Asset Allocation and Portfolio Management HW1

• Problem 1
  • Sufficiency
  • Necessity
• Problem 2
• Problem 3
  • $(a)\Rightarrow (b)$
  • $(c) \Rightarrow (a)$
  • $(b)$ and $(c)$ are equivalent
    • $(c) \Rightarrow (b)$
    • $(b) \Rightarrow (c)$
• Problem 4
  • Same decisions
  • Same certainty-equivalents

Problem 1

Sufficiency

If an agent is risk-averse, then

$\begin{aligned} &\mathbb{E}[u(\omega + \tilde{z})] \leq u(\omega) \quad \forall\omega,\forall\tilde{z} \text{ with } \mathbb{E}[\tilde{z}] = 0 \\ \Rightarrow &\mathbb{E}\left[ u(\omega + \tilde{z}) \right] \leq u(\omega + \mathbb{E}[\tilde{z}]) \\ \Rightarrow &\mathbb{E}\left[ u(\omega + \tilde{z}) \right] \leq u( \mathbb{E}[\omega +\tilde{z}]) \\ \text{Let $X = \omega +\tilde{z}$,} \\ \Rightarrow &\mathbb{E}\left[ u(X) \right] \leq u( \mathbb{E}[X]) \\ \Rightarrow &u() \text{ is concave.} \end{aligned}$

Necessity

If $u()$ is concave, then

$\begin{aligned} &\mathbb{E}\left[ u(X) \right] \leq u( \mathbb{E}[X]) \\ \text{Let $X = \omega +\tilde{z}$,} & \text{ where $\mathbb{E}[\tilde{z}] = 0$, $\omega$ is non-random} \\ \Rightarrow &\mathbb{E}\left[ u(\omega + \tilde{z}) \right] \leq u( \mathbb{E}[\omega +\tilde{z}]) \\ \Rightarrow &\mathbb{E}\left[ u(\omega + \tilde{z}) \right] \leq u(\omega + \mathbb{E}[\tilde{z}]) \\ \Rightarrow &\mathbb{E}[u(\omega + \tilde{z})] \leq u(\omega) \quad \forall\omega,\forall\tilde{z} \text{ with } \mathbb{E}[\tilde{z}] = 0 \\ \Rightarrow &\text{This agent is risk-averse.} \end{aligned}$

Problem 2

$\begin{aligned} &u(w) = − exp(−κw)/κ \\ \Rightarrow & u^\prime(w) = exp(−κw)\\ \Rightarrow & u^{\prime\prime}(w) = -κexp(−κw)\\ \end{aligned}$

So

$A(w) = -\frac{u^{\prime\prime}(w)}{u^\prime(w)} = \kappa \tag{2.1}$

And we know

$\begin{aligned} &\mathbb{E}\left[ u(w+\tilde{z}) \right] =\mathbb{E}\left[ u(\tilde{w}) \right] \quad \text{$\tilde{w}$ is normally distributed with mean $\mu$ and variance $\Sigma^2$} \\ \Rightarrow &\mathbb{E}\left[ u(\tilde{w}) \right] = -\frac{1}{\kappa}\mathbb{E}\left[ exp(−κ\tilde{w}) \right] \\ \Rightarrow & u(\tilde{w})= -\frac{1}{\kappa}exp(−κ\mu + \frac{\kappa^2\sigma^2}{2})\\ \end{aligned}$

So

$u(\tilde{w})= -\frac{1}{\kappa}exp(−κ(\mu - \frac{\kappa\sigma^2}{2})) \tag{2.2}$

And we know

$u(w-\Pi) = − exp(−κ(w-\Pi))/κ \tag{2.3}$

Combined $(2.2)$ and $(2.3)$ we can get

$\Pi = \frac{\kappa\sigma^2}{2} \tag{2.4}$

Plug $(2.1)$ into $(2.4)$ we can get

$\Pi = \frac{\sigma^2}{2}A(w)$

So the Arrow-Pratt approximation is exact.

Problem 3

$(a)\Rightarrow (b)$

It is equivalent to prove that not (b) implies not (a).

not (b) is

$\exists w, \text{ s.t. }A_v(w) < A_u(w)$

$\text{Let the point s.t.} A_v(w) < A_u(w) \text{ be }x \text{. i.e. } A_v(x) < A_u(x)$.

