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I have observed that IV is increasing with time to maturity by using market prices and plotting IV (from Black-Scholes) against log-moneyness, $\log(S_t/K)$. $S_t$ being the price of the stock at time $t$ and $K$ being the strike.

Using Martingales we can prove that the call-option's payoff function - i.e. $\max(S_t-K, 0)$ - is a submartingale under the $Q$-measure. Now this article from Columbia says that the call-price as a function of time to expiry, that is $C_t(T)$, must be not-decreasing to avoid arbitrage, which can be shown using standard martingale results - but why is that?

What are the calculations performed by "standard martinale results" which imply that if the call price was decreasing as a function of $T$ then there would be an arbitrage?

The argument that I do not understand is highlighted here: enter image description here

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For $r=q=0$ and $t\leq T'\leq T$:

$$ C_t(T)=E_{t}[(S_T -K)^+] = E_{t}[E_{T'}[(S_T -K)^+] \geq E_t[(S_{T'} -K)^+]=C_t(T'),$$

where we used the tower property of conditional expectation and the sub-martingality of $(S_{T'}-K)^+$ they mentioned (which is a consequence of Jensen inequality for conditional expectation).

A calendar spread (one long call with expiry $T$ and one short call with expiry $T'$) with negative price would violate the above inequality.

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  • $\begingroup$ Thanks for showing that calculations! Would you mind explaining why this then results in the statement "the term structure of implied volatility cannot be too inverted"? $\endgroup$
    – Landscape
    Jun 9, 2021 at 6:36
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    $\begingroup$ If the pricing of the calls is done via standard Black-Scholes functional $BS_t$, then $ C_t(T) = BS_t(T; \sigma)$ and $C_t(T') = BS_t(T'; \sigma') $ for some vol parameters $\sigma$ and $\sigma'$. You can experiment with very low $\sigma$ and very high $\sigma'$ to see if the inequality above gets broken. $\endgroup$
    – ir7
    Jun 9, 2021 at 14:32
  • $\begingroup$ I'll try calculating some different values. What exactly is $B$ in your formula? I've only seen $B$ in the notation of the 'Risk Free Assest' (i.e. the bank account). $\endgroup$
    – Landscape
    Jun 9, 2021 at 19:23
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    $\begingroup$ BS stands for the standard Black-Scholes formula itself. $\endgroup$
    – ir7
    Jun 9, 2021 at 19:27

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