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Show that American call and put prices are increasing in maturity $T$.

Does this mean I need to show that as $T$ increases than the American call and put prices increase as well? If so, how do I go about showing this?

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For American options, the longer the maturity, the more choices for the optimal exercises time, then the option value is bigger. For example, consider maturities $T_1$ and $T_2$, for the same option except for different maturities. Any optimal exercise time within $[0, T_1]$ is a possible exercise time within $[0, T_2]$, with a better time possibly falls in $(T_1, T_2]$.

Formally, the value of an American option, with maturity $T$, is given by \begin{align*} \sup_{\tau\in \mathcal{T}_{[0,T]}}E\Big(e^{-r\tau}\max\big(\phi(S_{\tau}-K, 0) \big)\Big), \end{align*} where $\mathcal{T}_{[0,T]}$ is the set of stopping times with values in $[0, T]$. Here, $\phi=1$ for a call and $-1$ for a put. Note that, for $0 < T_1 < T_2$, $\mathcal{T}_{[0, T_1]} \subset \mathcal{T}_{[0,T_2]}$. Therefore, \begin{align*} \sup_{\tau\in \mathcal{T}_{[0,T_1]}}E\Big(e^{-r\tau}\max\big(\phi(S_{\tau}-K, 0) \big)\Big) \le \sup_{\tau\in \mathcal{T}_{[0,T_2]}}E\Big(e^{-r\tau}\max\big(\phi(S_{\tau}-K, 0) \big)\Big). \end{align*} That is, American call and put prices are increasing in maturity $T$.

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american options are at least as expensive as their european counterparts. So it's enough to argue that european options increase in value as time to expiry prolongs, given other metrics remain the same. This is because of the "time value" of options.

On the other hand, longer time give you more opportunity to early exercise, which adds in zero or positive addition values. Thus, the options have more value.

Two factors come together, the american options increase in value as time to expiry increases.

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