- $\mathbb{E}(V_{i} | X_{1}, \ldots, X_{i-1}) = 0$
Let us introduce some notation, $X_{1:i} = X_{1}, \ldots, X_{i}$.
\begin{align*}
\mathbb{E}(V_{i} | X_{1:i-1}) &= \mathbb{E}\left( \left[\mathbb{E}(g | X_{1:i}) - \mathbb{E}(g | X_{1:i-1})\right] | X_{1:i-1}\right) \\\\
&=\mathbb{E}\left( \mathbb{E}(g | X_{1:i}) | X_{1:i-1}\right)- \mathbb{E}\left( \mathbb{E}(g | X_{1:i-1}) | X_{1:i-1} \right)
\end{align*}
Next apply the definitions of iterated expectations to $\mathbb{E}\left( \mathbb{E}(g | X_{1:i}) | X_{1:i-1}\right)$
First let us work with inner expectation.
\begin{align*}
\mathbb{E}(g | X_{1:i}) &= \int_{x_{i+1:n}} g f_{X_{1:n} | X_{1:i}} \mathrm{d}x_{i+1:n}
\end{align*}
Since $X_{i}$ are independent, we can reduce the joint pdf to
\begin{align*}
f_{X_{1:n} | X_{1:i}} &= \frac{\prod_{j=1}^{n} f_{X_{j}}(x_{j})} {\prod_{j=1}^{i} f_{X_{j}}(x_{j})} \\\\
&= \prod_{j=i+1}^{n} f_{X_{j}}(x_{j})
\end{align*}
Substituting the joint pdf in the inner expectation integral we get,
\begin{align*}
\mathbb{E}(g | X_{1:i}) &= \int_{x_{i+1:n}} g \prod_{j=i+1}^{n} f_{X_{j}}(x_{j})\text{dx}_{i+1:n}
\end{align*}
Observe that $\mathbb{E}(g | X_{1:i})$ is some function of $X_{1:i}$ only. Thus while doing the outer expectation, the conditional pdf will be $f_{X_{1:i} | X_{1:i-1}}$
Now the outer expectation:
\begin{align*}
\mathbb{E}\left( \mathbb{E}(g | X_{1:i}) | X_{1:i-1} \right)
&= \int_{x_{i}} \left ( \int_{x_{i+1:n}} g \prod_{j=i+1}^{n} f_{X_{j}}(x_{j}) \mathrm{d}x_{i+1:n} \right) \frac{\prod_{j=1}^{i} f_{X_{j}}(x_{j})}{\prod_{j=1}^{i-1} f_{X_{j}}(x_{j})} \mathrm{d}x_{i}\\\\
& = \int_{x_{i:n}} g \prod_{j=i}^{n} f_{X_{j}}(x_{j}) \mathrm{d}x_{i:n} \\\\
& = \mathbb{E}(g | X_{1:i-1})
\end{align*}
Using the same argument used above one can prove,
\begin{align*}
\mathbb{E}\left( \mathbb{E}(g | X_{1:i-1}) | X_{1:i-1} \right)
& = \mathbb{E}(g | X_{1:i-1})
\end{align*}
which leads to
\begin{align*}
\mathbb{E}(V_{i} | X_{1}, \ldots, X_{i-1}) &= \mathbb{E}\left( \mathbb{E}(g | X_{1:i}) | X_{1:i-1}\right)- \mathbb{E}\left( \mathbb{E}(g | X_{1:i-1}) | X_{1:i-1} \right) \\\\
&= \mathbb{E}(g | X_{1:i-1}) - \mathbb{E}(g | X_{1:i-1}) \\\\
&= 0
\end{align*}
- $\mathbb{E}(e^{t V_{i} } | X_{1}, \ldots, X_{i-1}) \le e^{t^2c_{i}^{2}/8}$
The above inequality will follow from Hoeffding's lemma if we can prove $(V_{i} | X_{1}, \ldots, X_{i-1})$ is both upper and lower bounded. We have already proved $\mathbb{E}(V_{i} | X_{1}, \ldots, X_{i-1}) = 0$.
