# Binomial tree prices the American put

When we use the binomial tree to price the American put, we should compare the discounted value from last nodes and the intrinsic value at each node.

But I confuse that, discounted value from last nodes is the value of European put at this node, and the value of European put is always greater than its intrinsic value, how does it occur the intrinsic value is greater than discounted value?

For the statement the value of European put is always greater than its intrinsic value, I get from the book Problems and Solutions in Mathematical Finance Equity Derivatives. Volume 2 page 74

• "The value of European put is always greater than its intrinsic value" - this is not true. Consider a one year European put struck at 150, with spot price 100, volatility 20% and interest rate 5%. The intrinsic value is \$50 and the total option value is \$43.04 – Chris Taylor Sep 19 '17 at 10:53
• @ChrisTaylor yeah, I doubt this statement too, but pls see the update – A.Oreo Sep 19 '17 at 10:58
• @Quantuple but how to explain the contradiction in the binomial tree of American Option – A.Oreo Sep 19 '17 at 12:06

Assume deterministic interest rates and no dividends to keep notations uncluttered.

Because the discounted value of all self-financing portfolio is a martingale under $\Bbb{Q}$ one can express the European put price as a risk-neutral expectation as follows: $$P(S_0;K,T)=\frac{1}{B_T} \Bbb{E}^\Bbb{Q}_0\left[\text{max}(K−S_T,0)\right] \tag{1}$$ Since $f : x \to \max(0, x)$ is a convex function, one can apply Jensen's inequality to the RHS of $(1)$. Further using the fact that $S_t/B_t$ is also a $\Bbb{Q}$ martingale under our working assumptions, one gets: \begin{align} P(S_0;K,T) &\geq \text{max}\left( \Bbb{E}^\Bbb{Q}_0\left[\frac{K}{B_T} −\frac{S_T}{B_T} \right], 0 \right) \\ &\geq {\color{green}{\max\left( \frac{K}{B_T} - S_0, 0 \right)}} \end{align} Hence we have an equality relating the put price (in blue below) to the RHS above (in green below), which is not the intrinsic value (in orange below). This is illustrated taking the example given in @Chris Taylor's comment

the result is simply not true. If it were true American puts would not be early exercised and they would be worth the same as Europeans.

The proof is flawed. The set -up of $\pi_t$ does not make sense. We cannot take $$\max(S_t-K,0)$$ as part of a self-financing portfolio. We could take bonds worth that but they wouldn't be worth $$\max(S_T-K,0)$$ at the end.

• Pls see Quantuple's answer above, I can not see anywhere unreasonable。 – A.Oreo Sep 19 '17 at 12:07
• The result only says the put value is higher than intrinsic value, but never say put value is submartingale, so I think there is no connection to the early exercising of American put. – A.Oreo Sep 19 '17 at 12:12
• @A. Oreo, my result is precisely along the lines of what Mark Joshi tells you. You are reading it wrong: it's not the intrinsic value at the end (i.e. $VI(K)$) but rather $VI(K/B_T)$... – Quantuple Sep 19 '17 at 12:37
• but if we use the inequality of convex function: $cf(x) \geq f(c x),$ then we have $E[c f(x)]\geq E[f(c x)] \geq f(E(c x)).$ And we take $f(x) = \max(K -x,0),\ c = \dfrac{1}{B_t},$ we have put value $>\max(K - S_t,0).$ – A.Oreo Sep 19 '17 at 13:00
• Where did you get that inequality?! Consider $c=2$ and $f : x \to x^2$, do you really believe that $2x^2 \geq 4x^2, \forall x$ ? No at the end you have $P \geq VI(K/B_T)$, while for positive interest rates $VI(K) > VI(K/B_T)$, but this gives you no way to relate $P$ to $VI(K)$. – Quantuple Sep 19 '17 at 13:29