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Let's assume the usual Black Scholes assumptions hold. My question is related to an answer on this question. There, the weights ($\Delta_t^1$,$\Delta_t^2$) are derived which form a locally risk free portfolio $$X_t =\Delta_t^1 S_t + \Delta_t^2C_t$$ with $$\Delta_t^1 = -\frac{\frac{\partial C}{\partial S} B_t}{C_t - \frac{\partial C} {\partial S}S},\quad \Delta_t^2 =\frac{B_t}{C_t - \frac{\partial C}{\partial S}S}$$ It is emphasized that the strategy $(−∂C∂S, 1)$ is not self financing. I have no doubt about the derivation. Rather, I'm interested to know, in the context of dynamic delta hedging, is it actually valid to set $\Delta_t$ (the amount of the underlying to buy or sell) to $∂C∂S$ instead of $\Delta_t^1$ as shown above? For example, in this paper on page 4 the authors investigate delta hedging strategies by setting $\Delta_t^1$ to $-∂C∂S$ (e.g. $N(d_1)$ for a European call option) and explicitly claim the self financing property. So how would one actutally compute the correct $\Delta_t$ for a delta hedging strategy in a BS world?

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  • $\begingroup$ See answer by the same user to the following question: Derivation of BS PDE problem using Delta hedging. $\endgroup$ – Daneel Olivaw Jun 3 '17 at 14:59
  • $\begingroup$ Thx for the link. To clarify, say, if I were to do a delta hedging simulation and therefore would have to compute the amount of the underlying to sell or buy on some discrete points using only $\Delta_t$ = $∂C∂S$, in general, I would obtain delta neutrality but not a self financed or locally risk free position? Wouldn't it be more correct then to use $\Delta_t^1$? $\endgroup$ – Tim Jun 3 '17 at 15:40
  • $\begingroup$ If you differentiate $X_t$ in that case $-$ hedging strategy $(-\partial C/\partial S,1)$ $-$ you get: $\frac{\partial X}{\partial S} = -\frac{\partial \Delta}{\partial S}S - \Delta + \frac{\partial C}{\partial S} = -\frac{\partial^2 C}{\partial S^2}S$ so you will be indeed $\Delta$-neutral but not $\Gamma$-neutral if I am not mistaken. $\endgroup$ – Daneel Olivaw Jun 3 '17 at 15:51
  • $\begingroup$ I see but my crucial question is about how to compute the $\Delta_t$ for delta hedging. Browsing the web reveals many sources (such as the linked paper above) which use $\partial C/\partial S$ and claim self-financing. I may miss something here. $\endgroup$ – Tim Jun 3 '17 at 16:07
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    $\begingroup$ The correct approach is the one described in the questions you and I linked to. You can also check the comments from users Gordon and zer0hedge at the bottom of the answer I linked to, it seems this mistake is common $-$ I also made it. Note that is in part due to the fact that the pricing PDE does not depend on the weight $\Delta_t^2$: this is because as you can observe $\Delta_t^1$ is a function of $\Delta_t^2$, hence in the process of deriving the PDE $\Delta_t^2$ ends up being cancelled: $ dX = rXdt \Leftrightarrow \Delta^2f(C,S)=r\Delta^2g(C,S)dt $. $\endgroup$ – Daneel Olivaw Jun 3 '17 at 16:26
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Main references

As explained in my comments, the correct approach to derive the hedging portfolio would be the one described in Gordon's answers to the following questions:

The hedging portfolio $C_t-(\partial C/\partial S)S_t$ is not self-financing

We can check that the hedging portfolio $(w_C,w_S)=(1,-\partial C/\partial S)$ is not self-financing. Letting $X_t$ be the portfolio value, we have $-$ dropping time subscripts:

$$ \begin{align} dX & = dC+d(w_SS) \\[6pt] & =dC+\left(Sdw_S+w_SdS+dw_SdS\right) \end{align} $$

The differential of the option is:

$$dC = \frac{\partial C}{\partial t}dt+\frac{\partial C}{\partial S}\mu S dt + \frac{\partial C}{\partial S}\sigma SdW+\frac{1}{2}\frac{\partial^2C}{\partial S^2}\sigma^2S^2dt$$

We differentiate the weight $w_S$:

$$ \begin{align} dw_S & =\frac{\partial w_S}{\partial t}dt+\frac{\partial w_S}{\partial S}dS+\frac{1}{2}\frac{\partial^2 w_S}{\partial S^2}dS^2 \\[6pt] & = \frac{\partial w_S}{\partial t}dt + \frac{\partial w_S}{\partial S}\mu Sdt+\frac{\partial w_S}{\partial S}\sigma SdW+\frac{1}{2}\frac{\partial^2 w_S}{\partial S^2}\sigma^2S^2dt \\[6pt] & = -\left(\frac{\partial^2C}{\partial S\partial t}dt + \frac{\partial^2C}{\partial S^2}\mu Sdt+\frac{\partial^2C}{\partial S^2}\sigma SdW+\frac{1}{2}\frac{\partial^3C}{\partial S^3}\sigma^2S^2dt\right) \end{align} $$

