# Derivative: Delta of a Down and Out Call Option with Barrier=Debt(K)

I am trying to compute the derivative of this function with respect to V0:

This is the price of a down and out call option, assuming the barrier equal to the level of debt K. In other terms, I need to compute the Delta of this DOC Option, in the case of Barrier=K (neither Barrier higher than K nor Barrier lower than K) and I cannot find this case anywhere in the literature. Furthermore, the derivative of the first two terms of the equation equals N(d1), the delta of a plain vanilla call option. Therefore, I just need the derivative of what is in the parenthesis [...] with respect to V_0.

Can someone of you help me?

Anything will be really appreciated!

• Your question is not readable. Can you please use Latex? Apr 21 '16 at 14:05
• Can you read it now? I apologize, but I am not able to write it in Latex, can you still help me? Apr 21 '16 at 16:06
• See Page 8 in people.maths.ox.ac.uk/howison/barriers.pdf. Your formula does not look correct to me. Apr 21 '16 at 18:21
• what I did was taking the formula of a down and out call option in the case of the barrier below the strike K and, as I am going for the case where barrier=K, I substituted the barrier with K. papers.ssrn.com/sol3/… page 15 Apr 22 '16 at 8:37
• Your formula is not the same as Formula (3) on Page 15 of the paper you referred to. Please double check and revise. Is the + sign before the last term a typo? Apr 22 '16 at 12:54

Let \begin{align*} C(S, K, t) = SN(d_1) - e^{-rt}KN(d_2) \end{align*} denote the Black-Scholes call option price with initial asset value $S$, strike $K$, and maturity $t$. Note that \begin{align*} \frac{\partial C}{\partial S} = N(d_1). \end{align*}

For the above barrier option, note that \begin{align*} E_0 &= V_0 N(d_1)-e^{-rt}KN(d_2) -\bigg[V_0 \Big(\frac{K}{V_0}\Big)^{2\lambda}N(d_1^B) -e^{-rt}K \Big(\frac{K}{V_0}\Big)^{2\lambda-2} N(d_2^B) \Big) \bigg]\\ &=C(V_0, K, t) - \Big(\frac{K}{V_0}\Big)^{2\lambda-2}\bigg[\frac{K^2}{V_0}N(d_1^B) -e^{-rt}K N(d_2^B) \Big) \bigg]\\ &=C(V_0, K, t) - \Big(\frac{K}{V_0}\Big)^{2\lambda-2} C\Big(\frac{K^2}{V_0}, K, t \Big). \end{align*} Therefore, \begin{align*} \frac{\partial E_0}{\partial V_0} &=N(d_1) + (2\lambda-2)\frac{K^{2\lambda-2}}{V_0^{2\lambda-1}}C\Big(\frac{K^2}{V_0}, K, t \Big)-\Big(\frac{K}{V_0}\Big)^{2\lambda-2}\frac{\partial C\Big(\frac{K^2}{V_0}, K, t \Big)}{\partial V_0}\\ &=N(d_1) + (2\lambda-2)\frac{K^{2\lambda-2}}{V_0^{2\lambda-1}}C\Big(\frac{K^2}{V_0}, K, t \Big)+\Big(\frac{K}{V_0}\Big)^{2\lambda}N(d_1^B), \end{align*} since \begin{align*} \frac{\partial C\Big(\frac{K^2}{V_0}, K, t \Big)}{\partial V_0} &= \frac{\partial C\Big(\frac{K^2}{V_0}, K, t \Big)}{\partial \frac{K^2}{V_0}}\frac{\partial \frac{K^2}{V_0}}{\partial V_0}\\ &=-\Big(\frac{K}{V_0}\Big)^{2}N(d_1^B). \end{align*}

• What would the value of the debt at time 0, $D_0$, be here? Assuming $V_0=60$, $K=50$, $\sigma=0.4$, $r=0.01$, and $t=1$, we have $E_0 \approx 10.253$. We should have $V_0=E_0+D_0$, which in this case yields $D_0 = 49.747$. But the implied return on this debt is $−\log\left(49.747/50\right) \approx 0.0051 < r=0.01$, which seems too low. Dec 30 '20 at 17:57