Asset Pricing Theory — Homework 1 (Solutions)

Due: October 5, 23:59
Format: Show all steps. You may use LaTeX/Markdown and attach figures.

1 Problem 1 — SDF geometry in $\mathbb{R}^2$

Consider a two-period, two-state economy with equally likely states. There is a single asset with payoff vector \[ x=(2,1) \] and price $p=1.5$. No arbitrage holds.

1.1 (1) Find one SDF that prices the asset

Pricing uses the physical measure with equal probabilities: \[ p=\mathbb{E}[m x]=\tfrac12(2m_1+m_2)=1.5 \;\;\Longleftrightarrow\;\; 2m_1+m_2=3 . \] Any $m$ on the line $2m_1+m_2=3$ is an SDF for the span of traded payoffs (here just $x$).
A convenient choice is the non-negative solution \[ m=(1.2,\,0.6),\quad \text{since } 2(1.2)+0.6=3. \] (Generally $m=(t,\,3-2t)$ with $t\in[0,1.5]$ also satisfies $m\ge 0$.)

1.2 (2) Plot the payoff line and the SDF line

Payoff line (in payoff space): \[ \mathrm{span}\{x\}=\{t(2,1):t\in\mathbb{R}\}, \] a line through the origin with slope $1/2$.
SDF line (in SDF space): \[ \{m\in\mathbb{R}^2:\tfrac12(2m_1+m_2)=1.5\} \;\equiv\; \{(m_1,m_2): m_2=3-2m_1\}, \] a line with intercept $3$ and slope $-2$. Under no-arbitrage we additionally look at the non-negative segment $m_1\in[0,1.5]$, $m_2\in[0,3]$.

(When plotting, keep payoff and SDF spaces separate; they are dual under the pricing inner product below.)

1.3 (3) Parallel/orthogonal decomposition of an SDF relative to $x$

Use the pricing inner product \[ \langle a,b\rangle=\tfrac12(a_1 b_1 + a_2 b_2). \] For any $m$ on the SDF line, define \[ m_{\parallel}=\frac{\langle m,x\rangle}{\langle x,x\rangle}\,x, \qquad m_{\perp}=m-m_{\parallel}. \] Compute the ingredients: \[ \langle m,x\rangle=\tfrac12(2m_1+m_2)=1.5=p,\qquad \langle x,x\rangle=\tfrac12(2^2+1^2)=\tfrac12(5)=2.5. \] Hence \[ m_{\parallel}=\frac{1.5}{2.5}\,x=0.6\,(2,1)=(1.2,0.6), \qquad m_{\perp}=m-m_{\parallel}. \] For the particular $m=(1.2,0.6)$ chosen above, $m_{\perp}=(0,0)$. For any other SDF on the line, $m_{\perp}\neq 0$ but satisfies $\langle m_{\perp},x\rangle=0$.

1.4 (4) Only the parallel component prices the asset

\[ p=\langle m,x\rangle=\langle m_{\parallel}+m_{\perp},x\rangle =\langle m_{\parallel},x\rangle+\underbrace{\langle m_{\perp},x\rangle}_{=0} =\langle m_{\parallel},x\rangle. \] Thus $m_{\perp}$ has no effect on the price of $x$; only the component parallel to $x$ matters. Geometrically, prices depend on the projection of $m$ onto $\mathrm{span}\{x\}$.

2 Problem 2 — Risk-neutral probabilities in a trinomial one-step model

At date 1 the risky asset takes values $\{uS,\,S,\,dS\}$ (with $u>1>d>0$). The risk-free asset has price $1$ and payoff $1+r_f$. There is a European call at-the-money (strike $K=S$) with price $C$.

Let the risk-neutral probabilities be $q_u,q_m,q_d$ for $\{uS,\,S,\,dS\}$.