Hence:

$$ \begin{align} & Sdw_S = -\left(\frac{\partial^2C}{\partial S\partial t}Sdt + \frac{\partial^2C}{\partial S^2}\mu S^2dt+\frac{\partial^2C}{\partial S^2}\sigma S^2dW+\frac{1}{2}\frac{\partial^3C}{\partial S^3}\sigma^2S^3dt\right) \\[6pt] & w_SdS = -\left(\frac{\partial C}{\partial S}\mu Sdt+\frac{\partial C}{\partial S}\sigma SdW\right) \\[6pt] & dw_SdS = -\frac{\partial^2C}{\partial S^2}\sigma^2S^2dt \end{align} $$

Terms cancel and we obtain:

$$ \begin{align} dX = \frac{\partial C}{\partial t}dt & - \frac{\partial^2C}{\partial S\partial t}Sdt - \frac{\partial^2C}{\partial S^2}\mu S^2dt \\[6pt] & - \frac{\partial^2C}{\partial S^2}\sigma S^2dW - \frac{1}{2}\frac{\partial^3C}{\partial S^3}\sigma^2S^3dt - \frac{1}{2}\frac{\partial^2C}{\partial S^2}\sigma^2S^2dt \end{align} $$

Hence we conclude that the self-financing condition $dX = dC+w_SdS$ is not verified: indeed the term in $\partial^2C/\partial S \partial t$ would not appear if it was self-financing, in which case it would read:

$$ \begin{align} dX & = \left(\frac{\partial C}{\partial t}dt+\frac{\partial C}{\partial S}\mu S dt + \frac{\partial C}{\partial S}\sigma SdW+\frac{1}{2}\frac{\partial^2C}{\partial S^2}\sigma^2S^2dt\right)-\frac{\partial C}{\partial S}\left(\mu Sdt + \sigma SdW \right) \\[6pt] & = \frac{\partial C}{\partial t}dt + \frac{1}{2}\frac{\partial^2C}{\partial S^2}\sigma^2S^2dt \end{align} $$

Independence of the pricing PDE and the option weight $w_C$

Note that the confusion around the hedging portfolio is in part due to the fact that the pricing PDE does not depend on the weight of the option $w_C$. As stated in your question, the correct stock weight $w_S$ is:

$$w_S = -w_C\frac{\partial C}{\partial S}$$

Recall that after having cancelled the random terms in $dX_t$ through the choice of $(w_C, w_S)$, we get:

$$ dX_t = w_C \left(\frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2S^2 \frac{\partial^2 C}{\partial S^2}\right)dt$$

From the risk-free return constraint, we then obtain $-$ dropping time subscripts:

$$ \begin{align} & dX = rXdt \\[6pt] \Leftrightarrow \quad & w_C \left(\frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2C}{\partial S^2}\right)dt = r\left(w_SS+w_CC \right)dt \\[6pt] \Leftrightarrow \quad & w_C \left(\frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2C}{\partial S^2}\right) = rw_C \left(-\frac{\partial C}{\partial S}S + C\right) \end{align} $$

Hence the derivative weight $w_C$ can be cancelled.

A note on hedging portfolios

Note the difference between our approach and the one described in your paper:

  • Here, we hold a portfolio of options and stocks and we require this portfolio to return the risk-free rate;
  • In your paper, we hold a portfolio made up on an option combined with stocks and riskless bonds and we require its value to be $0$.

$$\underbrace{w_C(t)C_t + w_S(t)S_t = B(t)}_{(1) \, \text{Our hedging portfolio}} \quad \Longleftrightarrow \quad \underbrace{C_t + w_S(t)S_t + w_B(t)B(t) = 0}_{(2) \, \text{Your paper's hedging portfolio}}$$

The self-financing condition is different in both cases:

$$ \begin{align} & (1) \; : \; C_tdw_C(t) + dw_C(t)dC_t + S_tdw_S(t) + dw_S(t)dS_t = 0 \\[12pt] & (2) \; : \; S_tdw_S(t) + dw_S(t)dS_t + B_tdw_B(t) + dw_B(t)dB_t = 0 \end{align} $$

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  • $\begingroup$ A lil late thank you for this, I really appreciate! $\endgroup$ – Tim Jun 25 '17 at 22:14
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for a self financing portfolio, you have a holding in stocks and one in bonds. If we want to do a hedging simulation, at the start of each step, work out the total value of the hedger's holding (excluding the thing being hedged). Treat this as a cash sum. Buy $N(d_1)$ units of the stock. Use the rest of the money (which may be negative) to buy riskless bonds. Hold this across the step. Repeat.

This is trivially self-financing and you will find that the variance of the net position goes to zero as the step size goes to zero.

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