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What are the upper and lower bound of American call and put options for an underlying with continuous dividend yield?

For European options, the bounds are known as

\begin{align*} [S_te^{-d\tau}-Ke^{-r\tau}]^+<C_t<S_te^{-d\tau}\\ [Ke^{-r\tau}-S_te^{-d\tau}]^+<P_t<Ke^{-r\tau} \end{align*}

However, I could not find a clear result regarding American options with dividend yield.

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  • $\begingroup$ your bounds are so wide that it is the same bounds $\endgroup$
    – M. Jeunesse
    Commented Aug 29, 2016 at 15:36
  • $\begingroup$ @MJ73550 It might be true but I would like a proof for that ; ) $\endgroup$
    – emcor
    Commented Aug 29, 2016 at 16:42
  • $\begingroup$ $S(\tau)$ in American option with dividend yield has asymptotic behavior, then how do you want bound for it. $\endgroup$
    – user16651
    Commented Aug 29, 2016 at 16:44
  • $\begingroup$ It really depends on how tight you'd like the bounds to be. Surely you must agree that the price of an American option is greater than or equal to the price of its European counterpart. The question is does that suit you in terms of lower bound? $\endgroup$
    – Quantuple
    Commented Aug 29, 2016 at 21:53

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For the lower bound, since american call option (resp. put) is bigger than european call option (resp. put). So your lower bounds for european options hold also for american options.

For the upper bound, there is a slight difference. (sorry for my too quick comment).

Here $S_0,T$ and $K$ are positive real numbers.

Let $C^A_T$ be the american call option of maturity $T$:

$$K<K' \Rightarrow (x-K)^+\geq (x-K')^+ \Rightarrow C^A_T(S_0,K)\geq C^A_T(S_0,K') $$

so $C^A_T(S_0,K)\leq C^A_T(S_0,0)=S_0$

Let $P^A_T$ be the american put option of maturity $T$:

Assuming you are in an exponential model (like BS) $x\to P^A_T(x,K)$ is non-increasing. Thus, $P^A_T(S_0,K)\leq P^A_T(0,K)=K$

So you get:

$$(S_0e^{-dT}-Ke^{-rT})^+\leq C^A_T(S_0,K) \leq C^A_T(S_0,K) \leq S_0$$ and $$(Ke^{-rT}-S_0e^{-dT})^+ \leq P^A_T(S_0,K)\leq P^A_T(S_0,K) \leq K$$

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  • $\begingroup$ The upper bound for the call cannot be $S_0$, since a call option forgoes all dividends on the stock. Therefore $C_t\leq S_0e^{-dT}$. The lower bounds are in my opinion just the actual intrinsic values, as one can exercise at any time now. However, with dividends there might be a tradeoff between exercising to receive the dividend yield vs. waiting to receive the risk-free rate? $\endgroup$
    – emcor
    Commented Aug 30, 2016 at 12:06
  • $\begingroup$ For European Options, I am ok. Can you explicit how you "receive" the risk-free rate ? $\endgroup$
    – M. Jeunesse
    Commented Aug 30, 2016 at 12:24
  • $\begingroup$ By not exercising the option, you can save interest on the strike price (or not have to take a loan with interest). $\endgroup$
    – emcor
    Commented Aug 30, 2016 at 12:47
  • $\begingroup$ are you ok with the fact that at time $0$, $C^A_T(S_0,K=0) = S_0$ ? $\endgroup$
    – M. Jeunesse
    Commented Aug 30, 2016 at 13:21
  • $\begingroup$ Yes I admit this would be correct.I believe the lower bounds would be $\max(S_0e^{-dT}-Ke^{-dT},S-K,0)\leq C_T^A$ (and accordingly for puts), as it is either optimal to wait (i.e. European bound) or exercise. E.g. for $D=1, R=2$ you get $S-1-(K-2)>S-K$ $\endgroup$
    – emcor
    Commented Aug 31, 2016 at 14:03

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