# Help to understand a Log-linearization and subsequent Differentiation with respect to time

At one point there are a couple of simple passages that however I did not fully understand.

In particular, I can't figure out how a piece of the first term of (3) disappeared in the log linearization. And also how all three terms appear in the differentiation with respect to time (4).

Can anyone help me understand the steps? I offer a virtual beer, points, and much appreciation. Thank you!

This is my guess:

We start with the following equation: $$\pi_t = \frac{(1-\mu)(1-\lambda)}{\mu}\left[\lambda \left(\frac{k_{et}}{s_t}\right)^\rho + (1-\lambda)\right]^{(\sigma-\rho)/\rho}\left(\frac{h_{ut}}{h_{st}}\right)^{(1-\sigma)}\left(\frac{\psi_{st}}{\psi_{ut}}\right)^\sigma$$

Taking logs gives: $$\ln(\pi_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) + \frac{\sigma - \rho}{\rho}\ln\left[\lambda \left(\frac{k_{et}}{s_t}\right)^\rho + (1-\lambda)\right] + (1-\sigma)\ln h_{ut} - (1-\sigma) \ln h_{st} + \sigma\ln \psi_{st} - \sigma \ln \psi_{ut}$$

Take the approximation $$\ln(1 + x ) \approx x$$ to simplify the second term to: $$\frac{\sigma - \rho}{\rho} \left[-\lambda + \lambda \left(\frac{k_{et}}{s_t}\right)^\rho\right] = -\lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k_{et}}{s_t}\right)^\rho$$ This gives: $$\ln(\pi_t) = \ln\left(\frac{(1-\mu)(1-\lambda)}{\mu}\right) - \lambda \frac{\sigma - \rho}{\rho} + \frac{\sigma - \rho}{\rho}\lambda \left(\frac{k_{et}}{s_t}\right)^\rho\\ + (1-\sigma)\ln h_{ut} - (1-\sigma) \ln h_{st} + \sigma\ln \psi_{st} - \sigma \ln \psi_{ut}$$

Now differatiating with respect to time gives on the left hand side $$g_{\pi_t}$$. The first two terms on the right hand side disappear. For the third term we get: \begin{align*} &\frac{\sigma - \rho}{\rho} \lambda \rho \left(\frac{k_{et}}{s_t}\right)^{\rho - 1} \frac{\dot k_t s_t - k_t \dot s_t}{(s_t)^2},\\ &= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k_{et}}{s_t}\right)^\rho \frac{s_t}{k_t} \frac{\dot k_t - k_t g_{s_t}}{s_t},\\ &= \frac{\sigma - \rho}{\rho} \lambda \left(\frac{k_{et}}{s_t}\right)^\rho (g_{k_t} - g_{s_t}) \end{align*} I think $$g_{s_t} = g_{h_{st}} + g_{\psi_{st}}$$. The other terms give: $$(1-\sigma) (g_{h_{ut}} - g_{h_{st}}) + \sigma (g_{\psi_{st}} - g_{\psi_{ut}})$$

• Yep, sorry I'll try to correct.
– tdm
Oct 30 at 11:52