Reading N(d₂) in Black–Scholes

HardFinance & Derivatives~4m

In the Black–Scholes call price $c = S\,N(d_1) - Ke^{-rT} N(d_2)$ , exactly one of the following statements about $N(d_1)$ and $N(d_2)$ is true in general (for $\sigma>0$ , $T>0$ ). Which one?

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Use $d_1 = d_2 + \sigma\sqrt{T}$ and that $N$ is strictly increasing.
Which of the two is the delta, and which is the risk-neutral exercise probability?

Answer

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N(d₁) > N(d₂) for all σ>0, T>0, since d₁ = d₂ + σ√T and N is increasing

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Asked at: Jane Street, Two Sigma

Reading N(d₂) in Black–Scholes

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