# How is the Chooser Option's value computed in this example?

In preparation for my finals, I am attempting a question on chooser options. One question asks

A European chooser option on an index ETF paying a yield of 3.0% with strike \$64 has a maturity of T2 = 21 months and a choice regarding the type of the option must be made after T1 = 12 months. The risk-free rate is 7%, stock return volatility is assumed to be 33% per year and currently a share costs$61. What is the value of the option?

One T2-Call = \$10.4641; 0.9778 times T1-Put (\$7.0157) with adjusted strike 62.1085 for a total cost of \$17.32. Here is my interpretation (most likely incorrect but necessary to illustrate my problem) : First and foremost, I do not understand how we can value such an option today given the information. What I do know is that at time t = 1 (yrs), the value of the option is $$V(1) = \mathrm{max}(c,p).$$ At t = 1, both options have the same strike price (\$64) and remaining maturity (0.75 yrs.). It can be shown through the put-call parity that $$V(T1) = \mathrm{max}(c,c+e^{-r\cdot 0.75}K-S_{1}e^{-q\cdot 0.75}) \\ = c+e^{-q\cdot 0.75}\mathrm{max}(0,Ke^{-(r-q)\cdot 0.75}-S_{1}).$$

The constants are given by

• $q$ = dividend yield = 3%
• $r$ = risk-free rate = 7%
• $K$ = strike price = 64
• $S_{1}$ = spot price at time 1 = unknown

Now in order to calculate the value of the call, I require the spot price at t = 1. This is my first problem since I have been given no such information.

How can I move forward from here, find the value of the chooser option at t = 1, and furthermore its value at t = 0 (if that's what the original question requires)?

• You did pretty well and are in fact almost finished IMHO. I did not do the computations, but just take the (discounted) expectation of the expression you wrote for $V(T1)$. First term is a T2 call. Second is $X$ units of a T1 put of adjusted strike $K e^{...}$. I just think you messed up the discount factors. I'll try to post a real answer shortly – Quantuple Jun 16 '16 at 7:55

Although the answer of @SRKX is right on spot, I was already writing a solution along the lines of how you had specifically approached the problem. I think it might still be useful to you, so here it goes

The price of the chooser option, as seen of today $t=0$ is by definition \begin{align} V_0 &= \underbrace{e^{-r T_2}}_{\text{Payoff dicount factor}} \underbrace{\mathbb{E}\left[\ \ \underbrace{\max\left( \mathbb{E}[(S_{T_2}-K)^+ \vert \mathcal{F}_{T_1}], \mathbb{E}[(K-S_{T_2})^+ \vert \mathcal{F}_{T_1} ] \right)}_{\text{Expected payoff at $T_2$ as seen of $T_1$}} \ \ \vert \mathcal{F}_0 \ \ \right]}_{\text{Expected payoff at $T_2$ as seen of $t=0$}} \\ &= e^{-r T_2} \mathbb{E}_0\left[ \max\left( \mathbb{E}_{T_1}[(S_{T_2}-K)^+], \mathbb{E}_{T_1}[(K-S_{T_2})^+] \right) \right] \end{align} If you're not familiar with the notation $\mathcal{F}_t$ used for filtrations, you can interpret it as "all the information we know at time $t$". The notation $\mathbb{E}_t[.]$ simply figures that the expectation is taken conditionally on the knowledge of $\mathcal{F}_t$. Naturally all of these expectations are taken under the risk-neutral measure $\mathbb{Q}$.

By definition, we also have that the price of European call/put options is given by $$C(T_1,S_{T_1};K,(T_2-T_1)) = e^{-r(T_2-T_1)} \mathbb{E}_{T_1}[(S_{T_2}-K)^+] \tag{def 1} := C_{12}$$ $$P(T_1,S_{T_1};K,(T_2-T_1)) = e^{-r(T_2-T_1)}\mathbb{E}_{T_1}[(K-S_{T_2})^+] \tag{def 2} := P_{12}$$ where $C(t,S_t;K,\tau)$ (resp. $P(t,S_t;K,\tau)$) denotes the price of a European call (resp. put) option as seen of time $t$, given the spot value $S_t$, the strike price $K$ and the time to expiry $\tau$.