2.1 (1) Risk-neutral pricing relations

Total probability: $q_u+q_m+q_d=1$.
Stock martingale condition: \[ S=\frac{1}{1+r_f}\big(q_u\,uS+q_m\,S+q_d\,dS\big) \;\Longleftrightarrow\; 1+r_f=q_u\,u+q_m+q_d\,d. \]
Call price (since $K=S$): payoff is $(uS-S)_+=(u-1)S$ in the up state, and $0$ otherwise: \[ C=\frac{1}{1+r_f}\,q_u\,(u-1)S. \]

2.2 (2) Solution for $(q_u,q_m,q_d)$

From the call price, \[ \boxed{\; q_u=\frac{C(1+r_f)}{(u-1)S}\; }. \] Use the stock martingale condition with $q_m=1-q_u-q_d$: \[ 1+r_f=q_u u+(1-q_u-q_d)+q_d d =1+q_u(u-1)+q_d(d-1). \] Solve for $q_d$: \[ \boxed{\; q_d=\frac{\,r_f - q_u(u-1)\,}{\,d-1\,}\; }. \] Then \[ \boxed{\; q_m=1-q_u-q_d \; }. \]

Admissibility (no-arbitrage) conditions. For a valid risk-neutral measure, \[ q_u,q_m,q_d\in[0,1]. \] Given $d<1$, the denominator $d-1<0$; thus $q_d\ge 0$ requires $r_f - q_u(u-1)\le 0$, i.e. \[ q_u\ge \frac{r_f}{u-1} \quad\Longleftrightarrow\quad C \;\ge\; \frac{r_f}{1+r_f}\,S \;\;\text{(after substituting $q_u$)}. \] Combined with $q_u\le 1$, we also need \[ 0 \;\le\; C \;\le\; \frac{(u-1)S}{1+r_f}. \] Finally check $q_m=1-q_u-q_d\ge 0$ after plugging the expressions above; this places the usual additional (model-consistent) bounds on $(u,d,r_f,C)$.

(Remark: In a complete market you would typically have two traded derivatives (or one plus the stock) to pin down all three $q$’s; here the ATM call and the stock/risk-free pair suffice because the middle-state payoff of the call is zero.)

