Please prove the Theorem 2 mathematically. Theorem 2 in Merton (1973): Consider an American call option on a non-dividend-paying stock over the lifetime of the option. Then option will never be exercised early. Put itanother way, it is never optimal to early'fpremature exercise an American call option on a non-dividendpaying stock. Put it another way, the American option on a non-dividend- paying stock is worth more alive than being exercised before the expiration date. By saying alive, we mean not exercised. Early f premature exercising means exercising before expiration date, so time to maturity of the option, 1' I} D. Strike price is always greater than 0. We also assume risk-free rate, r I} 0. One way to prove this is to use Lemma 4 Lower 13on for American Options directly. C: 2 minim, St K) C: is the value of the option alive and maI[U, S: K) is payoff of exercising the American call immediately, i.e., the intrinsic value of the option. Therefore, the value of the American call option alive is always no less than the payoff of exercising it immediately, at any time. (Otherwise, buying the option and immediately exercising it, constitutes an arbitrage.) So this option will never be exercised early. But this is not a formal mathematical proof. This question ask for a formal mathematical proof. Please refer to the hints below. Hints: I A formal mathematical proof follows directly from Lemma 2, and Theorem 1. a You can start by nding the lower bounds of the value of the American call on a non- dividendpaying stock. Those are the lower bounds of the value of the American call that is alive. - You can then nd the payoff of the American call if exercised immediately. I Then you should be able to show that the American call on a non-dividendpaying stock is worth more alive than being exercised