Understanding HJB equation for the infinite horizon consumption control problem

Question

Context
Given the following maximization problem as well as wealth dynamics $$\max\mathbb{E}\left[\int_0^{\infty} \frac{1}{\gamma} e^{-\beta t} c_t^\gamma \mathrm{d} t\right]$$ $$\mathrm{d} X_t=X_t\left(r+\pi_t(\alpha-r)\right) \mathrm{d} t-c_t \mathrm{~d} t+\pi_t X_t \sigma \mathrm{d} W_t, \quad X_0=x>0$$ we need to prove that Hamilton–Jacobi–Bellman (HJB) equation is given by

$$\sup _{c \geq 0, \pi}\left\{\frac{1}{\gamma} c^\gamma-\beta V(x)+x(r+\pi(\alpha-r)) V^{\prime}(x)-c V^{\prime}(x)+\frac{1}{2} \pi^2 x^2 \sigma^2 V^{\prime \prime}(x)\right\}=0$$ where $0<\gamma<1, \beta>0, c_t \geq 0, X_t \geq >0, t\geq 0$

My problem
I know that the HJB equation in the general form is given as $$\frac{\partial V}{\partial t}(t, x)+\sup _{u \in U}\left(F(t, x, u)+\mathcal{A}^u V(t, x)\right)=0 \\ V(T,x) = \Phi(x)$$ which is roughly $$\frac{\partial V}{\partial t}(t, x)+\sup _{u \in U}\left(F(t, x, u)+(\text{(values in front of dt )}) V(t, x)' + \frac 12 (\text{(values in front of dWt )}^2) V(t, x)''\right)=0$$

In our case $\Phi = 0$ as there is an infinite horizon and $F(t, x, u) = \frac{1}{\gamma} e^{-\beta t} c_t^\gamma $.

Even though I agree with the term $$x(r+\pi(\alpha-r)) V^{\prime}(x)-c V^{\prime}(x)+\frac{1}{2} \pi^2 x^2 \sigma^2 V^{\prime \prime}(x)$$ in the equation we need to proof, the equation does not match completely :

• No $\frac{\partial V}{\partial t}(t, x)$ term in front
• In $\sup(\cdot)$, $F(t, x, u)$ does not contain the exponent (though I think it might be due to plugging in $e^{-\infty} = 1$)
• Not clear why $-\beta V(x)$ appears

Answer 1

This is a discounted optimal control problem. Define for simplicity

$$dX_t=f(X_t,c_t,\pi_t)dt+g(X_t,\pi_t)dW_t$$

where our control inputs are both $c_t$ and $\pi_t$. The HJB equation is given by

$$ \frac{\partial V}{\partial t}(t, x)+\sup _{c\ge0,\pi}\left\{\frac{\partial V}{\partial x}(t, x)f(x,c,\pi)+\dfrac{1}{2}\frac{\partial^2 V}{\partial x^2}(t, x)g(X_t,\pi)^2+\frac{1}{\gamma} e^{-\beta t} c^\gamma\right\}=0. $$

Now, if we let $V(t,x)=e^{-\beta t}W(x)$, the above expression becomes

$$ e^{-\beta t}\left(-\beta W(x)+\sup _{c\ge0,\pi}\left\{W'(x)f(x,c,\pi)+\dfrac{1}{2}W''(x)g(X_t,\pi)^2+\frac{1}{\gamma} c^\gamma\right\}\right)=0 $$ and the exponential term can be dropped as it is always nonzero to get the desired result

$$ \sup _{c\ge0,\pi}\left\{-\beta W(x)+W'(x)f(x,c,\pi)+\dfrac{1}{2}W''(x)g(X_t,\pi)^2+\frac{1}{\gamma} c^\gamma\right\}=0. $$

So, to summarize:

• The term $\frac{\partial V}{\partial t}(t, x)$ becomes $-\beta W(x)$.
• The exponential term is factored out.