Expression for dual of subgaussian norm Here is the simplest statement of my question:

Let $Y$ be a centered real random variable and define $$\|Y\|_* = \sup \left\{ \mathbb{E}[X \cdot Y] ~:~ \forall t \in \mathbb{R} ~~ \mathbb{E}[e^{tX}] \le e^{t^2/2}\right\},$$
  where the supremum is over real random variables $X$ that may depend on $Y$.

Is there a closed-form expression for $\|Y\|_*$? (Or a good closed-form approximation.)

Here is a more detailed statement of my question:

Define a norm on the space of random variables by $$\|X\| := \inf \left\{ \max\{|\mu|,|\sigma|\} : \mu,\sigma \in \mathbb{R},~~\forall t \in \mathbb{R} ~~~ \mathbb{E}\left[e^{tX}\right] \leq e^{t\mu+t^2\sigma^2/2} \right\}.$$

If $\|X\|$ is finite, then $X$ is said to be subgaussian. The norm is scaled to have the property $\|\mathcal{N}(\mu,\sigma^2)\|= \max\{|\mu|,|\sigma|\}$. By Hoeffding's lemma, we have $$\|X\| \leq \|X\|_\infty := \inf\{\tau:\mathbb{P}[|X|\leq\tau]=1\},$$ i.e., bounded random variables are also subgaussian.

I'm interested in the dual norm, defined by $$\|Y\|_* := \sup \left\{ \mathbb{E}[X \cdot Y] : X \text{ is a random variable satisfying } \|X\| \leq 1 \right\}.$$
  Of course, $X$ and $Y$ are not independent in the above supremum.

Is there a simple expression for the dual norm $\|\cdot\|_*$? I would like to be able to calculate $\|\cdot\|_*$ and the definition above is not useful. Even a good approximation to the dual norm would be helpful.


