Pricing and upper price bounds of relax certificates

Abstract

Relax certificates are written on multiple underlying stocks. Their payoff depends on a barrier condition and is thus path-dependent. As long as none of the underlying assets crosses a lower barrier, the investor receives the payoff of a coupon bond. Otherwise, there is a cash settlement at maturity which depends on the lowest stock return. Thus, the products consist of a knock-out coupon bond and a knock-in claim on the minimum of the stock prices. In a Black-Scholes model setup, the price of the knock-out part can be given in closed (or semi-closed) form in the case of one or two underlyings only. With the exception of the trivial case of one underlying, the price of the knock-in minimum claim always has to be calculated numerically. Hence, we derive semi-closed form upper price bounds. These bounds are the lowest upper price bounds which can be calculated without the use of numerical methods. In addition, the bounds are especially tight for the vast majority of relax certificates which are traded at a discount relative to the corresponding coupon bond. This is also illustrated with market data.

Appendices

Appendix A: First hitting time: one-dimensional case

To derive the distribution of the first hitting time in the one-dimensional case, we use results given in He et al. (1998).They consider the probability density and distribution function of the maximum or minimum of a one-dimensional Brownian motion with drift. Along the lines of He et al. (1998), we define

$$ \underline{X}_t:=\mathop {\min}\limits_{0\leq s\leq t} X_s \qquad \overline{X}_t:=\max_{0\leq s\leq t} X_s $$

where X _t  = αt + σW _t ,  t ≥ 0 and α, σ are constants. W is a Brownian motion defined on some probability space.

Proposition 7

LetG(x, t;α) andg(y, x, t;α₁) be defined as

$$ \begin{aligned} G(x,t;\alpha)&:=N\left(\frac{x-\alpha t}{\sigma \sqrt{t}}\right)-e^{\frac{2\alpha x}{\sigma^2}}N\left(\frac{-x-\alpha t}{\sigma \sqrt{t}}\right),\\ g(y,x,t;\alpha_1)&:=\frac{1}{\sigma\sqrt{t}} N^\prime \left(\frac{x-\alpha_1 t}{\sigma\sqrt{t}}\right) \left(1-e^{-\frac{4x^2-4x4y} {2\sigma^2t}}\right) \end{aligned} $$

where N denotes the cumulative distribution function of the standard normal distribution andN′(z) the density of the standard normal distribution. Forx ≥ 0, it holds

$$ P(\overline{X}_t\leq x)=G(x,t;\alpha),\qquad f_{\tau_m^{(1)}}^P=g(y,x,t;\alpha_1)dy. $$

Forx < 0, it holds

$$ P(\underline{X}_t \geq x) =G(-x,t;-\alpha),\qquad P(X_1(t)\in dy,\underline{X}_1(t)\geq x)=g(-y,-x,t;-\alpha_1)dy. $$

Proof

C.f. Theorem 1 of He et al. (1998). and the proof given here. $\square$

Corollary 2

Let

$$ S_t = S_0 e^{\left(\mu -\frac{1}{2} \sigma^2\right)t + \sigma W_t} $$

where μ, σ (σ > 0) are constants. W is a Brownian motion defined on some probability space. For the first hitting time$\tau_m:={\rm inf}\{t\geq0 \vert S_t\leq m\}, \, (m < S_0)$it holds that

$$ \begin{aligned} P(\tau_m \leq t) &= N\left(\frac{ln\frac{m}{S_0}-\left(\mu-\frac{1}{2}\sigma^2\right)t} {\sigma\sqrt{t}}\right) +e^{2 \frac{\mu-\frac{1}{2}\sigma^2} {\sigma^2}ln \frac{m}{S_0}} N\left(\frac{ln\frac{m} {S_0}+\left(\mu-\frac{1}{2}\sigma^2\right)t} {\sigma\sqrt{t}}\right),\\ f_{\tau_m^{(1)}}^P &=\frac{-ln \frac{m}{S_0}} {\sqrt{2\pi\sigma^2t^3}}e^{-\frac{1}{2}\frac{\left(ln\frac{m} {S_0}-(\mu-\frac{1}{2}\sigma^2)t\right)^2}{\sigma^2t}}dt. \end{aligned} $$

