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

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?

• Two questions, is the price of a call option in the BS universe monotonic in volatility and what happens as vol tends to 0? – river_rat May 22 at 21:56
• You cannot derive a model independent result using a model. I mean, you can do it but it doesn't prove anything beyond the B-S world. – Arshdeep May 23 at 7:15

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
1. $$f$$ is a monotonically increasing function in $$\sigma$$, by say calculating its derivative wrt to $$\sigma$$,
2. 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}