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期权Delta（Δ）

Delta

专题
Finance / 金融
难度
L4

题目详情

A. 无分红欧式看涨的 delta 是什么？如何推导？

B. ATM（平值）看涨在 r=0r=0 时 delta 约是多少？临近到期会怎样？

C. 你多头一份看涨，并用空头标的做动态 delta 对冲。如果股价跳涨，如何调仓？

D. 估计小 τ\tau、小 rr 下平值看涨的近似价格。

英文原题

Δ=∂f∂S\Delta = \frac{\partial f}{\partial S}

For a European call on a stock with dividend yield yy:
Δ=e−yτ N(d1).\Delta = e^{-y\tau} \,N(d_1).

For a European put on a stock with dividend yield yy:
Δ=− e−yτ [ 1−N(d1)].\Delta = -\,e^{-y\tau}\,\bigl[\,1 - N(d_1)\bigr].

A. What is the delta of a European call on a non-dividend paying stock? How do you derive it?

B. Estimate the delta of an at-the-money call on a stock with zero dividend. What happens to that delta as it approaches maturity?

C. You long a GM call and decide to dynamically hedge by shorting GM shares. If GM’s stock jumps up, how do you rebalance your hedge?

D. Estimate an at-the-money call’s value for a small τ\tau and low r.r.

解析

A. 无分红欧式看涨：

Δ=∂c∂S=N(d1).\Delta=\frac{\partial c}{\partial S}=N(d_1).

对 BS 公式对 SS 求导即可。

B. 平值 S≈KS\approx K 且 r≈0r\approx 0 时，通常 Δ≈0.5\Delta\approx 0.5。当 τ→0\tau\to 0，平值极限仍趋近 0.5；深度实值趋近 1，深度虚值趋近 0。

C. 若 S↑S\uparrow，d1↑d_1\uparrow，则 Δ↑\Delta\uparrow，要保持 delta 中性需空更多标的（增加空头股数）。

D. 常见短期限近似：

c≈0.4 σSτ.c\approx 0.4\,\sigma S\sqrt{\tau}.

英文解析

AnswerA:
Δ=N(d1).\Delta = N(d_1).

The standard call formula is
c=S N(d1)  −  K e−rτ N(d2),d1=ln⁡ ⁣(SK)+(r+σ22) τστ,d2=d1−στ.c = S\,N(d_1) \;-\; K\,e^{-r\tau}\,N(d_2), \quad d_1 = \frac{\ln\!\bigl(\tfrac{S}{K}\bigr) +(r + \tfrac{\sigma^2}{2})\,\tau} {\sigma\sqrt{\tau}}, \quad d_2 = d_1 - \sigma\sqrt{\tau}.
Differentiate cc w.r.t. SS and you find ∂c∂S=N(d1).\frac{\partial c}{\partial S}=N(d_1).

AnswerB:
At-the-money (ATM): S=K.S=K. Typically Δ≈0.5\Delta\approx 0.5 if r=0.r=0. As τ→0,\tau\to0, indeed Δ→0.5.\Delta\to0.5. For deep in-the-money, Δ→1;\Delta\to1; for deep out-of-the-money, Δ→0.\Delta\to0.

AnswerC:
Δ=N(d1).\Delta= N(d_1). If S↑,S\uparrow, d1↑,d_1\uparrow, so Δ↑.\Delta\uparrow. You must short more GM shares to remain delta neutral.

AnswerD:
A short, near-ATM approximation:
c≈0.4 σ S τ.c \approx 0.4\,\sigma\,S\,\sqrt{\tau}.