Proof

Note that

$$ \tau_m:=\inf\left\{t\geq0 \left\vert S_t\leq m\right\} = \inf\left\{t\geq 0\left\vert \ln \frac{S_t}{S_0}\leq \ln\frac{m} {S_0}\right\}.\right.\right. $$

Let X _t denote the logarithm of the normalized asset price, i.e. $X_t:=\ln\frac{S_t}{S_0}=\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma W_t$ and set $\alpha = \mu-\frac{1}{2}\sigma^2.$ The stopping time τ _m is related to the first hitting time of a one-dimensional Brownian motion with drift α. With $x:=\ln\frac{m}{S_0}<0$ it follows

$$ P(\tau_m\leq t)=P(\underline{X}_t\leq x)=1-P(\underline{X}_t\geq x). $$

According to Proposition 7, we have

$$ \begin{aligned} 1-P(\underline{X}_t\geq x) &= 1-G(-x,t;-\alpha) = 1-N\left(\frac{-x+\alpha t}{\sigma\sqrt{t}}\right) + e^{\frac{2\alpha x}{\sigma^2}} N\left(\frac{x+\alpha t} {\sigma\sqrt{t}}\right) \\ &=N\left(\frac{x-\alpha t}{\sigma\sqrt{t}}\right) +e^{\frac{2\alpha x}{\sigma^2}}N\left(\frac{x+\alpha t} {\sigma\sqrt{t}}\right). \end{aligned} $$

Inserting α and x gives the distribution function. To derive the density function, define

$$ f(t):=N\left(\frac{x-\alpha t}{\sigma\sqrt{t}}\right) +e^{\frac{2\alpha x} {\sigma^2}}N\left(\frac{x+\alpha t} {\sigma\sqrt{t}}\right). $$

This implies

$$ \begin{aligned} f'(t)&=N'\left(\frac{x-\alpha t} {\sigma\sqrt{t}}\right) \times\left(\frac{-\alpha\sigma\sqrt{t}-\frac{\sigma} {2\sqrt{t}}(x-\alpha t)} {\sigma^2t}\right) \\ &\quad + e^{\frac{2\alpha x} {\sigma^2}}\times N'\left(\frac{x+\alpha t}{\sigma\sqrt{t}}\right) \times\left(\frac{\alpha\sigma\sqrt{t}-\frac{\sigma} {2\sqrt{t}}(x+\alpha t)}{\sigma^2t}\right) \end{aligned} $$

Using $N'(x)=\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}x^2},$ we get

$$ \begin{aligned} e^{\frac{2\alpha x}{\sigma^2}}N^{\prime}\left(\frac{x+\alpha t}{\sigma\sqrt{t}}\right) &= \frac{1}{\sqrt{2 \pi}}e^{\frac{2\alpha x}{\sigma^2}-\frac{1}{2}\left(\frac{x+\alpha t}{\sigma\sqrt{t}}\right)^2} = \frac{1}{\sqrt{2 \pi}}e^{\frac{1}{\sigma^2}[2\alpha x-\frac{1}{2t}(x+\alpha t)^2]} \\ &= \frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}\frac{(x-\alpha t)^2}{\sigma^2t}} = N^{\prime}\left(\frac{x-\alpha t}{\sigma \sqrt{t}}\right). \end{aligned} $$

Inserting this in the above equation for f′(t) gives

$$ f'(t) = N'\left(\frac{x-\alpha t}{\sigma\sqrt{t}}\right) \times \frac{-\sigma}{2\sqrt{t}\sigma^2 t}\left(x-\alpha t+x+\alpha t\right) = \frac{-x}{\sqrt{2\pi \sigma^2 t^3}}e^{-\frac{1} {2}\frac{(x-\alpha t)^2}{\sigma^2 t}}. $$

Using $\alpha=\mu-\frac{1}{2}\sigma^2$ and $x=\ln\frac{m} {S_0}$ gives the result. $\square$

Appendix B: First hitting time: two dimensional case

The distribution of the first hitting time of a two-dimensional arithmetic Brownian motion is given in He et al. (1998) and Zhou (2001):

