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Let $f$ be Riemann integrable such that $\int_a^b f(t) \ dt = 1$ and $f \geq 0$ on $[a,b]$. If $\sigma \in C^2$ is convex, show

$$\sigma\left(\int_a^b tf(t) \ dt\right)\leq\int_a^b f(t)\sigma(t) \ dt $$

and, in addition, discuss when equality holds given the tighter condition $f>0$.

I'm assuming this is related to probability theory, since $f$ meets the conditions to be a probability density function.

I was able to prove this using Jensen's inequality; however, my professor insists that isn't used. I tried a few other approaches, like fixing one variable and differentiating with respect to the other, but this doesn't seem to lead anywhere. Please help!

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  • $\begingroup$ Try to incorporate the proof of Jensen's inequality itself. $\endgroup$
    – rookie
    Mar 8 '17 at 6:18
  • $\begingroup$ Many methods of proving Jensen's is available here (en.m.wikipedia.org/wiki/Jensen's_inequality). Basically one needs to use the convexity of $\sigma$. $\endgroup$
    – rookie
    Mar 8 '17 at 6:22
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Of course it's just Jensen's inequality. Since you don't want to use that, here is a direct proof by integrating by parts several times.

By integration by parts, $$\int_a^b t f(t) \, \mathrm{d}t = b - \int_a^b F(t) \, \mathrm{d}t$$ where $F(t) = \int_a^t f(s) \, \mathrm{d}x.$

Consider the function $$g(x) = \sigma \Big( x - \int_a^x F(t) \, \mathrm{d}t \Big).$$ It is differentiable with $$g'(x) = (1 - F(x)) \sigma'\Big(x - \int_a^x F(t) \, \mathrm{d}t\Big).$$ Since $\sigma'$ is monotone increasing and $F(t) \ge 0$, we can bound $$g'(x) \le (1 - F(x)) \sigma'(x)$$ and integrating that inequality over $[a,b]$ gives $$\sigma \Big( \int_a^b t f(t) \, \mathrm{d}t \Big) \le \sigma(b) - \int_a^b F(x) \sigma'(x) \, \mathrm{d}x = \int_a^b f(x) \sigma(x) \, \mathrm{d}x,$$ using integration by parts again.

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