$\text{Let } v(x) = t \quad \Rightarrow x = v^{-1}(t)$

First, We want to show that $u(v^{-1}(\bullet)) \text{ is strictly concave at } t$

$\begin{aligned} &A_v(x) < A_u(x) \\ \Rightarrow &A_v(x) - A_u(x) < 0\\ \Rightarrow &-\frac{v^{\prime\prime}(x)}{v^\prime(x)} + \frac{u^{\prime\prime}(x)}{u^\prime(x)} < 0 \\ \Rightarrow &\frac{d}{dw}log\frac{u^\prime(x)}{v^\prime(x)} < 0 \\ \Rightarrow &log\frac{u^\prime(x)}{v^\prime(x)} \text{ is decreasing} \\ \Rightarrow &\frac{u^\prime(x)}{v^\prime(x)} \text{ is decreasing} \\ \Rightarrow &\frac{u^\prime(v^{-1}(t))}{v^\prime(v^{-1}(t))} = \frac{d}{dt}u(v^{-1}(t)) \text{ is decreasing} \\ &\text{Since } \frac{d}{dt}u(v^{-1}(t)) =u^\prime(v^{-1}(t))[\frac{d(v^{-1}(t))}{dt}] =u^\prime(v^{-1}(t))[\frac{1}{v^\prime(v^{-1}(t))}] = \frac{u^\prime(v^{-1}(t))}{v^\prime(v^{-1}(t))} \\ \Rightarrow &\frac{d^2}{dt^2}u(v^{-1}(t)) < 0 \\ \Rightarrow &u(v^{-1}(\bullet)) \text{ is strictly concave at } t \end{aligned}$

According to Jensen’s inequality

$\mathbb{E}\left[ u(v^{-1}(t)) \right] < u(v^{-1}(\mathbb{E}\left[ t \right]))\tag{3.1}$

And we know

$\mathbb{E}\left[ u(w+\tilde{z}) \right] = u(w - \Pi_u)$

$\mathbb{E}\left[ v(w+\tilde{z}) \right] = v(w - \Pi_v)$

So

$\Pi_u = w-u^{-1}(\mathbb{E}\left[ u(w+\tilde{z}) \right])$

$\Pi_v = w-v^{-1}(\mathbb{E}\left[ v(w+\tilde{z}) \right])$

$\begin{aligned} \Pi_v - \Pi_u &= -v^{-1}(\mathbb{E}\left[ v(w+\tilde{z}) \right]) + u^{-1}(\mathbb{E}\left[ u(w+\tilde{z}) \right]) \\ &= -v^{-1}(\mathbb{E}\left[ v(w+\tilde{z}) \right]) + u^{-1}(\mathbb{E}\left[ u(w+\tilde{z}) \right]) \end{aligned}$

Let $v(w+\tilde{z}) = t$, i.e. the risk is $\tilde{z} = v^{-1}(t) - w$. And we know $\mathbb{E}\left[ \tilde{z} \right] = v^{-1}(t) - w = 0$, so

$w = v^{-1}(t)$

$\tilde{z} = 0$

Then

$\Pi_v - \Pi_u = -v^{-1}(\mathbb{E}\left[ t \right]) + u^{-1}(\mathbb{E}\left[ u(v^{-1}(t)) \right]) \tag{3.2}$

Substituting (3.1) into (3.2), we obtain

$\begin{aligned} &\Pi_v - \Pi_u < -v^{-1}(\mathbb{E}\left[ t \right]) + u^{-1}(u(v^{-1}(\mathbb{E}\left[ t \right]))) \\ \Rightarrow &\Pi_v - \Pi_u < -v^{-1}(\mathbb{E}\left[ t \right]) + v^{-1}(\mathbb{E}\left[ t \right])\\ \Rightarrow &\Pi_v - \Pi_u < 0 \\ \Rightarrow &\Pi_v < \Pi_u \end{aligned}$

So

$\Pi_v < \Pi_u \quad \text{when } w = v^{-1}(t), \tilde{z} = 0 \tag{3.3}$

(a) implies that $\Pi_v \geq \Pi_u$ for any risk. so (3.3) implies not (a)

