Deriving the European Call Lower Bound from Black–Scholes

Summary

The document outlines a way to derive the European call lower bound directly from the Black–Scholes formula, without relying on put–call parity. Hold the spot price, strike, interest rate, and maturity fixed, and view the call price as a function of volatility. The suggested argument shows that the price increases with volatility, then evaluates its limit as volatility approaches zero.

In that limit, the terms inside the Black–Scholes normal distribution functions are governed by whether spot exceeds the discounted strike, equals it, or falls below it. This produces the lower-bound expression involving the positive part of spot minus discounted strike. The excerpt gives the outline and key limiting cases, rather than a full derivative calculation. It assumes the standard Black–Scholes setup and does not discuss dividends, alternative carry assumptions, or extensions to other option styles.

Key ideas

• Treat the Black–Scholes call price as a function of volatility while fixing the other inputs.
• Show that the call price is nondecreasing in volatility.
• Evaluate the zero-volatility limit using the sign of spot relative to the discounted strike.
• The limiting price yields the European call lower bound.

Tags

# European call option lower bound derivation by Black-Scholes formula


# European call option lower bound derivation by Black-Scholes formula












Derive the lower bound of european call options: $$C(S, t)\geq[S-e^{-r(T-t)}K]^+$$

I know how to derive it using put-call parity, but is there any way to derive from Black-Scholes formula?

## Answer by ir7 (score 1)

https://quant.stackexchange.com/a/64179

Hint:

Think of BS formula as a function of $\sigma>0$, $f(\sigma)$, with all other relevant parameters ($S$, $K$, $r$, $t$, $T$) fixed constants. Then show that

- $f$ is a monotonically increasing function in $\sigma$, by say calculating its derivative wrt to $\sigma$,

- and calculate $$\lim_{\sigma \rightarrow 0^+} f(\sigma).$$

Note that the main piece of calculation in (2) contains the 'switch' $ \ln\frac{S}{{\rm e}^{-r(T-t)}K}$ related to the right hand side of your inequality:

\begin{align}&\lim_{\sigma \rightarrow 0^+}\frac{\ln \left( \frac{S}{{\rm e}^{-r(T-t)}K} \right)\pm\frac{\sigma^2}{2}(T-t)}{\sigma\sqrt{T-t}} \\&=\begin{cases} \infty & ,\; \; \; \ln \left( \frac{S}{{\rm e}^{-r(T-t)}K} \right)>0\\ -\infty &, \; \; \; \ln \left( \frac{S}{{\rm e}^{-r(T-t)}K} \right) <0 \\ 0 &, \; \; \;\ln \left( \frac{S}{{\rm e}^{-r(T-t)}K} \right)=0 \\\end{cases}\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.