local volatility - Downward sloping smile in normal model

We consider an stock price $S$ following a normal model: $dS_t = \sigma dW_t$

We can write this as $\frac{dS_t}{S_t}=\frac{\sigma}{S_t}dW_t$

Hence we can see that $S$ follows a "log-normal" diffusion with a local volatility function $c(S)=\frac{\sigma}{S}$ which is downward sloping.

My question is: can we deduce that the log normal smile implied by this model will be downward sloping as well ? That is to say, if we have a local volatility function which is decreasing as a function of $S$, will the lognormal implied vol be decreasing as a function of the strike ?

Thanks !

Answer

Since $S_T = S_0 + \sigma W_T$, \begin{align*} C &:= E\left((S_T-K)^+ \right)\\ &= E\left((S_0+\sigma W_T-K)^+ \right)\\ &=\int_{\frac{K-S_0}{\sigma \sqrt{T}}}^{\infty}(S_0+\sigma\sqrt{T} x-K) \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\\ &=(S_0-K)\Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right)+\frac{\sigma\sqrt{T}}{\sqrt{2\pi}}e^{-\frac{(S_0-K)^2}{2\sigma^2 T}}, \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable. Then, \begin{align*} \frac{dC}{d K} &= -\Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right) <0. \end{align*} On the other hand, let $\sigma_I(K)$ be the log-normal implied volatility, that is, \begin{align*} C = C(K, \sigma_I(K)). \end{align*} Then \begin{align*} \frac{dC}{d K} &=\frac{\partial C}{\partial K} + \frac{\partial C}{\partial \sigma_I}\frac{\partial \sigma_I}{\partial K}. \end{align*} Here, \begin{align*} \frac{\partial C}{\partial K} = - \Phi(d_2), \end{align*} where \begin{align*} d_2 = \frac{\ln\frac{S_0}{K} - \frac{1}{2}\sigma_I^2 T}{\sigma_I \sqrt{T}}. \end{align*} Since \begin{align*} \lim_{K\rightarrow \infty}\frac{S_0-K}{\ln \frac{S_0}{K}} = \infty, \end{align*} we can expect that, for $K$ sufficiently large, \begin{align*} d_2 > \frac{S_0-K}{\sigma \sqrt{T}}. \end{align*} That is, \begin{align*} \frac{\partial C}{\partial \sigma_I}\frac{\partial \sigma_I}{\partial K} &= \Phi(d_2) - \Phi\left(\frac{S_0-K}{\sigma \sqrt{T}}\right) > 0. \end{align*} Then, \begin{align*} \frac{\partial \sigma_I}{\partial K} > 0, \end{align*} and the implied volatility is a increasing function of such strike levels. In conclusion, the implied volatility does not have to be a decreasing function of the strike.