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Consider a world of complete information with two agents X and Y and two time periods 1 and 2.

Person X only lives in second period.

Person Y lives in 1st and 2nd periods both.

X and Y are each endowed with and exogenous income I which can be allocated between consumption in both periods.

$s_Y$ = Y’s saving and $0\le s_Y \le I$

$c_{1Y} $ and $c_{2Y} $ are the agent Y’s first and second period consumptions respectively.

The A gent X’s preferences are altruistic for Y. After observing the saving of Y, $sY$,X determine how much his endowment $tX$ to transfer to Y for $t_x\in [0,I]$

$c_{2X}$ the agent X’s consumption in period 2.

Utility functions Y and X are respectively

$$V_Y= ln(C_{1Y}) + bln(C_{2Y})$$

$$V_X=ln(C_{2X})+a *V_Y$$

where a is positive and $a*b\ge 1$ and $b\in (0,1)$

What is the su game perfect equilibrium levels for $t_X$ and $s_Y$?

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  • $\begingroup$ You should probably know by now that simply stating the problem without showing any effort at solving it is unlikely to get you an answer. $\endgroup$ – Herr K. May 8 '18 at 14:05
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If $S_Y$ is the amount saved by $Y$ in the first period, and $T_X$ is the amount transferred by $X$ to $Y$ in the second period, then $X$'s payoff as function of his choice $T_X$, taking as given $Y$'s choice $S_Y$ is $$\ln(I - T_X) + a\left(\ln(I-S_Y) + b\ln(S_Y+T_X)\right)$$

$X$ will choose $T_X$ by solving the following problem : $$\max_{T_X \geq 0} \ \ \ln(I - T_X) + a\left(\ln(I-S_Y) + b\ln(S_Y+T_X)\right)$$

First-order condition for optimality is:

$$\frac{1}{I-T_X} =\frac{ab}{S_Y + T_X} $$

Solving it for $T_X$ we get,

$$T_X = \frac{abI-S_Y}{ab+1}$$

This is the best response transfer function of $X$.

Now we solve $Y$'s utility maximization problem in period 1, taking as given $X$'s best response function of period 2.

\begin{eqnarray*} \max_{S_Y\geq 0} & & \ln(I - S_Y) + b\ln\left(S_Y + \frac{abI-S_Y}{ab+1}\right)\end{eqnarray*}

which can be rewritten as

\begin{eqnarray*} \max_{S_Y \geq 0} & & \ln(I - S_Y) + b\ln\left(I+S_Y\right) + b \ln\left(\frac{ab}{ab+1}\right)\end{eqnarray*}

Solving it we get $$S_Y^* = \max\left(\frac{(b-1)I}{b+1}, 0\right) $$

Since $b \in (0, 1)$, $$S_Y^* = 0$$

Consequently, $$T_X^* = \frac{abI}{ab + 1}$$

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  • $\begingroup$ Thank you so much. Really great and smart solution. And I understand what I need to do when I see such type of SPE questions $\endgroup$ – user315 May 8 '18 at 20:57
  • $\begingroup$ Dear Amit, we are studying this question with my friend, we have some solution. but we are stuck. Please help me. You solve very well. So I’m asking you for help economics.stackexchange.com/questions/21923/… thanks a lot. $\endgroup$ – user315 May 11 '18 at 0:09

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