= Solution
The increasing <quadratic variation> has a limit in $[0,\infty]$. Since $X_t$ has a finite limit, the exponentials have limits too, with value zero when the bracket is infinite. Expanding exponents proves
$$
\mathcal E(X)_t
=\mathcal E(rX)_t^{1/r^2}
\left(e^{rX_t/(r+1)}\right)^{(r^2-1)/r^2},
$$
because the coefficient of $X_t$ on the right is
$$
\frac1r+\frac{r}{r+1}\frac{r^2-1}{r^2}=1
$$
and the bracket coefficient is $-1/2$. Taking limits preserves the identity, including the zero case.
Let $A=\mathbb E[\mathcal E(X)_\infty]$, $B=\mathbb E[\mathcal E(rX)_\infty]$, and $D=\mathbb E[e^{rX_\infty/2}]$. The <Holder inequality> first yields
$$
A\leq B^{1/r^2}
\left(\mathbb E[e^{rX_\infty/(r+1)}]\right)^{(r^2-1)/r^2}.
$$
When $D$ is finite, apply the <Jensen inequality> to the concave function $z\mapsto z^{2/(r+1)}$:
$$
\mathbb E[e^{rX_\infty/(r+1)}]
=\mathbb E[(e^{rX_\infty/2})^{2/(r+1)}]
\leq D^{2/(r+1)}.
$$
Since $D>0$, rearrangement gives the <terminal scaling inequality for stochastic exponentials>
$$
\boxed{
\mathbb E[\mathcal E(rX)_\infty]
\geq
\bigl(\mathbb E[\mathcal E(X)_\infty]\bigr)^{r^2}
\bigl(\mathbb E[e^{rX_\infty/2}]\bigr)^{-2(r-1)}.}
$$
Here $2(r^2-1)/(r+1)=2(r-1)$. If the last exponential moment is infinite, the bound has no positive content; the finite-moment form is the one used below.
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