Suppose a random variable $x$ is zero-mean and the cumulant generating function is $$ K\left( t \right) =\log \mathbb{E}[e^{tx}]. $$ Given any positive constant $\tau > 0$, does this inequality $$ \left( \tau -t \right) K^\prime\left( t \right) +K\left( t \right) \ge \frac{t}{\tau}K\left( \tau \right) $$ hold for all $ 0 \le t \le \tau$? Here $K^\prime$ is the derivative function of $K$.

Suppose a random variable $X$ is zero-mean and the cumulant generating function is $$ K\left( t \right) =\log \mathbb{E}[e^{tX}]. $$ Given any positive constant $\tau > 0$, does this inequality $$ \left( \tau -t \right) K^\prime\left( t \right) +K\left( t \right) \ge \frac{t}{\tau}K\left( \tau \right) $$ hold for all $ 0 \le t \le \tau$? Here $K^\prime$ is the derivative function of $K$.

Suppose a bounded random variable $x$ is zero-mean and the cumulant generating function is $$ K\left( t \right) =\log \mathbb{E}[e^{tx}]. $$ Given any positive constant $\tau > 0$, does this inequality $$ \left( \tau -t \right) K^\prime\left( t \right) +K\left( t \right) \ge \frac{t}{\tau}K\left( \tau \right) $$ hold for all $ 0 \le t \le \tau$? Here $K^\prime$ is the derivative function of $K$.

Suppose a random variable $x$ is zero-mean and the cumulant generating function is $$ K\left( t \right) =\log \mathbb{E}[e^{tx}]. $$ Given any positive constant $\tau > 0$, does this inequality $$ \left( \tau -t \right) K^\prime\left( t \right) +K\left( t \right) \ge \frac{t}{\tau}K\left( \tau \right) $$ hold for all $ 0 \le t \le \tau$? Here $K^\prime$ is the derivative function of $K$.