Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 2 / a / Solution

Let $m=\mathbb{E}X$. First, $m$ lies in the open interval $I$. For example, if $I$ has a finite lower endpoint $\ell$, then $X-\ell >0$ almost surely, and its expected value is strictly positive; hence $m>\ell$. The corresponding argument at a finite upper endpoint gives the other strict inequality. Integrability ensures $m$ is finite when either endpoint is infinite.

The tangent affine function at $m$ gives the supporting inequality for a differentiable convex function. Under the stipulated condition $f^{\prime \prime }>0$, contact with that tangent occurs only at $m$: defineThen $h(m)=0$, and $h(t)>0$ for every $t\ne m$. One direct proof uses $h^{\prime }(t)=f^{\prime }(t)-f^{\prime }(m)$: $f^{\prime }$ is strictly increasing, so $h$ decreases towards $m$ on the left and increases away from $m$ on the right. Thereforewith equality exactly when $h(X)=0$ almost surely, equivalently when $X=m$ almost surely. Conversely, an almost surely constant $X=m$ plainly gives equality. This proves the strict Jensen equality criterion.

No extra assumption that $f(X)$ is an integrable random variable is needed to interpret this argument. Its negative part is bounded by the absolute value of the integrable random variable $f(m)+f^{\prime }(m)(X-m)$, since $h(X)\ge 0$. Thus its expected value is defined in $(-\infty ,+\infty ]$; if it equals the finite number $f(m)$, then $\mathbb{E}[h(X)]=0$ and the required integrability follows automatically. Equality holds if and only if $X=\mathbb{E}X$ almost surely.