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:


What are the price boundaries for a bull call spread?


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?

  • 2
    $\begingroup$ 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 '19 at 17:51

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 $K_2 -K_1 \leq (K_2 -K_1)\frac{S_T}{K_2}$

  • 2
    $\begingroup$ 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 '19 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.