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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.

The conditions on the cumulative distribution function imply $0<X_n\le 2$ almost surely. With ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$, adjoin $Y_0=1$ and the trivial ${\mathcal{H}}_0$. The random variables $Y_n$ are nonnegative, adapted, and bounded by $2^n$ at each fixed time. Independence and $\mathbb{E}{X}_{n+1}=1$ giveThus this is the product martingale from independent mean-one factors, with $\mathbb{E}{Y}_n=1$. The almost sure supermartingale convergence theorem states that a nonnegative supermartingale converges almost surely to a finite integrable random variable. Applied to this nonnegative martingale, it shows $Y_n\to Y_{\infty }<\infty$ almost surely. Every convergent real sequence is bounded, so ${\sup }_{n\ge 1}{Y}_n<\infty$ almost surely.

The printed maximum can in fact be taken literally here. If $X_1=1$ almost surely, all products equal $1$. Otherwise the strict Jensen equality criterion, applied to the strictly convex function $t\mapsto -\sqrt{t}$ on $(0,\infty )$, gives $q=\mathbb{E}\sqrt{X_1}<1$. Independence yields $\mathbb{E}\sqrt{Y_n}=q^n$. By Markov's inequality, ${\sum }_n\mathbb{P}(Y_n>\epsilon )\le {\epsilon }^{-1/2}{\sum }_n{q}^n<\infty$ for every $\epsilon >0$. The First Borel-Cantelli lemma and the countable choices $\epsilon =1/m$ give $Y_n\to 0$ almost surely. Since $Y_1>0$, a sequence with this limit attains its positive supremum among finitely many initial terms. This is an instance of fractional moment decay of independent mean-one products.