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I'm currently stuck on trying to prove the following:

use a 1-step binomial model, we have two otherwise identical call options with expiration $T_1$ and $T_2$. Prove that $c_1 ≤ c_2$.

Here's the notations I've used:

$S_0$ = spot price

$u$ is when price moves up, S0 becomes $S_0\times u$; similar for $d$ but when price moves down

$S_u = S_0 \times u$

$S_d = S_0\times d$

$K$ = strike price for the option

$C_1$ & $C_2$ indicates the option price for the corresponding call

I've successfully proved the cases where $S_d < k < S_u$ and where $K > S_u > S_d$.

However, I'm stuck to prove for the case where $K < S_d < S_u$.

Jay
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Anna Lavender
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    You never define $c_1$ and $c_2$ unless they are supposed to be the same as $C_1$ and $C_2$. You assume that $T_1$ is before $T_2$ but do not say it. – Ross Millikan Apr 01 '21 at 15:01
  • Indeed. My first thought without even reading the rest was that the statement is obviously false, unless the $c$ value for all options are the same. Because for any two options unequal options, either $c_1 < c_2$ or $c_2 < c_1$. If $c_1 < c_2$, then switch out the indices. You put no restriction on the givens that prevents relabeling them. After switching, $c_2 < c_1$. Either way contradicts the statement. The lesson is, when stating a problem, you have to include all the relevant information. – Paul Sinclair Apr 01 '21 at 23:26

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