Show that under no arbitrage assumption, the PUT option price is a convex function of the underlying asset price. Is this true for the CALL option? What can you say about the delta of these options?
I am doing this for European call option you can easily replicate the same for the European put option. The Black-Scholes-Merton result is \begin{equation} c = S_0N(d_1) - Ke^{-rT}N(d_2) \tag{1} \end{equation} Where: \(c\) : Price of European Call option \(S_0\): Price of underlying asset today \(K\): Strike Price or Exercise Price \(r\): expected return, risk free return \(T\): time to maturity of the option \(\sigma \): stock volatility \(N(.)\): CDF of standard normal distribution \(d_1 = \dfrac{\ln(S_0e^{rT}/K) +\sigma ^2T/2 }{\sigma \sqrt{T}}\), \(d_2 = \dfrac{\ln(S_0e^{rT}/K) -\sigma ^2T/2 }{\sigma \sqrt{T}}\) Note that following holds \(d_1 = d_2 + \sigma \sqrt{T}\) \(N'(x) = \frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}) \quad \forall x\in \mathcal{R}\) By the above two observations lets first establish some relationship between \(N'(d_1)\) and \(N'(d_2)\). \begin{align} N'(d_1) &= N'(d_2+\sigma \sqrt{T}) \\ &= \frac{1}{\sqrt{2\pi}}\exp\left (-\frac{(d_2+\sigma \sqrt{T})^2}{2}\right ) \\ &= \frac{1}{\sqrt{2\pi}}\exp\left (-\frac{d_2^2}{2}-\frac{\sigma^2 T}{2}-d_2\sigma \sqrt{T}\right )\\ &= \frac{1}{\sqrt{2\pi}}\exp\left (-\frac{d_2^2}{2}\right )\exp\left (-\left (\frac{\sigma^2 T}{2}+d_2\sigma \sqrt{T}\right )\right)\\ &= N'(d_2)\exp\left (-\left (\frac{\sigma^2 T}{2}+d_2\sigma \sqrt{T}\right )\right) \tag{2} \end{align} Now by using the expression of \(d_2\) above we the following \begin{align} -\left (\frac{\sigma^2 T}{2}+d_2\sigma \sqrt{T}\right ) &= -\ln\left(\frac{S_0e^{rT}}{K}\right) \\ \Leftrightarrow \exp\left (-\left (\frac{\sigma^2 T}{2}+d_2\sigma \sqrt{T}\right )\right) &= \frac{K}{S_0e^{rT}} \end{align} By substituting the above expression in (2) we get \begin{align} S_0N'(d_1) = Ke^{-rT}N'(d_2) \tag{3}\end{align} Now let's show that \(c\) is convex in \(S_0\) i.e. the price of underlying asset \begin{align*} \frac{\partial c}{\partial S_0} &= S_0N'(d_1)\frac{\partial d_1}{\partial S_0} + N(d_1) - ke^{-rT}N'(d_2)\frac{\partial d_2}{\partial S_0}\\ &= S_0N'(d_1)\frac{\partial d_1}{\partial S_0} + N(d_1) - ke^{-rT}N'(d_2)\frac{\partial d_1}{\partial S_0}\\ &= N(d_1) + \frac{\partial d_1}{\partial S_0}(S_0N'(d_1)-Ke^{-rT}N'(d_2)) \\ &= N(d_1) \end{align*} The above expression is the delta of european call. \begin{align*} \frac{\partial ^2 c}{\partial S_0^2} &= N'(d_1)\frac{\partial d_1}{\partial S_0}\\ &= \frac{N'(d_1)}{S_0\sigma \sqrt{T}} \end{align*} Since the second partial derivative of \(c\) wrt \(S_0\) is always positive therefore \(c\) is convex in \(S_0\)
