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I am trying to understand how the shifted log-normal model works, in which we shift a log-normal model by a factor before the simulation so that interest rates don't turn negative during the simulation, and later adjust it back.

How can we be assured that this factor will make sure interest rates will never go negative during the simulation?

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Let us assume we are interested in some (forward) rate $F_t=F(t,T)$ which we assume is log-normally distributed: $$\text{d}F_t=\sigma F_t\text{d}W_t$$ However, we observe market rates can in practice be negative. In order to circumvent this issue, we would like to use shifted log-normal dynamics: $$\sigma(F_t+s)\text{d}W_t$$ where $s>0$ is the shift. We therefore define the shifted rate $F_t^s=f(F_t)=F_t+s$, which has the same dynamics than $F_t$ (apply Itô's Lemma to $f(F_t)$): $$\text{d}F_t^s=\sigma F_t^s\text{d}W_t$$ Yet we know the solution to the above SDE: $$\begin{align} F_t^s&=F_0^s\exp\left\{-\frac{\sigma^2}{2}t+\sigma W_t\right\} \\ &=(F_0+s)\exp\left\{-\frac{\sigma^2}{2}t+\sigma W_t\right\} \end{align}$$ Thus if $s>|F_0|$ where $|\cdot|$ is the absolute value we ensure $F_t^s>0$ for all $t$. Note that we of course have that: $$\begin{align} F_t &= F_t^s-s \\ &= -s+(F_0+s)\exp\left\{-\frac{\sigma^2}{2}t+\sigma W_t\right\} \end{align}$$ So that we can recover the proper rate once the shifted rate has been simulated. In particular note that the forwards are preserved under the forward measure $\mathcal{T}$, that is: $$E^\mathcal{T}\left(F_t^s-s\right)=F_0$$

It is the shifted rate $F_t^s$ that will not turn negative, not the rate $F_t$ itself: after all the reason shifted log-normal models were introduced is because we were observing negative rates on the market. A shifted log-normal model allows to represent rates that can be negative while preserving the preexisting modeling infrastructure based on log-normal dynamics.

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    $\begingroup$ Would be cool to show how a caplet with a negative strike can be priced using the above (it's just the Black76 with the strike shifted by the same constant "$s$" that is used to shift the forward rate, whilst the spot is also shifted: might be nice to tie it into what you wrote for everyone to see the "full solution" in practice) :) $\endgroup$ – Jan Stuller Mar 16 at 16:01
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Daniel Olivaw:

Sorry but your math is wrong. If $$dF(t)=\sigma(F(t)-\alpha)dW(t)$$ then $$F(t)=\alpha+(F(0)-\alpha)e^{-\frac{\sigma^2}{2}t+\sigma W(t)}$$

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    $\begingroup$ Hello @user53197. I think we’re looking at different things. You’re giving the solution to the base rate $F$, while I am looking at the shifted rate $F^s$ which per Ito’s Lemma has the dynamics specified in my post. You probably have a point in that my answer can lead to confusion though, I will try to review it. $\endgroup$ – Daneel Olivaw Mar 16 at 15:19

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