So we have proved that

$\text{not}(b) \Rightarrow \text{not}(a)$

which is equivalent to

$(a)\Rightarrow (b)$

$(c) \Rightarrow (a)$

Given $(c)$, we have

$v(w-\Pi_v) = \mathbb{E}\left[ v(w+\tilde{z}) \right] = \mathbb{E}\left[ \phi(u(w+\tilde{z})) \right]$

$\begin{aligned} \phi \text{ is concave} \Rightarrow \mathbb{E}\left[ \phi(u(w+\tilde{z})) \right] &\leq \phi(\mathbb{E}\left[ u(w+\tilde{z}) \right]) \\ &=\phi(u(w-\Pi_u)) \\ &=v(w-\Pi_u) \end{aligned}$

$\Rightarrow v(w-\Pi_v) \leq v(w-\Pi_u)$

And $v$ is utility function, so $v$ is increasing, which implies $\Pi_v \geq \Pi_u$ for any risk. So agent $v$ is more risk-averse than agent $u$.

$(b)$ and $(c)$ are equivalent

$u$ is increasing, i.e. $u^\prime \geq 0$.

$\begin{aligned} v(w) &= \phi(u(w)) \\ \Rightarrow v^{\prime}(w) &= \phi^{\prime}(u(w))u^\prime(w) \\ \Rightarrow v^{\prime\prime}(w) &= \phi^{\prime\prime}(u(w))(u^\prime(w))^2 +\phi^{\prime}(u(w))u^{\prime\prime}(w) \\ \Rightarrow A_v(w) &= -\frac {\phi^{\prime\prime}(u(w))(u^\prime(w))^2 +\phi^{\prime}(u(w))u^{\prime\prime}(w)} {\phi^{\prime}(u(w))u^\prime(w)} \\ &=A_u(w) - \frac{\phi^{\prime\prime}(u(w))u^\prime(w)} {\phi^{\prime}(u(w))} \end{aligned}$

$(c) \Rightarrow (b)$

$\phi^{\prime\prime} \leq0$ and $\phi^\prime > 0$ $\Rightarrow$ $\frac{\phi^{\prime\prime}(u(w))u^\prime(w)} {\phi^{\prime}(u(w))} \leq 0$. So $A_v(w) \geq A_u(w)$ for all $w$.

$(b) \Rightarrow (c)$

$A_v(w) \geq A_u(w)$ for all $w$ means $\frac{\phi^{\prime\prime}(u(w))u^\prime(w)} {\phi^{\prime}(u(w))} \leq 0$. $u$ and $v$ are utility function, so $\phi$ should be increasing function, i.e. $\phi^\prime > 0$. So we have $\phi^{\prime\prime} \leq 0$ i.e. Function v is a concave transformation of function u.

Problem 4

Same decisions

$v(x) = a + bu(x)$ for all $x$, for some pair of scalars $a$ and $b$, where $b$ >0.

$\begin{aligned} &v(x) > v(y)\\ \Leftrightarrow& a + bu(x) > a+bu(y) \\ \Leftrightarrow& u(x) > u(y) \end{aligned}$

This deduction also holds for equality.

$\begin{aligned} &v(x) = v(y)\\ \Leftrightarrow& a + bu(x) = a+bu(y) \\ \Leftrightarrow& u(x) = u(y) \end{aligned}$

So a decision-maker with utility function $v()$ makes the same decisions as a decision maker with utility function $u()$

Same certainty-equivalents

$\begin{aligned} &e \text{ is certainty-equivalent of $\tilde{z}$ for $v()$} \\ \Leftrightarrow &v(w + e) = \mathbb{E}\left[ v(w+\tilde{z}) \right]\\ \Leftrightarrow &a + bu(w+e) = \mathbb{E}\left[ a+bu(w+\tilde{z}) \right]\\ \Leftrightarrow &a + bu(w+e) = a+b\mathbb{E}\left[ u(w+\tilde{z}) \right]\\ \Leftrightarrow &u(w+e) = \mathbb{E}\left[ u(w+\tilde{z}) \right]\\ \Leftrightarrow &e \text{ is certainty-equivalent of $\tilde{z}$ for $u()$} \\ \end{aligned}$

So a decision-maker with utility function $v()$ has the same certainty-equivalents as a decision maker with utility function $u()$