I came across this question in a sheet of practice problems which has me a bit stumped.

A down-and-out call option with maturity T, strike K = 100 and barrier L = K coinciding with the strike, trades at the price 5 SEK. The underlying stock is dividend-free and trades at 120 SEK, and the down-and-out version of a zero-coupon T-bond with face value 50 SEK and barrier L trades at 27 SEK. What is the arbitrage-free value of the down-and-in version of the T-claim X = S(T) with barrier L?

I have tried to combine the In/Out parity inequality with the (barrier) put/call parity, except that I can't (as far as I can see) calculate the down and out put value required:

Letting $\Pi(t)$ denote the price of our claim $X$, with $\Pi_{LO}, \Pi_{LI}$ denoting the corresponding down and out at $L$ contract, and down and in respectively. Then

$$\Pi_{LI} = \Pi_{t} - \Pi_{LO}$$

where $\Pi_t$ is just the stock price $S(t)$, and using the put/call parity we have

$$P_{LO} = KB_{LO} + C_{LO} - S_{LO}$$ where $P,B,C,S$ denoted the put price, zero coupon bond price, EU call price and the stock price. In the question $KB_{LO}$ and $ C_{LO}$ are given, but not $P_{LO}$ so I would need to find it, which I am not sure is possible only given the above information.

Any ideas on how to approach this?

  • 1
    $\begingroup$ What is your interest rate? $\endgroup$ – Gordon Jan 3 '18 at 16:37

Let \begin{align*} \tau = \inf\{t: t \ge 0, S_t \le L \}. \end{align*} Then the down-out-call option has payoff \begin{align*} (S_T-K, 0)^+\pmb{1}_{\tau >T}, \end{align*} and the down-out version zero-coupon $T$-maturity bond has payoff \begin{align*} \pmb{1}_{\tau >T}. \end{align*} Moreover, for the down-in payoff $X$, since $L=K$, \begin{align*} X &= S_T \pmb{1}_{\tau \le T} \\ &=S_T - S_T \pmb{1}_{\tau >T}\\ &=S_T - (S_T-K) \pmb{1}_{\tau >T} - K \pmb{1}_{\tau >T}\\ &=S_T - (S_T-K)^+ \pmb{1}_{\tau >T} - K \pmb{1}_{\tau >T}. \end{align*} Assuming zero interest rate, then the value of $X$ is given by \begin{align*} 120-5 - 2 \times 27 = 61. \end{align*}

  • $\begingroup$ Thank you very much! Very clear solution. One question though, why is it $2\times 27$ in the last line? 61 is indeed the correct answer, but why isn't the bond price just 27? $\endgroup$ – Chris Jan 7 '18 at 17:34
  • $\begingroup$ Since $K=100$, and the price 27 is for notional amount of 50. $\endgroup$ – Gordon Jan 7 '18 at 18:08

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