Proposition 8

Let$X^{(j)}_t=\alpha_j t+\sigma_j W_t^{(j)}$ (j = 1, 2), where α _j and σ _j are constants. W⁽¹⁾,   W⁽²⁾are two correlated Brownian motions with < W⁽¹⁾, W⁽²⁾ >  _t  = ρ t. Then, the probability thatX⁽¹⁾ and X⁽²⁾will not hit the upper boundariesx⁽¹⁾ > 0 and x⁽²⁾ > 0 up to time t is given by

$$ \begin{aligned} Q\left(\overline{X}_t^{(1)}\leq x^{(1)},\overline{X}_t^{(2)}\leq x^{(2)}\right) &=\frac{2}{\alpha t}e^{a_1 x_1 + a_2 x_2 + b t} \sum_{n=1}^{\infty}{\rm sin}{\left(\frac{n\pi\theta_0}{\alpha}\right)}\\ &\quad e^{-\frac{r_0^2}{2t}}\int\limits_{0}^{\alpha}{\rm sin}{\left(\frac{n\pi\theta} {\alpha}\right)}g_n(\theta)d\theta \end{aligned} $$

where$\overline{X}_t:=\max_{0\leq s\leq t} X_s.$The parameters are defined by

$$\begin{array}{*{20}l} a_1 = \frac{-\alpha_1\sigma_2 + \rho\alpha_2\sigma_1} {(1-\rho^2)\sigma_1^2\sigma_2} & \quad a_2 =\frac{-\alpha_2\sigma_1 + \rho\alpha_1\sigma_2}{(1-\rho^2)\sigma_2^2\sigma_1} \\d_1= a_1\sigma_1+a_2\sigma_2\rho &\quad d_2= a_2\sigma_2\sqrt{1-\rho^2}\\\end{array} $$

and by

$$ \begin{aligned} b &= \alpha_1 a_1 + \alpha_2 a_2 + \frac{1}{2} \sigma_1^2 a_1^2 + \frac{1}{2} \sigma_2^2 a_2^2 + \rho \sigma_1 \sigma_2 a_1 a_2\\ \alpha&=\left\{\begin{array}{cc}{\rm tan}^{-1}\left(-\frac{\sqrt{1-\rho^2}}{\rho}\right) &{ if } \rho<0\\ \pi+ {\rm tan}^{-1}\left(-\frac{\sqrt{1-\rho^2}}{\rho}\right)& otherwise \\ \end{array}\right.\\ \theta_0&=\left\{\begin{array}{cc} {\rm tan}^{-1}\left( \frac{\frac{x2}{\sigma_2}\sqrt{1-\rho^2}} {\frac{x1}{\sigma_1}-\rho\frac{x2}{\sigma_2}}\right) &{ if (.)} >0\\ \pi+ {\rm tan}^{-1}\left(\frac{\frac{x2}{\sigma_2}\sqrt{1-\rho^2}}{\frac{x1}{\sigma_1}-\rho\frac{x2}{\sigma_2}}\right)& otherwise \\ \end{array}\right.\\ r_0&=\frac{x_2}{\sigma_2}/ sin(\theta_0). \end{aligned} $$

The functiong _n is defined as

$$ g_n(\theta)= \int\limits_0^\infty r e^{-\frac{r^2}{2 t}} e^{d_1 r \sin\left( \theta-\alpha\right) -d_2 r \cos\left( \theta-\alpha\right)} I_{\frac{n \pi}{\alpha}}\left( \frac{rr_0} {t} \right) dr. $$

I _v (z) is the modified Bessel function of order v.

