# Is there a constant that reverses Jensen's inequality?

The general Jensen's inequality states: $\varphi\left(\mathbb{E}[X]\right) \leq \mathbb{E}\left[\varphi(X)\right]$. I'm wondering if there is a constant $c$ (function of $\varphi$), such that $c\varphi\left(\mathbb{E}[X]\right) \geq \mathbb{E}\left[\varphi(X)\right]$?

More specifically I wan't to show $\log\mathbb{E}[e^X]\leq(e-1)\mathbb{E}[X]$. Here $(e-1)$ would be the $c$ mentioned above.

I've noticed that many other common inequalities have such 'reversing' constants (or at least a corresponding lower bound). Like $\log(1+x)\leq x$, and $\frac{x}{x+1}\leq \log(1+x)$.

• your inequality won't hold without extra assumptions on $X$. take $X$ to be $1$ or $-1$ with probabilities $1/2$ for example Commented Jun 2, 2014 at 16:31
• I was taking $X$ to be positive. I guess this settles whether there is a constant in general. Commented Jun 2, 2014 at 16:33
• Coud I ask have you figured out what kind of inequality we could have for $E[e^X]<?$
– null
Commented Apr 11, 2020 at 20:26
• @NathanExplosion Looking at it now, it is obvious that $E[\phi(X)]$ can be arbitrarily large. Even with $\phi(x) = x^2$ it might not exist. But for sub-gaussian $X$ I suppose we get $\log E[e^X] \le E[X] + V[x]/2$. Commented Apr 12, 2020 at 8:28
• @ThomasAhle thanks, so it really depends on the situation.
– null
Commented Apr 12, 2020 at 8:30

For the particular problem about the exponential function, let $X=-100$, $0$, or $100$ each with probability $\frac{1}{3}$. Then $E(X)=0$, but $E(e^X)$ is large.
• There are interesting questions underneath, such as finding natural conditions on the distribution of $X$ that give your desired inequality. Commented Jun 2, 2014 at 16:38