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I have a question regarding the application of a shift to the Black-Scholes formula for negative forward rates.

I am reading in the Brigo book that "increasing the shift $\alpha$ shifts the volatility curve down, whereas decreasing $\alpha$ shifts the curve up".

Does this not imply then that the probability distribution will consequently change and thus prices will be affected? If that is the case, then isn't applying a shift to the distribution causing misprices for e.g. caps/floors/swaptions ?

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Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\,\beta_t\, dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The caplet Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.

Applying a shift, the probability distribution has been changed, for example, lognormal distribution has been changed to a shifted lognormal distribution. However, the price will not change, while the implied volatility changes. In the current negative or small interest rate environment, people tend to quote an interest rate product by its price. Then given the price, an implied volatility is computed with a certain shift parameter; otherwise, it may not be feasible to find the implied volatility (e.g., the forward Libor rate $L(0; T, T+\Delta)$ may be negative). See also this paper.

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  • $\begingroup$ Thanks for your answer, very thorough and helpful. You describe how price is not affected, but what about the vega of the option? Delta and Gamma remain the same, but intuitively I guess that is not the case for vega? $\endgroup$ – Adam Mar 8 '16 at 9:38
  • $\begingroup$ @Adam: As hedge ratios are model dependent, they will all change when the lognormal model is changed to the shifted lognormal model. For example, though we can adjust the volatility so that the prices are matched, however, we do not match $d_1$, the delta hedge ratio $\Phi(d_1)$ is then not macthed. $\endgroup$ – Gordon Mar 8 '16 at 13:44
  • $\begingroup$ looking more closely at your answer, could you please ellaborate a bit more on how did you arrive to the conclusion that the derivative with respect to $\alpha$ results in vega times the derivative of volatility with respect to $\alpha$? To obtain $vega$ you would have had to derive the option's price with respect to $\sigma$, so I must have missed something. Moreover, how do you conclude that $d\hat{\sigma}/d\alpha < 0$ and from there that increasing the shift shifts the vol curve down? Thanks $\endgroup$ – Adam Mar 11 '16 at 11:46
  • $\begingroup$ @Adam: Since $\hat{\sigma}$ is a function of $\alpha$, the dderivative of the left-hand-side (lhs) of (3) is then given by $\frac{\partial lhs}{\partial \hat{\sigma}}\frac{\partial \hat{\sigma}}{\partial \alpha}$. Here $\frac{\partial lhs}{\partial \hat{\sigma}}$ is a (undiscounted) vega, which is always positive. Moreover, the derivative of the right-hand-side of (3) is negative, then the derivative $\frac{\partial \hat{\sigma}}{\partial \alpha}$ is negative. $\endgroup$ – Gordon Mar 11 '16 at 13:43

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