Problem 26.2 Easy Free 2 variations

At-the-Money Call Approximation

A stock trades at $S_0=\$100$ and follows geometric Brownian motion with zero interest rate. An at-the-money European call with three months to expiry has implied volatility $40\%$ . Using the mental approximation $C\approx 0.4\sigma S\sqrt{T}$ , estimate the call price without a calculator.

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. At the money with zero rates, the Black–Scholes price is $S$ times a probability difference that is linear in $\sigma\sqrt{T}$ to first order.

With $r=0$ and $K=S$ ,

d_{1,2}=\frac{\ln(S/K)\pm\tfrac12\sigma^2T}{\sigma\sqrt T}=\pm\tfrac12\sigma\sqrt T .

Hence

C=S\,N(d_1)-K\,N(d_2)=S\bigl(2N(\tfrac12\sigma\sqrt T)-1\bigr).

For small $x$ , $N(x)\approx\tfrac12+\varphi(0)\,x$ with $\varphi(0)=\frac{1}{\sqrt{2\pi}}$ . Therefore

C\approx \frac{\sigma S\sqrt T}{\sqrt{2\pi}}\approx 0.4\,\sigma S\sqrt T,

because $\frac{1}{\sqrt{2\pi}}\approx 0.3989$ .

Here $\sigma=0.4$ , $S=100$ and $T=\tfrac14$ , so $\sqrt T=\tfrac12$ . Then

C\approx 0.4\times 0.4\times 100\times\tfrac12 = 8 .

The exact Black–Scholes value is $100\bigl(2N(0.1)-1\bigr)\approx 7.97$ . The rule is accurate because $\sigma\sqrt T=0.2$ is small. It slightly overestimates, since $N$ is concave on $[0,\infty)$ .

\boxed{C\approx \$8}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 26.2a

An at-the-money European call on a stock with $S_0=12$ expires in three months and trades at $C_0=2.3$ . Using the approximation $C\approx 0.4\sigma S\sqrt{T}$ , estimate the annualized implied volatility $\sigma$ . Round to three decimal places.

Work it out, then compare with the answer.

Solution

Invert the rule $C\approx 0.4\,\sigma S\sqrt T$ . With $S=12$ and $T=\tfrac14$ , so $\sqrt T=\tfrac12$ :

\sigma\approx\frac{C}{0.4\,S\sqrt T}=\frac{2.3}{0.4\times 12\times 0.5}=\frac{2.3}{2.4}\approx 0.958 .

Using the exact constant $\frac{1}{\sqrt{2\pi}}$ in place of $0.4$ gives

\sigma=\frac{2.3\sqrt{2\pi}}{6}\approx 0.961 .

Solving the full Black–Scholes equation $2N(\sigma/4)-1=2.3/12$ gives $\sigma\approx 0.970$ . The linear rule loses some accuracy at volatilities this high.

\boxed{\sigma\approx 0.958}

Variation 26.2b

A stock trades at $S=20$ with volatility $\sigma=0.4$ . Using the approximation that an at-the-money call is worth about $0.4\sigma S\sqrt{T}$ and that an at-the-money put has roughly the same value, estimate the price of an at-the-money straddle (one call plus one put, same strike equal to $S$ ) expiring in $T=0.36$ years. Round to two decimal places.

Work it out, then compare with the answer.

Solution

Take $r=0$ , the standing assumption of this section. Put–call parity with $K=S$ then gives $C-P=S-K=0$ , so the at-the-money put equals the call exactly. The straddle is therefore worth twice the call.

With $\sigma=0.4$ , $S=20$ and $\sqrt T=0.6$ :

2\times 0.4\,\sigma S\sqrt T = 2\times 0.4\times 0.4\times 20\times 0.6 = 3.84 .

With the exact constant $\frac{1}{\sqrt{2\pi}}$ the estimate is $3.83$ . The exact Black–Scholes value is $40\bigl(2N(0.12)-1\bigr)\approx 3.82$ .

\boxed{\text{Straddle}\approx 3.84}

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