= Solution
{c}
Write $p(s)=p(T,s;K_3,T_3)$ and $c(s)=c(T,s;K_3,T_3)$. Because the right can be left unexercised, the <chooser with positive exercise cost> has payoff
$$
\boxed{H(s)=\max\{p(s)-K,\ c(s)-K,\ 0\}.}
$$
Let $\tau=T_3-T$ and $D=K_3e^{-r\tau}$. <Put-call parity> gives $c(s)-p(s)=s-D$. Thus the put is selected for $s<D$ and the call for $s>D$; the two are equal at $s=D$. Before truncation at zero, the payoff is decreasing on the left and increasing on the right.
At the meeting point, the <Black-Scholes formula> gives
$$
p(D)=c(D)=M=D\{2\Phi(\sigma\sqrt\tau/2)-1\}.
$$
If $K<M$, the payoff has a positive minimum $M-K$ at $D$. If $K=M$, it touches zero there. If $M<K<D$, each side has a distinct exercise threshold and the payoff is zero between them. If $K\geq D$, even the left-end put value cannot cover the exercise cost, and only the right call branch can be positive. The illustration shows the middle case, including the competing underlying option values.
\Image[/past-exam-of-the-martingales-and-financial-mathematics-course-of-the-university-of-oxford/2009-chooser.png]
{title=Chooser payoff with an exercise cost above its central option value}
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