Question on subgame perfect equilibrium

Question

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 $s_Y$?

Answer 1

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