Numeraire Pricing and Hedging in a One-Asset Diffusion Model

# Completeness and Hedging Question


# Completeness and Hedging Question












A question in some private notes I'm struggling to work through (exam. prep.). (iii) is where I hit a wall with my understanding & I'm lost thereafter. Any help/clarification gratefully received.

...

Consider a financial market with $d = 1$ risk security, whose price $S^1$ is determined by $$dS_t^1 = (1/{S_t^1})dt + dW_t^1$$

In addition, the risk-free rate is $r_f=0$, so the price of the money-market account is $S^0 = 1$.

i) Derive an expression for the market price of risk

ii) Derive a stochastic differential equation for the numeraire portfolio

iii) Use the real-world pricing forumla to derive an expression for the price of a zero-coupon bond (ZCB) with a face-value of $1 (Note: Notice the ZCB is an equity derivative in this model!)

iv) Derive expression for the weights of the hedge portfolio for the ZCB

v) Consider a portfolio consisting of a long position in the hedge portfolio above, and a short position in the money-market account, structured in such a way that its initial value is zero. What is the final payoff of this portfolio? What type of arbitrage is it?

vi) Does the model in question satisfy NA$_+$? Justify. vii) Does the model in question satisfy NA? Justify. viii) Does the model in question satisfy NUPBR? Justify. ix) Does the model in question satisfy NFLVR? Justify.

## Answer by user16651 (score 1)

https://quant.stackexchange.com/a/18525

$i,ii,iii)$ Define $x_t$ by $x_t=(S_t^1)^2$. Ito's formula give us \begin{align} & dx_t=d(S_t^1)^2=2S_t^1dS_t+\frac{1}{2}(2)d[S_t^1,S_t^1](t)\\ & dx_t=d(S_t^1)^2=2S_t^1(\frac{1}{S_t^1}+dW_t)+dW_tdW_t=3dt+2S_t^1dW_t\\ \end{align} then \begin{align} & dx_t=3\ dt+2\sqrt x_t\ dW_t\\ \end{align} The following portfolio is constructed: we buy a bond of dollar value $V_1$ with maturity $T_1$ and sell another bond of dollar value $V_2$ with maturity $T_2$. The portfolio value $\Pi$ is given by \begin{align} \Pi=V_1-V_2 \end{align} where the $t$ subscripts are omitted for convenience. Assuming the portfolio is self-financing, the change in portfolio value is \begin{align} d\,\Pi=dV_1-dV_2 \end{align} The strategy is similar to that for the Black-Scholes case. I apply Ito’s lemma to obtain the processes for $V_1$ , which allows us to find the process for $\Pi$ .To form the hedging portfolio, first apply Ito’s lemma to the value of the derivative,$V_1=V_1(x_t,t)$ We must differentiate $V_1$ with respect to the variables $t$ and $x$,The result is that $dV_1$ follows the process \begin{align} dV_1=\frac{\partial V_1}{\partial t}dt+\frac{\partial V_1}{\partial x}dx+\frac{1}{2}\frac{\partial^2 V_1}{\partial x^2}d[x,x](t) \end{align} in other words \begin{align} dV_1=(\frac{\partial V_1}{\partial t}+3\frac{\partial V_1}{\partial x}+2x_t\frac{\partial^2 V_1}{\partial x^2})dt+2\sqrt x_t \frac{\partial V_1}{\partial x}dW_t \end{align} I have use the fact that $d[x,x](t)=(2\sqrt x_t\ dW_t)(2\sqrt x_t\ dW_t)=4x_t dt$ , $dt\,dW_t=0$ and $dt\,dt=0$. Then portfolio value can be written \begin{align} &d\,\Pi=(\frac{\partial V_1}{\partial t}+3\frac{\partial V_1}{\partial x}+2x_t\frac{\partial^2 V_1}{\partial x^2})dt+2\sqrt x_t \frac{\partial V_1}{\partial x}dW_t\\ &\,\,\,\,\,\,\,\,\,\,-(\frac{\partial V_2}{\partial t}+3\frac{\partial V_2}{\partial x}+2x_t\frac{\partial^2 V_2}{\partial x^2})dt-2\sqrt x_t \frac{\partial V_2}{\partial x}dW_t\\ \end{align} In order for the portfolio to be hedged against movements Wiener process last two terms in this equation must be zero. This implies that the hedge parameters must be \begin{align} \frac{\partial V_1}{\partial x}=\frac{\partial V_2}{\partial x} \end{align} then \begin{align} &d\,\Pi=(\frac{\partial V_1}{\partial t}+2x_t\frac{\partial^2 V_1}{\partial x^2})dt\\ &\,\,\,\,\,\,\,\,\,\,-(\frac{\partial V_2}{\partial t}+2x_t\frac{\partial^2 V_2}{\partial x^2})dt\\ \end{align} The condition that the portfolio earn the risk-free rate, $S^0=1$, implies that the change in portfolio value is $d\,\Pi=S^0\,\Pi dt$ \begin{align} d\,\Pi=S^0\,\Pi\,dt=\Pi\,dt=(V_1-V_2)dt \end{align} By combination these equations we have \begin{align} \frac{\partial V_1}{\partial t}+2x_t\frac{\partial^2 V_1}{\partial x^2}-V_1=\frac{\partial V_2}{\partial t}+2x_t\frac{\partial^2 V_2}{\partial x^2}-V2 \end{align} The above relation is valid for arbitrary maturity dates $T_1$ and $T_2$, so this equation should be independent of maturity $T$.Then \begin{align} \frac{\partial V}{\partial t}+2x_t\frac{\partial^2 V}{\partial x^2}-V=\lambda \end{align}

## Answer by mbison (score 0)

https://quant.stackexchange.com/a/31939

If moneymarket rate is always 0, then the bond is also priced at 1 by no arbitrage. However, if you have to go through all the measure change exercise...

To price the zero-bond as a derivative: Let B(t,S) be the price of the bond as a function of S. We know that $B(T, S_T) =1$. To price any derivative we will use martingale pricing:

$\frac{B(t_0)}{N(t_0)} = E^Q \frac{B(T)}{N(T)}$ where N(t) is our numeraire.

We will take the moneymarket account to be our numeraire. Note that since the riskfree rate is always zero in this example we have $dN = 0 dt = 0$.

The dynamics of the discounted stock are given by: $d\frac{S}{N} = \frac{1}{N}dS - \frac{S}{N^2}dN = \frac{dS}{N} = dS$, since N(t) =1 for all t.

We change the measure from P to Q in order to make $d\frac{S}{N}$ a martingale. In this case it is equivalent to making $dS$ a martingale. We apply girsanov and endup with $dS = dW^Q$.

$B(t_0) = E^Q(B(T)) = \int 1 dW^Q = 1$

(chapter 11 of Bjorks book is good reference for girsanov).