这是一份liverpool利物浦大学ECON343的成功案例

Suppose the individual has labor income of $Y_{1}$ in the first period of life and zero in the second period. Second-period consumption is thus $(1+r)\left(Y_{1}-C_{1}\right) ; r$, the rate of return, is potentially random.
(i) Find the first-order condition for the individual’s choice of $C_{1}$.
(ii) Suppose $r$ changes from being certain to being uncertain, without any change in $E[r]$. How, if at all, does $C_{1}$ respond to this change?
Suppose the individual has labor income of zero in the first period and $Y_{2}$ in the second. Second-period consumption is thus $Y_{2}-(1+r) C_{1} \cdot Y_{2}$ is certain; again, $r$ may be random.

ECON343 COURSE NOTES ：

(i) Show that $\partial \ln K_{\mathrm{t}+1} / \partial \ln K_{\mathrm{t}}$ (holding $A_{t}, L_{t}, C_{t}$, and $G_{t}$ fixed) is (1+ $\left.r_{t+1}\right)\left(K_{\mathrm{t}} / K_{t+1}\right)$.
(ii) Show that this implies that $\partial \ln K_{t+1} / \partial \ln K_{t}$ evaluated at the balanced growth path is $\left(1+r^{}\right) / e^{n+g} \cdot 4^{47}$ Show that $$ \tilde{K}{t+1} \simeq \lambda{1} \tilde{K}{t}+\lambda{2}\left(\tilde{A}{t}+\tilde{L}{t}\right)+\lambda_{3} \tilde{G}{t}+\left(1-\lambda{1}-\lambda_{2}-\lambda_{3}\right) \tilde{C}{l} \text {. } $$ where $\lambda{1} \equiv\left(1+r^{}\right) / e^{n+g}, \lambda_{2} \equiv(1-\alpha)\left(r^{}+\delta\right) / \alpha e^{n+g}$, and $\lambda_{3}=-\left(r^{}+\delta\right)$ $(G / Y)^{} / \alpha e^{n+g}$; and where $(G / Y)^{}$ denotes the ratio of $G$ to $Y$ on the balanced growth path without shocks. (Hints: 1 . Since the production function is Cobb-Douglas, $Y^{}=\left(r^{}+\delta\right) K^{} / \alpha^{} 2$. On the balanced growth path, $K_{t+1}=e^{n+g} K_{t}$, which implies that $\left.C^{}=Y^{}-G^{}-\delta K^{}-\left(e^{n+g}-1\right) K^{*} .\right)$
Use the result in (b) and equations (4.43)-(4.44) to derive (4.52), where $b_{\mathrm{KK}}=\lambda_{1}+\lambda_{2} a_{\mathrm{LK}}+\left(1-\lambda_{1}-\lambda_{2}-\lambda_{3}\right) a_{\mathrm{CK}} b_{\mathrm{KA}}=\lambda_{2}\left(1+a_{\mathrm{LA}}\right)+\left(1-\lambda_{1}-\lambda_{2}-\lambda_{3}\right) a_{\mathrm{CA}}$, and $b_{\mathrm{KG}}=\lambda_{2} a_{\mathrm{LG}}+\lambda_{3}+\left(1-\lambda_{1}-\lambda_{2}-\lambda_{3}\right) a_{\mathrm{CG}}$.