Interpreting the Ratio of Normal CDFs in Black–Scholes Hedging

Summary

This note interprets the ratio of the standard normal cumulative probabilities Φ(d₂) and Φ(d₁) in the Black–Scholes call formula. The call value is expressed as the stock holding value, S₀Φ(d₁), minus the discounted-strike cash amount, Ke⁻ʳᵀΦ(d₂). Thus Φ(d₁) is the call's delta, while Φ(d₂) scales the discounted cash leg; neither probability alone is the entire hedge position.

For a portfolio of options, the note equates the aggregate share holding and short cash amount to the quantities supplied in the example, then rearranges their ratio to recover Φ(d₂)/Φ(d₁), accounting for strike and discounting. It reports a numerical value using the stated inputs. This interpretation is specific to the Black–Scholes framework and the given call hedge decomposition; the ratio is not a standalone market quantity independent of maturity, rates, and option parameters.

Key ideas

• In the Black–Scholes call formula, Φ(d₁) is the option delta and determines the share hedge per option.
• The discounted-strike term involving Φ(d₂) represents the short cash leg of the call decomposition.
• The ratio Φ(d₂)/Φ(d₁) can be inferred from the cash and share hedge totals after adjusting for strike and discounting.
• The numerical ratio applies to the stated example and its Black–Scholes assumptions.

Tags

# Ratio of gaussian CDFs in Black-scholes option pricing formula


# Ratio of gaussian CDFs in Black-scholes option pricing formula












What is meant by $\frac {\Phi (d_2)}{\Phi (d_1)}$ in the Black Scholes call option price?

I found it in a solution as $\frac{\text{short position in cash}}{(\text{number of shares})(\text{strike price discounted to time zero})}$

Reference can be found here

Q. Number 4 on page number 19, and Its solution on page number 26

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/22988

For a call option with price given by \begin{align*} c = S_0 \Phi(d_1) - K e^{-rT}\Phi(d_2), \end{align*} the delta hedge ratio $\Phi(d_1)$ is the number of shares to hold. That is, $S_0 \Phi(d_1)$ is the total holding share value for hedging, while $K e^{-rT}\Phi(d_2)$ is the total cash amount in short.

In the question, it says that, for $N$ options, 250,000 shares of the stock are hold, and the amount of $ £413,057$ is in short. The strike price is $K=2.0$. Therefore, \begin{align*} N \Phi(d_1) = 250000, \mbox{ and } N K e^{-rT}\Phi(d_2) = £413057. \end{align*} Consequently, \begin{align*} \frac{\Phi(d_2)}{\Phi(d_1)} &= \frac{N K e^{-rT}\Phi(d_2)}{N\Phi(d_1)}\frac{N}{N K e^{-rT}}\\ &=\frac{N K e^{-rT}\Phi(d_2)}{N\Phi(d_1) K e^{-rT}} \\ &=\frac{413057}{250000 \times 2.0 \times e^{-0.03 \times 0.5}}\\ &= 0.8386. \end{align*}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.