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I'm reading an interview book called A Practical Guide to Quantitative Finance Interview and I have some doubts regarding part of its solution and highlighted them in bold:

Question:

What are the price boundaries for a bull call spread?

Solution:

A bull call spread is a portfolio with two options: long a call $c_1$ with strike $K_1$ and short a call $c_2$ with strike $K_2$ and $(K_1<K_2)$. The cash flow a bull spread is summarized in the attached screenshottable for call spread cash flow.

Since $(K_1<K_2)$, the initial cash flow is negative. Considering that the final payoff is bounded by $K_2-K_1$, the price of the spread, $c_1-c_2$, is bounded by $e^{-rT}(K_2-K_1)$.

But it also says (here is where my doubt is), the payoff is also bounded by $\frac{(K_2-K_1)S(T)}{K_2}$, but why? Could anyone share some advice on it? Really appreciate it!

By the way, another two questions (might be a stupid one) that is not related to the above question:

Question A:

As we know, one assumption for Black Scholes equation is underlying asset follows geometric Brownian motion, but can we say it is equivalent to "underlying follows lognormal distribution"? In other words: geometric Brownian motion is equivalent to lognormal distribution

Question B

And in order to use Ito lemma for those products that are derivatives on underlying asset, we need to make sure the underlying asset follows geometric Brownian motion?

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    $\begingroup$ Question A : yes, a geometric Brownian motion is lognormally distributed. Question B: IT does not have to be a geometric brownian motion. We choose the latter because it guarantees that the underlying is postive, and it is relatively simple distribution . $\endgroup$
    – Canardini
    Dec 5, 2019 at 17:51

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If $S_T<K_1$, the payoff is zero, and we have $\frac{(K_2-K_1)S(T)}{K_2} \geq0$

If $K_1 \leq S_T<K_2$, the payoff is $(S_T -K_1)$. We have $$K_1K_2 \geq S_TK_1$$

and $$S_TK_2+K_1K_2 \geq S_TK_2+S_TK_1$$

Thus, $$S_TK_2-S_TK_1 \geq S_TK_2-K_1K_2$$

Finally, $$S_T(K_2-K_1) \geq (S_T-K_1)K_2$$ $$\frac{S_T(K_2-K_1)}{K_2} \geq (S_T-K_1)$$

If $S_T \geq K_2$,we have that $\frac{S_T}{K_2} \geq1$, and because the payoff is $K_2 -K_1$, therefore $K_2 -K_1 \leq (K_2 -K_1)\frac{S_T}{K_2}$

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    $\begingroup$ A simpler way to see the same results is to draw a straight line passing by the point $(0,0)$ and the point $(K_2,K_2-K_1)$. Obviously this line is always above the call spread payoff. The slope of the payoff is $\frac{K_2-K_1-0}{K2-0}$, you can conclude $\endgroup$
    – Canardini
    Dec 6, 2019 at 4:44

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