The volatility of the short rate starts at the constant $\sigma$, and then exponentially decreases to zero. This is evident from the equation

$$\mathrm{dr} = \lambda(t)\mathrm{dt} + \sigma e^{-\alpha t}\mathrm{dw}$$

, where the term $\sigma e^{-\alpha t}$ represents the volatility of the short rate. Here, $\sigma$ is the initial volatility, and $e^{-\alpha t}$ is the exponential decay factor. As time ($t$) increases, $e^{-\alpha t}$ decreases exponentially, thus causing the overall volatility to decrease towards zero. This behavior is consistent with the concept of time-dependent volatility, where volatility changes over time. In this case, the volatility starts at a constant value ($\sigma$) and decreases exponentially over time, eventually approaching zero.

Choice B is incorrect. The volatility of the short rate does not end at the constant $\lambda$. In Model 3, $\lambda$ represents the drift term and not the volatility. Furthermore, there is no indication in Model 3 that suggests that volatility decreases exponentially to zero from $\lambda$.

Choice C is incorrect. According to Model 3, the volatility of the short rate starts at a constant $\sigma$ and then decreases exponentially over time due to $e^{-\alpha t}$ factor in equation. It does not increase from zero to infinity as suggested by this choice.

Choice D is incorrect. As explained above, in Model 3, $\lambda$ represents drift term and not initial point for volatility of short rate which then decreases to zero.