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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?

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  • $\begingroup$ Two questions, is the price of a call option in the BS universe monotonic in volatility and what happens as vol tends to 0? $\endgroup$ – river_rat May 22 at 21:56
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    $\begingroup$ 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. $\endgroup$ – Arshdeep May 23 at 7:15
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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

  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}

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