Why is it that for European Puts on Non-Dividend-Paying Stocks, the lower-bound for price is $$p=Ke^{-rT}-S_0?$$
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2$\begingroup$ It is a consequence of a no arbitrage argument which you can easily workout. Hint: compare a strategy where you initially hold a put with a strategy where you initially hold -1 share of stock and an amount $Ke^{-rT}$ of cash. $\endgroup$ – Antoine Conze Jan 23 '19 at 15:43
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Let's consider the following 2 portfolios:
Portfolio A: one European put option plus one share,
Portfolio B: a zero coupon bond paying off $K$ at time $T$.
If $S_T<K$ then the option in portfolio A is exercised at $T$ and the portfolio is worth $K$.
If $S_T>K$, then the put option expires worthless and the portfolio is worth $S_T$ at this time. Hence, portfolio A is worth $max(S_T, K)$ at $T$. Portfolio B is worth $K$ in time $T$. Hence, portfolio A is always worth as much as, and can sometimes be worth more than more than B at $T$. It follows then that in the absence of arbitrage opportunities portfolio A must be worth at least as much as B today. Hence: $$p+S_0>= Ke^{-rT}$$ or $$p>=Ke^{-rT}-S_0$$
