It would nice if someone could help me with this problem.

I am looking at an improvement to the classical Jensen's Inequality: $$\int_\limits{}^{} \phi(x) \mu \mathrm{d}x \geq \phi\left[\int_\limits{}^{} x \mu \mathrm{d}x\right]$$

for a convex function $\phi$. The improvement (by a theorem) states that:

Consider $\mu$ a probability measure on $(0,\infty)$.

Let $\phi(x)$ be of the form $\sum_\limits{n=0}^{\infty} w_{n}x^{n}$ where $w_{n}$ is a non-negative sequence. Let $\mu_{1}$ be $\mu$ restricted, normalized to $(0,1)$ and let $\mu_{2}$ be $\mu$ restricted, normalized to $(1,~ R)$ where $R > 1$. Then:

$$\int_\limits{}^{} \phi(x) \mu \mathrm{d}x \geq \phi\left[\int_\limits{}^{} x \mu \mathrm{d}x\right] + \mu[(1,R)]\cdot t \cdot \tfrac{1}{2} \sum_\limits{n=2}^{\infty} w_{n}n(n - 1)$$


$$t = \int_\limits{}^{} x^2 \mu_{2} \mathrm{d}x - \left[\int_\limits{}^{} x\mu_{2} \mathrm{d}x\right]^{2}$$

The problem is I'm not sure if I'm calculating the new right hand element correctly. Consider the example $\phi(x) = \exp\left(\frac{1}{2}x\right)$ [which is convex] and $\mu(x) = \exp(-x)$, which is a probability measure on $(0,~ \infty)$. Supposed we let $R=\infty$. Then by Jensen's Inequality,

$$\begin{align} % \text{LHS} % &= \int_\limits{0}^{\infty} \exp\left(\frac{1}{2}x\right) ~ \exp(-x) ~ \mathrm{d}x % \\ &= 2 % \\ &> \text{RHS} % \\ &= \phi\left[\frac{1}{2}\int_\limits{0}^{\infty} x ~ \exp(-x) \mathrm{d}x\right] % \\ &= \exp\left(\frac{1}{2}\right) % \\ & \approx 1.65 % \end{align}$$

Now I need to calculate $\mu[(1,~R)]\cdot t \cdot \frac{1}{2} \sum_\limits{n=2}^{\infty} w_{n}n(n - 1)$, the improvement. If I let $w_{n} = \frac{1}{n!}\left(\frac{1}{2}\right)^n$, then:

$$\sum_\limits{n=2}^{\infty} w_{n}n(n - 1) = \frac{\sqrt{e}}{4}$$


$$\begin{align} % t &= \int_\limits{1}^{R=\infty} x^2 \exp(-x) \mathrm{d}x - \left[\int_\limits{1}^{R=\infty} x \exp(-x) \mathrm{d}x\right]^{2} % \\ &= \left(\frac{5}{e}\right)-\left(\frac{2}{e}\right)^{2} % \\ &\approx 1.298056 \end{align}$$

Therefore, the additional improvement is equal to:

$$\mu[(1,R)]\cdot t \cdot \tfrac{1}{2} \sum_\limits{n=2}^{\infty} w_{n}n(n - 1) = \mu[(1,R)]\cdot 1.298056 \cdot \tfrac{1}{2} \cdot \tfrac{\sqrt{e}}{4} = 0.2675 \mu[(1,R)]$$

So the improvement is given by $2 \geq 1.65 + 0.2675 \mu[(1,R)]$ if I haven't made a mistake.

I don't know how to calculate $\mu[(1,R)]$ in this example as I don't know what is meant by restricting and normalising a distribution to $(1,R)$. Does it mean we take $\mu[(1,R)] = 1$ ? If someone could help me via this example I would be grateful.

  • 1
    $\begingroup$ Where is $\mu_1$ in the formula that you cite? In my opinion, $\mu$ restricted and normalised to a borel set $A$ with $\mu(A)>0$ is the conditional probability measure $\mu_{A}$ defined by $\mu_A(B)=\mu(A\cap B)/\mu(A)$. This must be explained in the book where you found the formula. $\endgroup$ – MassiveJack Apr 3 '15 at 18:44
  • $\begingroup$ @MassiveJack $\mu_{1}$ does not appear in the formula but does appear in the proof of the theorem. Unfortunately it doesn't. I'm still not sure to be perfectly honest $\endgroup$ – user225927 Apr 4 '15 at 0:51

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