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Jensen's inequality is well known as

$$E\big[f(X)\big]\le f\big(E[X]\big)$$

where $X$ is a integrable random variable and $f: R\to R$ is a bounded concave function, see also http://en.wikipedia.org/wiki/Jensen%27s_inequality

Now I have a question about whether we may have a generalized result for a more abstract space. Let $\Omega:=D([0,1],R)$ be the space of all cadlag functions defined on $[0,1]$. Denote by $X$ the canonical process, i.e. $X_t(\omega)=\omega_t$. Let $f: \Omega\to R$ be a bounded concave function and $P$ be a martingale measure, i.e. $f\big(\alpha \omega+(1-\alpha)\omega'\big)\ge \alpha f(\omega)+(1-\alpha)f(\omega')$ for any $\omega, \omega'\in\Omega$, $\alpha\in [0,1]$ and $X=(X_t)_{0\le t\le 1}$ is a $P-$martingale.

Could we also show that

$$E^P\big[f(X)\big]\le f\big(E^P[X]\big),$$

where $E^P[X]\in \Omega$ is a constant function taking $E^{P}[X_0]$. Thx for the reply!

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  • $\begingroup$ Maybe there exists already such results. If someone knows about that please let me know. Thx a lot! $\endgroup$
    – CodeGolf
    Dec 4, 2014 at 15:37
  • $\begingroup$ Is the reasoning as same as that on $R$? $\endgroup$
    – CodeGolf
    Dec 4, 2014 at 16:46
  • $\begingroup$ Consider the composition of the two map: $x \mapsto x^2$ and $f$ $\endgroup$
    – Zbigniew
    Dec 21, 2014 at 1:37
  • $\begingroup$ Could you make the map $x\mapsto x^2$ clearer? Here $x$ stands for a path? $\endgroup$
    – CodeGolf
    Dec 22, 2014 at 12:37
  • $\begingroup$ For example $E^P\big[f(X)\big]$ is the composition of $f, \, \mathbb E$ and $x\mapsto x^p$. However $\big(E^P[X]\big)$ is the composition of $ \mathbb E, \, x\mapsto x^p$ and $f$. $\endgroup$
    – Zbigniew
    Dec 23, 2014 at 19:06

1 Answer 1

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Since $X=(X_t)_{t\in[0,1]}$ is a martingale, we have $E(X_t|X_0)=X_0=X_0\,\omega_1(t)$ for each $t\in[0,1]$, where $\omega_1$ denotes the function on $[0,1]$ with constant value $1$. So, by Jensen's inequality for conditional expectations (see e.g. [cond. Jensen's ineq.]), $$E(f(X)|X_0)\le f(E(X|X_0))=f(X_0\,\omega_1), $$ whence $Ef(X)=EE(f(X)|X_0)\le Ef(X_0\,\omega_1)$, as desired.

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