Below are various things I know or think about this question, which may be helpful for answering it.
My intuition is that $\|\cdot\|\approx\|\cdot\|_\infty$, as, in my experience, most properties of bounded random variables extend to subgaussian random variables. Since the $1$-norm is the dual of the $\infty$-norm, my intuition is that $\|\cdot\|_*\approx\|\cdot\|_1$.
This intuition can be made a bit more formal by looking at $p$-norms, as follows. It is easy to show that $$\|X\|_p := \mathbb{E}[|X|^p]^{1/p} \leq (\sqrt{p}+2) \cdot \|X\|$$ for all $p \in [1,\infty)$ and all subgaussian $X$. Thus, by Hölder's inequality, for all $p \in (1,\infty)$ and all random variables $X$ and $Y$, $$\mathbb{E}[X \cdot Y] \leq \|X\|_p \cdot \|Y\|_{1+\frac{1}{p-1}} \leq O(\sqrt{p}) \cdot \|X\| \cdot \|Y\|_{1+\frac{1}{p-1}}.$$ Hence $\|Y\|_* \leq O\left(\frac{1}{\sqrt{\varepsilon}}\right) \cdot \|Y\|_{1+\varepsilon}$ for all $\varepsilon > 0$. Since $\|X\| \leq \|X\|_\infty$, we also have $\|Y\|_* \geq \|Y\|_1$. 
Here is an example that "breaks" this intuition. However, it only slightly breaks it, which is why I think the intuition is still correct. Let $X$ be a standard Gaussian and $Y=\mathsf{sign}(X) \cdot e^{X^2/2}/(1+X^2)$. Then $\|X\|=1$, but $\|X\|_\infty = \infty$. And $\|Y\|_1 = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{1}{1+x^2} \mathrm{d}x = \sqrt{\frac{\pi}{2}}$, while $\|Y\|_* \geq \mathbb{E}[XY] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{|x|}{1+x^2} \mathrm{d}x = \infty$. However, note that $\mathbb{E}[|Y|\log|Y|]=\infty$, so one only needs something "slightly larger" than the $1$-norm for this example. My intuition is that, in general, $\|\cdot\|_*$ is only slightly larger than $\|\cdot\|_1$.
My guess is that the answer is something asymptotically like $\|Y\|_* \overset{?}{=} \mathbb{E}\left[|Y|\sqrt{\log(1+|Y|)} \right]$. Has anyone seen a norm like this before? (I can show that, if $\mathbb{E}\left[|Y|\sqrt{\log(1+|Y|)} \right]=\infty$, then $\|Y\|_*=\infty$.)
I can prove the following upper bound on the dual norm. This is the strongest bound I have been able to prove so far. $$\|Y\|_* \leq \sqrt{2} \mathbb{E}[|Y|] + 4\sqrt{\mathbb{E}[|Y|] \cdot \left(\mathbb{E}[|Y|\log|Y|] - \mathbb{E}[|Y|]\log\mathbb{E}[|Y|]\right)}.$$
Note that by Jensen's inequality and the convexity of $x \mapsto x \log x$, we have $\mathbb{E}[|Y|\log|Y|] \geq \mathbb{E}[|Y|]\log\mathbb{E}[|Y|]$. So the right hand side of the above bound is well-defined and non-negative. Multiplying $Y$ by a constant also multiplies the expression by that constant. So this expression is almost a norm, although I don't know if it satisfies the triangle inequality.
Furthermore, if $\mu=\mathbb{E}[Y]$, then $\|Y\|_* = |\mu| + \|Y-\mu\|_*$. This centering can also be combined with the above bound.
 A: I do not have an answer, but have derived a dual formulation that produces an upper bound. For fixed $y$, you are interested in:
\begin{align}
\max  &\int_\mathbb{R} xy f(x) dx \\
\text{s.t.} &\int_\mathbb{R} \exp(tx) f(x) dx \leq \exp(t^2 / 2) \quad\forall t\\
     &\int_\mathbb{R} f(x) dx = 1\\
     &f(x) \geq 0
\end{align}
This is a linear optimization problem with optimization variables $f(x)$. Its dual is:
\begin{align}
\min  &\int_\mathbb{R} \exp(t^2 / 2) g(t) dt + \mu \\
\text{s.t.} &\int_\mathbb{R} \exp(tx) g(t) dt + \mu \geq xy \quad\forall x \\
     &g(t) \geq 0, \mu \geq 0
\end{align}
The first constraint in the dual can be written as
$$\min_{x,\nu} \left\{ \int_\mathbb{R} \exp(t \nu(t)) g(t) dt - xy : \nu(t)=x \right\} \geq -\mu$$
Let's dualize the left part of this constraint via the Lagrangian:
$$L(x,\nu,\lambda) = \int_\mathbb{R} \exp(t\nu(t)) g(t) dt - xy + \int_\mathbb{R} \lambda(t) (x - \nu(t)) dt$$
You need $\int_\mathbb{R} \lambda(t) dt = y$ as otherwise the value is $-\infty$ (by letting $x \to \pm \infty$) and the constraint is violated, so the terms with $x$ vanish. The derivative of $L$ with respect to $\nu(t)$ is $\exp(t \nu(t)) t g(t) - \lambda(t)$. Therefore, $\lambda(t) \geq 0$ for $t>0$ and $\lambda(t) \leq 0$ for $t<0$ (as otherwise the value is $-\infty$), and settings the derivative to $0$ yields:
\begin{align}
\min  &\int_\mathbb{R} \exp(t^2 / 2) g(t) dt + \mu \\
\text{s.t.} &\int_\mathbb{R} \frac{\lambda(t)}{t} \left( 1 - \log\left(  \frac{\lambda(t)}{tg(t)} \right) \right)dt \geq -\mu \\
     & \int_\mathbb{R} \lambda(t) dt = y \\
     & t \lambda(t) \geq 0 \quad \forall t \\
     &g(t) \geq 0, \mu \geq 0, \lambda(t) \in \mathbb{R}.
\end{align}
You can substitute out $\mu$:
\begin{align}
\min  &\int_\mathbb{R} \exp(t^2 / 2) g(t) dt + \max\left\{0, \int_\mathbb{R} \frac{\lambda(t)}{t} \left( \log\left(  \frac{\lambda(t)}{tg(t)} \right) - 1 \right) dt \right\} \\
\text{s.t.} & \int_\mathbb{R} \lambda(t) dt = y \\
     & t \lambda(t) \geq 0 \quad \forall t \\
     &g(t) \geq 0, \lambda(t) \in \mathbb{R}.
\end{align}
Any feasible pair $(g(t), \lambda(t))$ yields an upper bound to the dual norm.
