Deriving a Martingale Condition for an American Put Bound
Summary
The document derives a positive exponent that makes the discounted negative power of a Black–Scholes stock price a martingale, as a step toward bounding the value of an American put. It rewrites the stock process using geometric Brownian motion and separates the expression into an exponential martingale and a deterministic time factor. For the whole process to remain a martingale, that factor must be constant, which gives the quadratic condition for the exponent.
The answer simplifies the positive solution to twice the risk-free rate divided by variance. The document states the proposed bound but does not complete its proof or explain how the martingale is used to establish the option-price inequality. Its derivation assumes the stated Black–Scholes model and nonnegative interest rate; it offers no numerical examples or discussion of other models.
Key ideas
- A negative power of the Black–Scholes stock price can be combined with discounting to construct a martingale.
- The exponential Brownian component is a martingale after including its variance correction.
- The remaining deterministic factor must have zero growth for the full process to be a martingale.
- The positive exponent satisfying the condition is twice the risk-free rate divided by volatility squared.
- The answer derives the exponent but does not show the full American put bound proof.
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Full text
# Black Scholes/American Put/Martingale Condition
# Black Scholes/American Put/Martingale Condition
Consider a Black Scholes model with $r \geq 0$. Show that the price of an American Put Option with maturity $T > 0$ is bounded by $\frac{K}{1 + \alpha} {(\frac{\alpha K}{1 + \alpha})}^{\alpha}{S_{0}^{-\alpha}}$. Hint: find $\alpha > 0$ such that $e^{-rt}S_{t}^{-\alpha}$ is a martingale.
I am trying to work through the hint and when rewriting $e^{-rt}S_{t}^{-\alpha}$ I get $e^{-rt}S_{t}^{-\alpha} = exp(-\alpha \sigma W_t - t(r + \alpha (r-\frac{{\sigma}^2}{2})))$
I know that in case of the GBM the drift has to equal zero for it to be a martingale, which property has to be fulfilled here?
Edit: the condition that my professor wrote:
$r + \alpha (r-\frac{{\sigma}^2}{2}) = \frac{\alpha^2 \sigma^2}{2}$
which leads to a quadratic equation for $\alpha$, but O do not quite understand where is equation comes from.
## Answer by Kurt G. (score 1, accepted)
https://quant.stackexchange.com/a/77211
Too long for a comment.
\begin{align} S_t&=S_0\,e^{rt+\sigma W_t-\sigma^2t/2}\,,&S_t^{-\alpha}=S_0^{-\alpha}\,e^{-\alpha\, r\,t\,-\,\alpha\,\sigma\, W_t\,+\,\alpha\,\sigma^2\,t/2}\,, \end{align} \begin{align} e^{-rt}S_t^{-\alpha}&=S_0^{-\alpha}e^{-(1+\alpha)\,r\,t\,-\,\alpha\,\sigma\, W_t\,+\,\alpha\,\sigma^2\,t/2}\\ &=S_0^{-\alpha}e^{-(1+\alpha)\,r\,t\,+\,(\alpha^2+\alpha)\,\sigma^2\,t/2}\,\underbrace{e^{-\,\alpha\,\sigma\, W_t\,-\,\alpha^2\sigma^2\,t/2}}_{\text{martingale}}\,. \end{align} For the whole expression to be a martingale we must have (your professor's quadratic equation): \begin{align} -(1+\alpha)r+(\alpha^2+\alpha)\sigma^2/2=0\,. \end{align} One solution of that is $\alpha=-1$ but your professor wants a positive solution. This is \begin{align} \alpha&=\frac{-\frac{\sigma^2}{2}+r+\sqrt{(\frac{\sigma^2}{2}-r)^2+2\sigma^2r}}{\sigma^2}\\ &=\frac{-\frac{\sigma^2}{2}+r+\frac{\sigma^2}{2}+r}{\sigma^2}=\frac{2r}{\sigma^2}\,. \end{align}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.