On the variance proxy of a positive (and bounded) sub-Gaussian variable Consider a random variable $X \ge 0$ which takes values in an interval $[0, b]$, and further
$$
\text{P}(X \ge t) \le C \exp\left(\frac{-t^{2}}{B}\right),
\quad
\forall t \ge 0,
$$
for given constants $C \gg 1$ and $B >0$.
Since $X$ is bounded, it is a sub-Gaussian variable, and its variance proxy can be upper bounded by $O\left((b-0)^{2}\right)$ based on the length of the interval.
Q1:
First, a clarification on the definition:
if we temporarily ignore the fact that $X$ is bounded (but taking into account that $X \ge 0$), then is the above tail bound enough to say that $X$ is sub-Gaussian? (E.g., does the value of $C$ matter?)
Q2:
Using the tail bound, is it possible to get a better upper bound on the variance proxy?
In particular, I saw a claim that based on the above tail bound, the moments of $X-\mathbb{E}[X]$ can upper be bounded by those of a Gaussian with variance $O(B \sqrt{\log{C}})$. Is that true?
Edit: To bound all moments of $X-\mathbb{E}[X]$ by those of a Gaussian with variance $\gamma$, I would need to show that $X-\mathbb{E}[X]$ is sub-Gaussian with variance proxy $\gamma > 0 $, i.e., that $\mathbb{E}[e^{s(X-\mathbb{E}[X])}] \le e^{s^{2}\gamma/2}$.
Motivated by Michael's answer, which gives an upper bound on the variance $\sigma^{2}$ of $X$, we could put the question this way: is there a straightforward connection between $\gamma$ and  $\sigma^{2}$? I see a related question here: Bound variance proxy of a subGaussian random variable by its variance
 A: Considering both bounds, we know that: 
$$P[X > t] \leq \left\{ \begin{array}{ll}
\min[1,C e^{-t^2/B}] &\mbox{ if $t \in [0,b)$} \\
0  & \mbox{ if $t\geq b$} 
\end{array}
\right. $$
This is the tightest bound since we can consider a random variable $W$ with $P[W>t]$ given exactly by the right-hand-side of the above inequality. Notice that $C e^{-t^2/B} \geq 1$ whenever $t \in [0, \sqrt{B\log(C)}]$. For simplicity assume that $b \geq \sqrt{B\log(C)}$. 
We know that, for general nonnegative random variables $Y$, we have $Y=\int_{0}^{\infty} 1\{Y>t\}dt$ and hence $E[Y]=\int_{0}^{\infty}P[Y>t]dt$. Thus, 
\begin{align}
E[X^2] &= \int_{0}^{\infty} P[X^2>t]dt \quad \mbox{[since $X^2 \geq0$]}\\
&= \int_0^{\infty} P[X > \sqrt{t}]dt \quad \mbox{[since $X \geq 0$]}\\
&=\int_0^{B\log(C)} \underbrace{P[X>\sqrt{t}]}_{\leq 1}dt + \int_{B\log(C)}^{b^2}\underbrace{P[X>\sqrt{t}]}_{\leq Ce^{-t/B}}dt \quad \mbox{[since $P[X>b]=0$]} \\
&\leq B\log(C) + -CBe^{-t/B}|_{B\log(C)}^{b^2}\\
&=B\log(C) + B - CBe^{-b^2/B}
\end{align}
This is the best upper bound on $E[X^2]$ since it holds with equality for the random variable $W$ defined above. A simpler bound is then $E[X^2] \leq B\log(C) + B$, and this holds regardless of the value of $b$. (It even holds when $0\leq b < \sqrt{B\log(C)}$, since reducing the value of $b$ cannot increase the bound.)
In particular, $\sigma=\sqrt{Var(X)} \leq \sqrt{E[X^2]} \leq \sqrt{B + B\log(C)}$. 
A: (I figured that the statement of my second question is true. I am sharing a proof sketch for future reference.)

Let $\mu = \mathbb{E}\left[{X}\right]$,
and define $Y = X- \mu$.
It suffices to show that
$$
\mathbb{E}
\left[
 \exp\left(\lambda \cdot Y\right)
\right]
\le
\exp\left(\frac{1}{2} \cdot \sigma^{2} \cdot \lambda^{2}\right),
\quad \lambda \in \mathbb{R},
$$
for $\sigma^{2} = O(B\sqrt{\log{C}})$.
(In the sequel, I assume that $\log{C} \ge 1$.)
First, we derive an upper bound on the absolute moments of $Y$:
\begin{align}
 \mathbb{E}
 \left[
  \left\lvert{Y}\right\rvert^{p}
 \right] 
 =
 \mathbb{E}
 \left[
  |{X} - \mu |^{p}
 \right] 
 \le
 \mathbb{E}\left[ \max\left\lbrace {X}, \mu\right\rbrace^{p} \right] 
 \le
 \mathbb{E}\left[ {X}^{p} \right] + \mu^{p},
\end{align}
where we have taken into account that $X\ge 0$ and in turn $\mu \ge 0$.
