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Define $\ln\frac{c_t}{c}=\hat{c_t}$ which means % deviation from steady state.
Then $\frac{1}{c_t +\alpha g_t}=\lambda_t$ is equivalent as
$\frac{1}{c e^{\hat{c_t}} + \alpha g e^{\hat{g_t}}}=\lambda e^{\hat{\lambda_t}}$
$c e^{\hat{c_t}} + \alpha g e^{\hat{g_t}}=\frac{1}{\lambda} e^{\hat{-\lambda_t}}$
$-\frac{c}{c +\alpha g} \hat{c_t} - \frac{\alpha g}{c+\alpha g}\hat{g_t}\approx \hat{\lambda_t}$
How to derive the last equation with approximation?

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1 Answer 1

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Taking a Taylor expansion of $e^\hat x$ gives around $0$ gives: $$ e^\hat x \approx e^0 + e^0 \hat x = 1 + \hat x. $$ Substitution gives: $$ \begin{align*} &(c + \alpha g) + c \hat c_t + \alpha g \hat g_t \approx \frac{1}{\lambda}(1 - \hat \lambda_t),\\ \iff & \lambda(c + \alpha g) + \lambda c \hat c_t + \lambda \alpha g \hat g_t \approx 1 - \hat \lambda_t,\\ \iff & 1 + \frac{c}{c + \alpha g} \hat c_t + \frac{\alpha g}{c + \alpha g} \hat g_t \approx 1 - \hat \lambda_t,\\ \iff & -\frac{c}{c + \alpha g} \hat c_t- \frac{\alpha g}{c + \alpha g}\hat g_t \approx \hat \lambda_t \end{align*} $$

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  • $\begingroup$ Why the $ \frac{1}{\lambda(c+\alpha g)}$ term doesn't affect RHS? $\endgroup$
    – guest
    Oct 16, 2021 at 11:20
  • $\begingroup$ $\lambda = \frac{1}{c + \alpha g} $ so $\frac{1}{\lambda(c + \alpha g)} = 1$. $\endgroup$
    – tdm
    Oct 18, 2021 at 5:26

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