac{\mu x \partial_xu(t,x)}{\sigma^2 x^2\partial_{xx}u(t,x)}$

$$0=\partial_t u(t,x)- C\partial_x u(t,x)-\frac{\mu^2}{2\sigma^2}\frac{(\partial_x u(t,x))^2}{\partial_{xx}u(t,x)}$$

We postulate $u(t,x)=f(t)(b+x)^{\gamma(t)}$

one has : $$ 0 = f'(t)(b+x)^{\gamma(t)}+\gamma'(t)(b+x)^{\gamma(t)-1} - C \gamma(t) (b+x)^{\gamma(t)-1}-\frac{\mu^2}{2\sigma^2}\frac{(b+x)^{\gamma(t) } }{\gamma(t)-1}$$

And finally we have : $$0=\gamma'(t) - C \gamma(t)\text{ and }\gamma(T)=\gamma$$ and $$f'(t)= \frac{\mu^2}{2\sigma^2(\gamma(t)-1)}\text{ and }f(T)=1$$

which leads to:

$$u(t,x)= (b+x)^{\gamma e^{-C(T-t)}}\left(1-\int_{t}^T\frac{\mu^2}{2\sigma^2(\gamma e^{-C(T-s)}-1)}ds\right)$$

Post in comments, if you see computation mistakes.