Question on subgame perfect equilibrium
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} &amp; &amp; \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} &amp; &amp; \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,
$$T_X^* = \frac{abI}{ab + 1}$$