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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!

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  • $\begingroup$ @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
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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 $P(X \ge K) = P(Y \ge K) = 1/2$.

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