Problem 1.

Because of the stochastic shock, we cannot define a steady state as a point. Rather, we define a steady-state distribution. To compute it, we need to introduce a notion of transition probabilities and the associated diffusion equations.
Definition 1. The transition probability of a Markov process $\left(X_{t}\right)$ is
$$P\left(s, X_{s}, t, B\right)=P\left(X_{t} \in B \mid X_{s}\right) \text { with prob. } 1$$
where $P\left(X_{t} \in B \mid X_{s}\right)$ is the conditional distribution function of a Borel set $B$ given $X_{s}$
The density of $P$ is the transitional density and it is denoted by $p(s, x, t, y)$ for transition from $\left(s, X_{s}=x\right)$ to $\left(t, X_{t}=y\right)$.

(a) Prove the following proposition:
Proposition 1. Suppose that $F$ is the unique bounded solution of the following PDE.
$$F_{s}+\mu F_{x}+\frac{1}{2} \sigma^{2} F_{x x}=0 \text { for } s \in[0, t), \quad F(t, x)=\mathbb{1}{B}(x)$$ Then, $F$ has the stochastic representation $$F(s, x)=E{s, x}\left[\mathbb{1}{B}\left(X{t}\right)\right]$$
where the process $\left(X_{t}\right){0 \leq u \leq s}$ is defined to be the solution of the following SDE: $$d X{u}=\mu\left(u, X_{u}\right) d u+\sigma\left(u, X_{u}\right) d W_{u}, \quad X_{s}=x$$
(b) Express $F(s, x)=E_{s, x}\left[\mathbb{1}{B}\left(X{t}\right)\right]$ using the transition probability.
(c) It is known that the converse of the above proposition holds. State the converse.
(d) From the converse in (c), prove the following:
Proposition 2. Let $X$ be a solution to the following SDE.
$$d X_{t}=\mu\left(t, X_{t}\right) d t+\sigma\left(t, X_{t}\right) d W_{t}, \quad X_{s}=x$$
Assume that $P(s, x ; t, d y)$ has a density $p(s, x ; t, y) d y$. Then, we have
$$\left(p_{s}+\mathcal{A} p\right)(s, x ; t, y)=0 \text { for }(s, x) \in(0, t) \times \mathbb{R}, \quad p(s, x ; t, y) \rightarrow \delta_{y} \text { as } s \rightarrow t$$
where $\mathcal{A}$ is the Itŏ operator.
This is called a Kolmogorov backward equation.
(e) Consider the following proposition.
Proposition 3. Let $X$ be a solution to the following SDE.
$$d X_{t}=\mu\left(t, X_{t}\right) d t+\sigma\left(t, X_{t}\right) d W_{t}, \quad X_{s}=y$$
Assume that $P(s, x ; t, d y)$ has a density $p(s, x ; t, y)$. Then, we have
$$p_{s}=\left(\mathcal{A}^{}\right) p(s, x, t, y) \text { for }(t, y) \in(0, T) \times \mathbb{R}, \quad p(s, x, t, y) \rightarrow \delta_{y} \quad \text { as } t \rightarrow s$$ where $\mathcal{A}^{}$ is the adjoint operator defined by
$$\left(\mathcal{A}^{*} F\right)(t, x)=-\frac{\partial}{\partial x}[\mu(t, x) F(t, x)]+\frac{1}{2} \frac{\partial^{2}}{\partial x^{2}}\left[\sigma^{2}(t, x) F(t, x)\right]$$

This equation is known as the Fokker-Planck equation. Explain that this is actually a Kolmogorov forward equation (that is the mirror image of the Kolmogorov backward equation). Derive this forward equation from the backward equation.

Problem 2.

Let $X_{t}$ be the solution of the following SDE:
$$d X_{t}=\beta\left(X_{t}\right) d t+\alpha\left(X_{t}\right)^{1 / 2} d W_{t}, \quad X_{0}=x_{0}$$
where $W_{t}$ is a Wiener process, $\alpha, \beta$ are $C^{2}$ functions on $[0, \infty)$, and $\alpha(x)>0$ on $(0, \infty)$, and $\alpha(0)=\beta(0)=0 .$
(a) Show that the evolution of per capita capital satisfies the above
$$d k_{t}=\left[s f\left(k_{t}\right)-\left(n-\sigma^{2}\right) k_{t}\right] d t-\sigma k_{t} d Z_{t}$$
(b) What is the corresponding Fokker-Planck forward equation?
(c) Define the steady-state distribution to be
$$\pi(x):=\lim _{t \rightarrow \infty} p(x, t)$$
What is the corresponding Fokker-Planck forward equation for $\pi(x)$ ?

Problem 3.

Note that the equation you derived in the above is the second-order linear homogenous differential equation with non-constant coefficients. Solve for $\pi$ to show that a (general) solution $\pi(x)$ can be written as follows:
$$\pi(x)=m_{1} I_{1}(x)+m_{2} I_{2}(x) \text { with } \int_{0}^{\infty} \pi(x) d x=1$$
where
\begin{aligned} &I_{1}(x)=\frac{1}{\alpha(x)} \exp \left[2 \int^{x} \frac{\beta(y)}{\alpha(y)} d y\right] \ &I_{2}(x)=I_{1}(x) \int_{0}^{x} \exp \left[2 \int_{0}^{y} \frac{\beta(s)}{\alpha(s)} d s\right] d y \end{aligned}
State the solution procedure so that anyone can understand it. The expression for $I_{2}$ differs from Equation (B.4) in page 390. Which one is correct, the above or Equation (B.4) or both?
Show that $m_{2}=0$ must hold.

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