Let $(X,Y)$ be a martingale on $\mathbb R$ and $\psi:\mathbb R\to\mathbb R$ be a convex function. Then it follows by Jensen's inequality that

$$\mathbb E[\psi(X)]~~\le~~ \mathbb E[\psi(Y)]$$

and if $\psi$ is strictly convex and $\mathbb E[\psi(X)]=\mathbb E[\psi(Y)]$, then $X=Y$ almost surely. Now let us consider a special convex function $\psi(x)=(x-K)^+$ (that is not strictly convex), my question is the following:

If $\mathbb P[X\neq Y]>0$ and $\mathbb E[(X-K)^+]=\mathbb E[(Y-K)^+]$, could we show that $X, Y\le K$ or $X, Y\ge K$?

Many thanks!

  • $\begingroup$ @NateEldredge Thanks for pointing out this. I have edited my question. You mean that, even $\mathbb P[X\neq Y]>0$ and $\mathbb E[(X-K)^+]=\mathbb E[(Y-K)^+]$, we can always contruct $X$ and $Y$ such that $\mathbb P[X\ge K]>0$ and $\mathbb P[X< K]>0$? $\endgroup$ – CodeGolf Nov 20 '16 at 19:45

Let $Z_1, Z_2$ be two independent fair coin flips, i.e. $P(Z_i = 1) = P(Z_i = -1) = 1/2$. Let $X = 2 Z_1$ and $Y = X + 1_{\{Z_1 = -1\}} Z_2$. Then $(X,Y)$ is a martingale. In words, a gambler bets \$2 on a coin flip. If she wins she stops playing. If she loses she bets \$1 on a second coin flip. Then $P(X \ne Y) = P(Z_1 = -1) = 1/2 \ne 0$.

Let $K=1$. Then $(X-1)^+ = (Y-1)^+$, i.e. $1$ if the first coin came up heads and $0$ otherwise. So $E[(X-K)^+] = E[(Y-K)^+]$. Yet $P(X \ge K) = P(Y \ge K) = 1/2$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.