Now,  $X^{p}$ is a nonnegative random variable, and in turn
\begin{align}
 \mathbb{E}\left[ X^{p} \right]
 &=
 \int_{0}^{\infty} \mathrm{P}\left( {X}^{p} \ge u\right) du
 =
 \int_{0}^{\infty} \mathrm{P}\left( {X}^{p} \ge t^{p}\right) \cdot {p} \cdot t^{p-1} dt
 \nonumber\\
 &=
 \int_{0}^{\infty} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
 \nonumber\\
 &=
 \underbrace{
  \int_{0}^{t_{0}} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
 }_{I_{1}}
 +
 \underbrace{
  \int_{t_{0}}^{\infty} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt,
 }_{I_{2}}
 \label{sug-gaussian:X-pth-moment-start-ub}
\end{align}
for any $t_{0} \ge 0$.
For $t_{0} = ({B}\sqrt{\log{C}})^{1/2}$, we have
\begin{align}
 I_{1}
 =
 \int_{0}^{t_{0}} \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
 \le
 \int_{0}^{t_{0}} {p} \cdot {t^{p-1}} dt
 =
 t_{0}^{p}
 =
 \left({B}\sqrt{\log{C}}\right)^{p/2}.
 \label{sug-gaussian:I1-ub}
\end{align}
For the second part,
let $f(t) = {C} \cdot e^{-t^{2}/B}$ and $g(t) = e^{-t^{2}/\left({B}\sqrt{\log{C}}\right)}$.
One can verify that $f(t) \le g(t)$ for $t \ge ({B}\sqrt{\log{C}})^{1/2}$ (assuming that $\log{C} \ge 1$).
Then, 
\begin{align}
 I_{2}
 &=
 \int_{t_{0}}^{\infty}
  \mathrm{P}\left( {X} \ge t\right) \cdot {p} \cdot {t^{p-1}} dt
 \le
 \int_{t_{0}}^{\infty}
 f(t) \cdot {p} \cdot {t^{p-1}} dt
 %\nonumber\\&
 \le 
 \int_{t_{0}}^{\infty}
 g(t) \cdot {p} \cdot {t^{p-1}} dt,
 \label{sug-gaussian:I2-ub}
\end{align}
It follows that for the particular choice of $t_{0}$,
\begin{align}
 I_{2}
 &\le 
 \int_{t_{0}}^{\infty}
  g(t) \cdot {p} \cdot {t^{p-1}} dt
 \le
 \int_{0}^{\infty}
  g(t) \cdot {p} \cdot {t^{p-1}} dt
 \nonumber\\
 &\le
 \int_{0}^{\infty}
  e^{-t^{2}/(B\sqrt{\log{C}})} \cdot {p} \cdot {t^{p-1}} dt
 =
 (B \sqrt{\log{C}})^{p/2}
 \cdot \frac{p}{2}
 \cdot \Gamma\left( \frac{p}{2}\right)
\end{align}
Combining the bounds on the two parts, we have 
\begin{align}
 \mathbb{E}\left[ X^{p} \right]
 \le
 \left({B}\sqrt{\log{C}}\right)^{p/2}
 \cdot
 \left(
  1 + \frac{p}{2} \cdot \Gamma\left( \frac{p}{2}\right)
 \right).
\end{align}
Specifically for $p=1$, we find
\begin{align}
 \mu
 =
 \mathbb{E}\left[ X\right]
 \le
 2 \cdot \left({B}\sqrt{\log{C}}\right)^{1/2}.
 \label{sub-gaussian:ub-on-mean}
\end{align}
Further, for any $p \ge 2$,
\begin{align}
 \mathbb{E}\left[ X^{p} \right]
 \le
 \left({B}\sqrt{\log{C}}\right)^{p/2}
 \cdot 2^{p}
 \cdot \left( \frac{p}{2}\right)^{p/2}.
\end{align}
Finally, combining the above we find that for any $p \ge 2$,
$$
 \mathbb{E}
 \left[
  \left\lvert{Y}\right\rvert^{p}
 \right]
 \le
    c^{\prime}
 \left({B}\sqrt{\log{C}}\right)^{p/2}
     \cdot 2^{p}
 \cdot \left( \frac{p}{2}\right)^{p/2},
$$
for some positive constant $c^{\prime}$.
Having the above upper bound on 
$\mathbb{E}\left[ |Y|^{p}\right]$, 
we can now show (see Lemma 5.5 in https://arxiv.org/pdf/1011.3027v7.pdf that
\begin{align}
 \mathbb{E}
 \left[
  \exp\left({\lambda}{Y}\right)
 \right]
 &=
 \exp(O\left((B\sqrt{\log{C}})^{2}\lambda^{2}\right))
\end{align}
which is the desired result.