--- title: "Asset Pricing Theory — Homework 1 (Solutions)" format: html toc: true number-sections: true --- **Due:** October 5, 23:59 **Format:** Show all steps. You may use LaTeX/Markdown and attach figures. --- ## Problem 1 — SDF geometry in $\mathbb{R}^2$ Consider a two-period, two-state economy with equally likely states. There is a single asset with payoff vector $$ x=(2,1) $$ and price $p=1.5$. No arbitrage holds. ### (1) Find one SDF that prices the asset Pricing uses the physical measure with equal probabilities: $$ p=\mathbb{E}[m x]=\tfrac12(2m_1+m_2)=1.5 \;\;\Longleftrightarrow\;\; 2m_1+m_2=3 . $$ Any $m$ on the line $2m_1+m_2=3$ is an SDF for the span of traded payoffs (here just $x$). A convenient choice is the non-negative solution $$ m=(1.2,\,0.6),\quad \text{since } 2(1.2)+0.6=3. $$ (Generally $m=(t,\,3-2t)$ with $t\in[0,1.5]$ also satisfies $m\ge 0$.) ### (2) Plot the payoff line and the SDF line - **Payoff line (in payoff space):** $$ \mathrm{span}\{x\}=\{t(2,1):t\in\mathbb{R}\}, $$ a line through the origin with slope $1/2$. - **SDF line (in SDF space):** $$ \{m\in\mathbb{R}^2:\tfrac12(2m_1+m_2)=1.5\} \;\equiv\; \{(m_1,m_2): m_2=3-2m_1\}, $$ a line with intercept $3$ and slope $-2$. Under no-arbitrage we additionally look at the non-negative segment $m_1\in[0,1.5]$, $m_2\in[0,3]$. *(When plotting, keep payoff and SDF spaces separate; they are dual under the pricing inner product below.)* ### (3) Parallel/orthogonal decomposition of an SDF relative to $x$ Use the pricing inner product $$ \langle a,b\rangle=\tfrac12(a_1 b_1 + a_2 b_2). $$ For any $m$ on the SDF line, define $$ m_{\parallel}=\frac{\langle m,x\rangle}{\langle x,x\rangle}\,x, \qquad m_{\perp}=m-m_{\parallel}. $$ Compute the ingredients: $$ \langle m,x\rangle=\tfrac12(2m_1+m_2)=1.5=p,\qquad \langle x,x\rangle=\tfrac12(2^2+1^2)=\tfrac12(5)=2.5. $$ Hence $$ m_{\parallel}=\frac{1.5}{2.5}\,x=0.6\,(2,1)=(1.2,0.6), \qquad m_{\perp}=m-m_{\parallel}. $$ For the particular $m=(1.2,0.6)$ chosen above, $m_{\perp}=(0,0)$. For any other SDF on the line, $m_{\perp}\neq 0$ but satisfies $\langle m_{\perp},x\rangle=0$. ### (4) Only the parallel component prices the asset $$ p=\langle m,x\rangle=\langle m_{\parallel}+m_{\perp},x\rangle =\langle m_{\parallel},x\rangle+\underbrace{\langle m_{\perp},x\rangle}_{=0} =\langle m_{\parallel},x\rangle. $$ Thus $m_{\perp}$ has no effect on the price of $x$; only the component parallel to $x$ matters. Geometrically, prices depend on the projection of $m$ onto $\mathrm{span}\{x\}$. --- ## Problem 2 — Risk-neutral probabilities in a trinomial one-step model At date 1 the risky asset takes values $\{uS,\,S,\,dS\}$ (with $u>1>d>0$). The risk-free asset has price $1$ and payoff $1+r_f$. There is a European call **at-the-money** (strike $K=S$) with price $C$. Let the risk-neutral probabilities be $q_u,q_m,q_d$ for $\{uS,\,S,\,dS\}$. ### (1) Risk-neutral pricing relations - **Total probability:** $q_u+q_m+q_d=1$. - **Stock martingale condition:** $$ S=\frac{1}{1+r_f}\big(q_u\,uS+q_m\,S+q_d\,dS\big) \;\Longleftrightarrow\; 1+r_f=q_u\,u+q_m+q_d\,d. $$ - **Call price (since $K=S$):** payoff is $(uS-S)_+=(u-1)S$ in the up state, and $0$ otherwise: $$ C=\frac{1}{1+r_f}\,q_u\,(u-1)S. $$ ### (2) Solution for $(q_u,q_m,q_d)$ From the call price, $$ \boxed{\; q_u=\frac{C(1+r_f)}{(u-1)S}\; }. $$ Use the stock martingale condition with $q_m=1-q_u-q_d$: $$ 1+r_f=q_u u+(1-q_u-q_d)+q_d d =1+q_u(u-1)+q_d(d-1). $$ Solve for $q_d$: $$ \boxed{\; q_d=\frac{\,r_f - q_u(u-1)\,}{\,d-1\,}\; }. $$ Then $$ \boxed{\; q_m=1-q_u-q_d \; }. $$ **Admissibility (no-arbitrage) conditions.** For a valid risk-neutral measure, $$ q_u,q_m,q_d\in[0,1]. $$ Given $d<1$, the denominator $d-1<0$; thus $q_d\ge 0$ requires $r_f - q_u(u-1)\le 0$, i.e. $$ q_u\ge \frac{r_f}{u-1} \quad\Longleftrightarrow\quad C \;\ge\; \frac{r_f}{1+r_f}\,S \;\;\text{(after substituting $q_u$)}. $$ Combined with $q_u\le 1$, we also need $$ 0 \;\le\; C \;\le\; \frac{(u-1)S}{1+r_f}. $$ Finally check $q_m=1-q_u-q_d\ge 0$ after plugging the expressions above; this places the usual additional (model-consistent) bounds on $(u,d,r_f,C)$. *(Remark: In a complete market you would typically have two traded derivatives (or one plus the stock) to pin down all three $q$'s; here the ATM call and the stock/risk-free pair suffice because the middle-state payoff of the call is zero.)* ---

Asset Pricing Theory — Homework 1 (Solutions)

1 Problem 1 — SDF geometry in \(\mathbb{R}^2\)

1.1 (1) Find one SDF that prices the asset

1.2 (2) Plot the payoff line and the SDF line

1.3 (3) Parallel/orthogonal decomposition of an SDF relative to \(x\)

1.4 (4) Only the parallel component prices the asset

2 Problem 2 — Risk-neutral probabilities in a trinomial one-step model

2.1 (1) Risk-neutral pricing relations

2.2 (2) Solution for \((q_u,q_m,q_d)\)