Proof

Cf. Proposition 1 of Zhou (2001) and the proof given there. $\square$

Corollary 3

For two stocksS^(k)andS^(l)with volatilities σ _k and σ _l and correlation ρ_k,l, the distribution function of the first hitting time min{τ_m,k, τ_m,l} under the risk-neutral measure is given by

$$ Q\left({\rm min}\{\tau_{m,k},\tau_{m,l}\} \leq t \right)\\ =1-\frac{2}{\alpha t}e^{a_k \ln\left(\frac{S_0^{(k)}}{m}\right) + a_l \ln\left(\frac{S_0^{(l)}}{m}\right)+ b t} \sum_{n=1}^{\infty}{\rm sin}{\left(\frac{n\pi\theta_0}{\alpha}\right)} \cdot e^{-\frac{r_0^2}{2t}}\int\limits_{0}^{\alpha}{\rm sin}{\left(\frac{n\pi\theta}{\alpha}\right)}g_n(\theta)d\theta $$

where

$$ \begin{aligned} a_k &= \frac{\left(r-0.5\sigma_k^2\right)\sigma_l - \rho_{k,l} \left(r-0.5\sigma_l^2\right)\sigma_k}{\left(1-\rho_{k,l}^2\right)\sigma_k^2\sigma_l}\qquad \qquad a_l = \frac{\left(r-0.5\sigma_l^2\right)\sigma_k - \rho_{k,l} \left(r-0.5\sigma_k^2\right)\sigma_l} {\left(1-\rho_{k,l}^2\right)\sigma_l^2\sigma_k}\\ d_k&= a_k\sigma_k+a_l\sigma_l\rho_{k,l} \qquad\qquad d_l= a_l\sigma_l\sqrt{1-\rho_{k,l}^2} \end{aligned} $$

and

$$ \begin{aligned} b &= -\left(r-0.5\sigma_k^2\right) a_k - \left(r-0.5\sigma_l^2\right) a_l + \frac{1}{2} \sigma_k^2 a_k^2 + \frac{1}{2} \sigma_l^2 a_l^2 + \rho_{k,l} \sigma_k \sigma_l a_k a_l \\ g_n(\theta) &= \int\limits_0^\infty r e^{-\frac{r^2}{2 t}} e^{d_k r \sin\left( \theta-\alpha\right) -d_l r \cos\left( \theta-\alpha\right)} I_{\frac{n \pi} {\alpha}}\left( \frac{rr_0} {t} \right) dr. \end{aligned} $$

I _v (z) is the modified Bessel function of order v. α,   θ₀, andr₀ are given in Proposition 8 in Appendix B for the case wherek = 1 andl = 2.

Proof

The stock prices are given by

$$ S_t^{(j)} = S_0 e^{\left(r-\frac{1}{2}\sigma_j^2\right)t+\sigma_j W^{(j)}_t} \quad j=k,l. $$

The first hitting time of the lower boundary $m_j< S_0^{(j)}$ by the geometric Brownian motion $S_t^{(j)}$ is

$$ \tau_m^{(j)} := \inf\left\{t\geq 0 \left\vert S_t^{(j)}\leq m_j\right\} \right. = \inf\left\{t\geq 0 \left\vert -\ln\frac{S_t^{(j)}}{S_0^{(j)}}\geq \ln\frac{S_0^{(j)}} {m_j}\right\}\right.. $$

With the definition of the arithmetic Brownian motion

$$ X_t^{(j)}:= - \ln \frac{S_t^{(j)}}{S_0^{(j)}}=-\left(r-\frac{1} {2}\sigma_j^2\right)t-\sigma_j W^{(j)}_t, $$

the first hitting time can be rewritten as $\tau_m^{(j)}=\inf\left\{t\geq 0 \left\vert X_t^{(j)} \geq \ln\frac{S_0^{(j)}} {m_j}\right\}.\right.$ Using the relation $\{\tau_{m,j} > t\}=\left\{\overline{X}_t^{(j)} < \ln\frac{S_0^{(j)}}{m_j}\right\}$ we can conclude

$$ Q\left(\min\{\tau_{m,k},\tau_{m,l}\}>t \right) = Q\left( \overline{X}_t^{(k)} < \ln\frac{S_0^{(k)}}{m_k}, \overline{X}_t^{(l)} < \ln\frac{S_0^{(l)}}{m_l} \right). $$

Since both, $\ln\frac{S_0^{(k)}}{m_k}>0$ and $\ln\frac{S_0^{(l)}}{m_l}>0,$ the result follows from Proposition 8. $\square$