# Price and constant hedging portfolio for straddle: $X=|S(T)-K|$

wondering if somebody could check my answer for a homework question!

Given a straddle, characterized by its pay-off at maturity $X=|S(T)-K|$, I am asked to find the price of the (simple) claim at any time $t\in [0,T]$. Starting at $$V(t)=e^{-r(T-t)}\mathbb{E_Q}[|S(T)-K||\mathcal{F}_t]$$ with $S(t)=S_0\exp{((r-\frac{1}{2}\sigma^2)t+\sigma W^\mathbb{Q}(t))}$, I finally end up with the value process: $$V(t)=2S(t)\Phi(d_1)-2Ke^{-r(T-t)}\Phi(d_2)-S(t)+Ke^{-r(T-t)}$$ where $$d_1=\frac{\ln(S(t)/K)+(r+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}},\quad d_2=d_1-\sigma\sqrt{T-t}$$

Now, it is suggested that there exists a constant hedging portfolio for this claim. The portfolio consists not only of stocks and bonds, but also contains European call options. That is, I wish to find a portfolio with value process $\hat{V}(t)=x_tS(t)+y_tB(t)+z_tC(t)$, where $(x_t,y_t,z_t)$ remains constant for $t\in[0,T]$. My gut tells me that $(x_t,y_t,z_t)=(-1,Ke^{-rT},2)$ as the value process of the claim can be written as: $$V(t)=Ke^{-r(T-t)}-S(t)+2C(t)$$ with $C(t)$ the price for a call option (by the Black-Scholes formula). I can "semi-justify" this by defining $V(t)=v(t,x)+2C(t)$ with $v(t,x)=Ke^{-r(T-t)}-x$, so that $v$ satisfies the Black-Scholes partial differential equation and thus the portfolio with $$x_t=v_x(t,S(t))=-1, \qquad y_t=e^{rt}(v(t,S(t))-x_tS(t))=Ke^{-rT}$$ hedges the claim $K-S(t)$. Then I just add $2C(t)$ to both the value process of the portfolio and $v(t,S(t))$.

So! My two questions here are:

1. Have I calculated the value process $V(t)$ for the claim correctly?
2. Have I correctly justified the hedging portfolio for the claim?

just take a call and a put struck at $K$ and add them together. For the hedge just add the hedges together as well.
• In other words use the identity $\|a-b\| = Max(0,a-b)+Max(0,b-a)$ with $a=S(T)$ and $b=K$ – Alex C Jun 5 '16 at 11:47
Another approach as follow. The $T$-Straddle option $X$, i.e. X=\left\{ \begin{align} & K-S(T)\quad ,\quad 0<S(T)\le K \\ & S(T)-K\quad ,\quad S(T)>K \\ \end{align} \right. has then following contract function \Phi (x)=\left\{ \begin{align} & K-x\quad ,\quad 0<x\le K \\ & x-K\quad ,\quad x>K \\ \end{align} \right. which can be decomposed into the following basic contract functions written $$\Phi (x)=K\times {{\Phi }_{B}}(x)-1\times {{\Phi }_{S}}(x)+2\times {{\Phi }_{C,K}}(x)$$ such that \begin{align} & {{\Phi }_{B}}(x)=1 \\ & {{\Phi }_{S}}(x)=x \\ & {{\Phi }_{C,K}}(x)=\max \{x-K\,,0\} \\ \end{align} then $$\Pi (t,X)=K{{e}^{-r(T-t)}}-{{S}_{t}}+2{{C}_{BC}}({{S}_{T}},K)$$ Note \begin{align} & E_{t}^{Q}\left[ {{\Phi }_{B}}(x) \right]=E_{t}^{Q}\left[ 1 \right]={{e}^{-r(T-t)}} \\ & E_{t}^{Q}\left[ {{\Phi }_{S}}(x) \right]=E_{t}^{Q}\left[ {{S}_{t}} \right]={{S}_{t}} \\ & E_{t}^{Q}\left[ {{\Phi }_{C,K}}(x) \right]=E_{t}^{Q}\left[ {{({{S}_{T}}-K)}^{+}}]=C_{BC}(S_T,K \right) \\ \end{align}