# A generalization of Jensen's Inequality

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!

• Maybe there exists already such results. If someone knows about that please let me know. Thx a lot! – CodeGolf Dec 4 '14 at 15:37
• Is the reasoning as same as that on $R$? – CodeGolf Dec 4 '14 at 16:46
• Consider the composition of the two map: $x \mapsto x^2$ and $f$ – Zbigniew Dec 21 '14 at 1:37
• Could you make the map $x\mapsto x^2$ clearer? Here $x$ stands for a path? – CodeGolf Dec 22 '14 at 12:37
• 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 $f$. – Zbigniew Dec 23 '14 at 19:06

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.