8
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Maybe there exists already such results. If someone knows about that please let me know. Thx a lot! $\endgroup$ – CodeGolf Dec 4 '14 at 15:37
  • $\begingroup$ Is the reasoning as same as that on $R$? $\endgroup$ – CodeGolf Dec 4 '14 at 16:46
  • $\begingroup$ Consider the composition of the two map: $x \mapsto x^2$ and $f$ $\endgroup$ – Zbigniew Dec 21 '14 at 1:37
  • $\begingroup$ Could you make the map $x\mapsto x^2$ clearer? Here $x$ stands for a path? $\endgroup$ – CodeGolf Dec 22 '14 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 '14 at 19:06
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

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.