Therefore, $$V_0 = e^{-r T_2} \mathbb{E}_0\left[ \max\left( \frac{C_{12}}{e^{-r(T_2-T_1)}}, \frac{P_{12}}{e^{-r(T_2-T_1)}} \right) \right]$$ Yet by call-put parity: $$C_{12} - P_{12} = e^{-r(T_2-T_1)}( S_1 e^{(r-q)(T_2-T_1)} - K )$$ so that we can further write (similarly to what you did) \begin{align} V_0 &= e^{-r T_2} \mathbb{E}_0\left[ \max\left( \frac{C_{12}}{e^{-r(T_2-T_1)}}, \frac{C_{12}}{e^{-r(T_2-T_1)}} - (S_1e^{(r-q)(T_2-T_1)} - K) \right) \right] \\ &= e^{-r T_2} \mathbb{E}_0\left[ \left( \frac{C_{12}}{e^{-r(T_2-T_1)}} + \max\left( 0, K - S_1e^{(r-q)(T_2-T_1)} \right) \right) \right] \\ &= \mathbb{E}_0\left[ e^{-rT_1} C_{12} \right] + \mathbb{E}_0\left[ e^{-rT_2} \max\left( 0, K - S_1e^{(r-q)(T_2-T_1)} \right) \right] \tag{1} \end{align}

Now using $(\text{def } 1)$ the first term of $(1)$ becomes: $$\mathbb{E}_0 \left[ e^{-rT_1} C_{12} \right] = \mathbb{E}_0 \left[ e^{-rT_2} \mathbb{E}_{T_1}[(S_{T_2}-K)^+] \right] = C(0,S_0;K,T_2)$$ by the tower property of conditional expectations.

Similarly, the second term of $(1)$ can on the other hand be expressed as: \begin{align} \mathbb{E}_0\left[ e^{-rT_2} \max\left( 0, K - S_1e^{(r-q)(T_2-T_1)} \right) \right] &= \mathbb{E}_0\left[ \max\left( 0, Ke^{-rT_2} - S_1e^{-rT_1-q(T_2-T_1)} \right) \right] \\ &= e^{-q(T_2-T_1)} \mathbb{E}_0\left[ e^{-r{T_1}} \max\left( 0, Ke^{-(r-q)(T_2-T_1)} - S_1 \right) \right] \\ &= e^{-q(T_2-T_1)} P(0,S_0; Ke^{-(r-q)(T_2-T_1)}, T_1) \end{align} So that $(1)$ becomes $$V_0 = C(0,S_0;K,T_2) + \underbrace{e^{-q(T_2-T_1)}}_{= 0.9778} P(0,S_0; \underbrace{Ke^{-(r-q)(T_2-T_1)}}_{= 62.1085}, T_1)$$ hence a $T_2$ call struck at $K$ + 0.9778 units of a $T_1$ put with adjusted strike 62.1085.

• This is unbelievable. Thank you so much. The mathematics is incredibly elegant and much more sophisticated than anything I've seen in Hull. One small question: in the tedious process of calculating the value of the call and put with the parameters in your last equation, do you get C = \$10.4641 and P = \$7.0157 as displayed in my original question? I seem to get different figures, using Excel and a web calculator (C = 10.5338, P = 8.0410) – Gustavo Louis G. Montańo Jun 16 '16 at 9:22
• Glad I could help. Actually I did not do the computations (except for the underbraced part). I get $C=\$10.534$and$P=\$7.0469$. I think you may be getting $8.04$ for the put because you did not adjust the strike? Anyway, it's not exactly the same that what's displayed in your original question. So either we both did the same mistake (cause our call prices match), which seems to me very unlikely, or there is a problem in the initial question. – Quantuple Jun 16 '16 at 9:59
• Ah, yes. I made a small mistake. No matter, I think your solution is perfect. Thanks for your help, Quantuple. – Gustavo Louis G. Montańo Jun 16 '16 at 10:24

You can refer to one of my previous answers here for a detailed development.

There are actually two ways you can price this: - the price of a call plus a put with adjusted strike (like above) - a put plus the price of a call with an adjusted strike (like in my answer).

The only difference is whether you do $\max( a, b ) = b + ( a - b )^+$, or $\max( a, b ) = a + ( b - a )^+$.

• Thank you so much SRKX. You and Quantuple have made the question incredibly clear! I also appreciate your inputs in my last 2 recent questions. Thanks a lot, pal! – Gustavo Louis G. Montańo Jun 16 '16 at 9:23