I have a market with safe rate r and risky asset S $$ \frac{dS_t}{S_t}=(r+Y_t)dt+\sigma dW_t \quad \quad (1)$$ $$ dY_t = - \lambda Y_t +dB_t \quad \quad (2)$$ where W, B are Brownian Motions with correlation $\rho$.

I am deriving the HJB equation for the utility maximization problem $$\max_{X} E[logX_T] \quad \quad (3)$$

The V function depends on t, X and Y. HJB is going to be the drift of the dV function derived using Ito formula. Hence the initial dV function will be of the following form $$dV(t,X_t,Y_t)=V_t dt + V_x dX_t + V_ydY_t + \frac{1}{2} \big{(} V_{xx} d \langle X \rangle_t + 2 V_{xy} d \langle X,Y \rangle_t + V_{yy} d \langle Y \rangle_t \big{)} \quad \quad (4)$$

so I supposed to get the below HJB equation $$\sup_{\pi} \big{(} V_t + V_x (r+y \pi)x - \lambda y V_y + \frac{1}{2} (V_{xx}\sigma^2 \pi^2 x^2 + 2V_{xy} \sigma \rho \pi x + V_{yy} ) \big{)} =0 \quad \quad (5)$$ using the follwing dynamics $$dX_t=X_t(r+ \pi y )dt + \pi \sigma X_t dW_t \quad \quad (6)$$ $$$$

Now, I don't quite follow how the $dX_t$ is created and how the $\pi$ comes into play here. My guess is that it represents portfolio weight and $dX_t$ represents the change in the capital.

In the theory (study notes) I have a general formula for one variable and it looks like $$dV(X_t^{\Pi},t)=V_t + X_t^{\Pi} (r + \Pi' \mu)V_x + \frac{\Pi_t' \Sigma \Pi_t}{2} (X_t^{\Pi})^2 V_{xx})dt + V_x X_t^{\Pi} \Pi_t' \sigma dW_t \quad \quad (7)$$

The use of the $\pi, \Pi$ components confuses me. Could anybody clarify of the logic behind the use of $\pi$ and $dX_t$ derivation please?

  • $\begingroup$ What is $X_t$? It appears that you have used without definition. $\endgroup$
    – Gordon
    May 16, 2016 at 13:08
  • $\begingroup$ $X_t$ is the stochastic wealth, it seems that it was derived as $$\frac{dX_t}{X_t} =\pi (\frac{dS_t}{S_t} - r_t dt) + r dt = \pi ((r+Y_t)dt+\sigma dW_t - r_t dt) + r dt = (r+\pi Y_t)dt + \pi \sigma dW_t) $$, but I am still confused with the way the $\Pi'$ and $\Pi$ are used in (7) $\endgroup$
    – Michal
    May 16, 2016 at 13:29

1 Answer 1


We assume that \begin{align*} \frac{dX_t}{X_t} &= (r+\pi Y_t)dt + \pi\sigma dW_t,\tag{1}\\ dY_t &= -\lambda Y_t + dB_t.\tag{2} \end{align*} From $(2)$, \begin{align*} Y_t = Y_0 e^{-\lambda t}+ e^{-\lambda t}\int_0^t e^{\lambda u} dB_u. \end{align*} Moreover, from $(1)$, \begin{align*} \ln X_T &= \ln X_0 + (r-\frac{1}{2}\pi^2\sigma^2)T + \pi \int_0^TY_t dt + \pi\sigma W_T\\ &=\ln X_0 + (r-\frac{1}{2}\pi^2\sigma^2)T +\pi Y_0 \int_0^Te^{-\lambda t} dt \\ &\qquad\quad \ \ \, +\pi\int_0^Te^{-\lambda t}\int_0^t e^{\lambda u} dB_u dt + \pi\sigma W_T. \end{align*} Then \begin{align*} E(\ln X_T) &= \ln X_0 + (r-\frac{1}{2}\pi^2\sigma^2)T +\pi Y_0 \int_0^Te^{-\lambda t} dt \\ &=\ln X_0 + (r-\frac{1}{2}\pi^2\sigma^2)T + \frac{\pi Y_0}{\lambda}(1-e^{-\lambda T})\\ &=\ln X_0 - rT -\frac{1}{2}\sigma^2T \left(\pi - \frac{Y_0}{\lambda\sigma^2T}(1-e^{-\lambda T})\right)^2 +\frac{Y_0^2}{\lambda^2\sigma^2T}(1-e^{-\lambda T})^2. \end{align*} That is, \begin{align*} \max_{\pi}E(\ln X_T) &= \ln X_0 - rT + \frac{Y_0^2}{\lambda^2\sigma^2T}(1-e^{-\lambda T})^2, \end{align*} which is achieved at \begin{align*} \pi = \frac{Y_0}{\lambda\sigma^2T}(1-e^{-\lambda T}). \end{align*}


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