<p>An explanation behind this could be that, as you mention</p>

<blockquote>
  <p>[...] the payoff of an American put option <i><strong>at maturity</strong></i> is given by
  $$\max\left\{K-S(T),0\right\}$$</p>
</blockquote>

<p>and </p>

<blockquote>
  <p>[...] sources [...] say that $$(S,t)=\sup_\tau E[e^{-r(\tau-t)}\max\left\{K-S(\tau),0\right\}]$$</p>
</blockquote>

<p>Thus the difference between the two appears to be the discounting factor considering how far (in time) we are from the exercising/stopping date (which is the same as the maturity date if $\tau=T$), which implies to compute (continuously) the present value of this futur payoff. If  $(\tau-t \rightarrow0) \land (\tau=T \lor t \rightarrow T)$, both equations become equal.</p>

<p>If one decides (for some reasons) that the option will be exercised at maturity, i.e. $\tau=T$ and that "now", $t$, is "turning into this futur date $T$", i.e. $t \rightarrow T$, one gets</p>

<p>$T-t \rightarrow 0 \implies \sup_T e^{-r(T-t)} \rightarrow 1$ </p>

<p>$\implies \sup_T E[e^{-r(T-t)}\max\left\{K-S(T),0\right\}]$</p>

<p>$\rightarrow  E[\max\left\{K-S(T),0\right\}]\rightarrow\max\left\{K-S(T),0\right\}$</p>

<p>Where the last above row shows that the expected payoff becomes certain.</p>

CC BY-SA 3.0
2017-05-03T11:40:10.460Z
why would they become equal?
user13132
CC BY-SA 3.0
2017-05-03T11:54:06.260Z
I made some corrections, I meant that it some sources ....$$(S,t)=\sup_\tau E[e^{-r(\tau-t)}\max\left\{K-S(\tau),0\right\}]$$. Will this make any difference?
user13132
CC BY-SA 3.0
2017-05-03T13:40:28.343Z
Yes this makes a difference, even if this difference becomes smaller as $t$ becomes $T$. The expectation "operator", $E$, stands for the fact that we are not sure of the payoff at $T$, and thus, that we are not working with a deterministic payoff, but with a stochastic one, whose probability distribution is assumed known so as to be capable of computing its average, i.e. $E[max\{K−S(T),0\}]$. Finally, as $t$ becomes $T$, all the possible values of this payoff concentrate on a restricted set of values, and ultimately, this set of values contains only one value.
keepAlive
CC BY-SA 3.0
2017-05-03T11:08:09.030Z
2017-05-03T12:36:33.727Z
2017-05-03T12:36:33.727Z
keepAlive
<p>I am reading about the American option and sources got me confused in the part where the American put option is considered on a non-dividend paying asset.</p>

<p>I understand that the payoff of an American put option at maturity is given by $$\max\left\{K-S(T),0\right\}$$ where $K$ is the exercise price and $S(T)$ is the price of the underlying at maturity.</p>

<p>Some sources state that considering n-arbitrage, the American put option problem aims to find the option value $$P(S,t)=\sup_\tau EG_\tau$$ where $G_\tau$ is the gain function and E is expectation. </p>

<p>In my understanding, the gain function is equivalent to the payoff function, hence $$V(S,t)=\sup_\tau E[max\left\{K-S(\tau),0\right\}]$$ for all stopping times $\tau$.</p>

<p>However, I also stumbled upon sources that say that $$(S,t)=\sup_\tau E[e^{-r(\tau-t)}\max\left\{K-S,0\right\}]$$ for all stopping times $\tau$.</p>

<p>What could be an explanation behind this?</p>

<asset-pricing>
expected payoff of an American put option
user13132
# expected payoff of an American put option

(Asked by: username_0 on 2017-05-03)

I am reading about the American option and sources got me confused in the part where the American put option is considered on a non-dividend paying asset.

I understand that the payoff of an American put option at maturity is given by $$\max\left\{K-S(T),0\right\}$$ where $K$ is the exercise price and $S(T)$ is the price of the underlying at maturity.

Some sources state that considering n-arbitrage, the American put option problem aims to find the option value $$P(S,t)=\sup\_\tau EG\_\tau$$ where $G\_\tau$ is the gain function and E is expectation.

In my understanding, the gain function is equivalent to the payoff function, hence $$V(S,t)=\sup\_\tau E[max\left\{K-S(\tau),0\right\}]$$ for all stopping times $\tau$.

However, I also stumbled upon sources that say that $$(S,t)=\sup\_\tau E[e^{-r(\tau-t)}\max\left\{K-S,0\right\}]$$ for all stopping times $\tau$.

What could be an explanation behind this?

**Tags:** asset-pricing

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## 1 answers

### Answer by username_1 on 2017-05-03. Score: 1 | Accepted answer

An explanation behind this could be that, as you mention

> [...] the payoff of an American put option ***at maturity*** is given by
> $$\max\left\{K-S(T),0\right\}$$

and

> [...] sources [...] say that $$(S,t)=\sup\_\tau E[e^{-r(\tau-t)}\max\left\{K-S(\tau),0\right\}]$$

Thus the difference between the two appears to be the discounting factor considering how far (in time) we are from the exercising/stopping date (which is the same as the maturity date if $\tau=T$), which implies to compute (continuously) the present value of this futur payoff. If $(\tau-t \rightarrow0) \land (\tau=T \lor t \rightarrow T)$, both equations become equal.

If one decides (for some reasons) that the option will be exercised at maturity, i.e. $\tau=T$ and that "now", $t$, is "turning into this futur date $T$", i.e. $t \rightarrow T$, one gets

$T-t \rightarrow 0 \implies \sup\_T e^{-r(T-t)} \rightarrow 1$

$\implies \sup\_T E[e^{-r(T-t)}\max\left\{K-S(T),0\right\}]$

$\rightarrow E[\max\left\{K-S(T),0\right\}]\rightarrow\max\left\{K-S(T),0\right\}$

Where the last above row shows that the expected payoff becomes certain.

> **Comments:**
> **username_0**: why would they become equal?  
> **username_0**: I made some corrections, I meant that it some sources ....$$(S,t)=\sup_\tau E[e^{-r(\tau-t)}\max\left\{K-S(\tau),0\right\}]$$. Will this make any difference?  
> **username_1**: Yes this makes a difference, even if this difference becomes smaller as $t$ becomes $T$. The expectation "operator", $E$, stands for the fact that we are not sure of the payoff at $T$, and thus, that we are not working with a deterministic payoff, but with a stochastic one, whose probability distribution is assumed known so as to be capable of computing its average, i.e. $E[max\{K−S(T),0\}]$. Finally, as $t$ becomes $T$, all the possible values of this payoff concentrate on a restricted set of values, and ultimately, this set of values contains only one value.  

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