The following proof is taken from John Duchi's wonderful notes on the inequalities. Let
\begin{align*}
U_{i} &= \sup_{u} \quad \mathbb{E}(g | X_{1:i−1}, u) − \mathbb{E}(g | X_{1:i−1}) \\\\
L_{i} &= \inf_{l} \quad \mathbb{E}(g | X_{1:i−1}, l) − \mathbb{E}(g | X_{1:i−1})
\end{align*}
Now
\begin{align*}
U_{i} - L_{i}
&\le \sup_{l,u} \,
\mathbb{E}(g | X_{1:i−1}, u) - \mathbb{E}(g | X_{1:i−1}, l) \\\\
&\le \sup_{l,u}
\left (
\int_{x_{i+1}:n}
[
g(X_{1:i-1}, u, x_{i+1:n}) - g(X_{1:i-1}, l, x_{i+1:n})
]
\prod_{j=i+1}^{n} f_{X_{j}}(x_{j})
\mathrm{d}x_{i+1:n}
\right )
\\\\
&\le
\int_{x_{i+1}:n}
\sup_{l,u} \;
\left (
g(X_{1:i-1}, u, x_{i+1:n}) - g(X_{1:i-1}, l, x_{i+1:n})
\right )
\prod_{j=i+1}^{n} f_{X_{j}}(x_{j})
\;
\mathrm{d}x_{i+1:n}
\\\\
&\le
\int_{x_{i+1}:n}
c_{i}
\prod_{j=i+1}^{n} f_{X_{j}}(x_{j})
\;
\mathrm{d}x_{i+1:n}
\\\\
&= c_{i}
\end{align*}
The third line follows from Jensen's inequality since $\sup$ is convex and the fourth line from the assumptions in the McDiarmid Inequality. Hence $L_{i} \le V_{i} \le U_{i}$ and we can apply Hoeffding's lemma to get
\begin{align*}
\mathbb{E}(e^{t V_{i} } | X_{1}, \ldots, X_{i-1}) &\le e^{t^2c_{i}^{2}/8}
\end{align*}
- $\mathbb{E} \left( e^{t \sum_{i=1}^{n} V_{i}} \right) =
\mathbb{E} \left( e^{t \sum_{i=1}^{n-1} V_{i}}
\mathbb{E} \left( e^{tV_{n}} | X_{1}, \ldots, X_{n-1} \right) \right)$
A straightforward application of iterated expectation will lead us to the above result.
\begin{align*}
\mathbb{E} \left( e^{t \sum_{i=1}^{n} V_{i}} \right)
&=
\mathbb{E} \left(
\mathbb{E} \left( e^{t \sum_{i=1}^{n} V_{i}} | X_{1}, \ldots, X_{n-1} \right)
\right) \\\\
&=
\mathbb{E} \left(
e^{t \sum_{i=1}^{n-1} V_{i}}
\mathbb{E} \left( e^{t V_{n}} | X_{1}, \ldots, X_{n-1} \right)
\right) \\\\
\end{align*}
The inner expectation is wrt $X_{n}$ while the outer expectation is wrt $X_{1:n-1}$
On applying the Hoeffding's inequality $n$ times repeatedly, we get:
\begin{align*}
\mathbb{E} \left( e^{t \sum_{i=1}^{n} V_{i}} \right)
&=
\mathbb{E} \left(
e^{t \sum_{i=1}^{n-1} V_{i}}
\mathbb{E} \left( e^{t V_{n}} | X_{1}, \ldots, X_{n-1} \right)
\right) \\\\
&\le
\mathbb{E} \left(
e^{t \sum_{i=1}^{n-1} V_{i}}
\exp \left( t^{2} c_{n}^{2}/8 \right)
\right) \\\\
&\le \exp\left( \frac{1}{8} \sum_{i=1}^{n} t^{2} c_{i}^{2} \right)
\end{align*}
Now we are ready to prove McDiarmid's inequality,
\begin{align*}
\mathbb{P} \left( g(X_{1}, \ldots, X_{n}) - \mathbb{E}(g(X_{1}, \ldots, X_{n})) \ge
\epsilon \right)
&= \mathbb{P} \left( \sum_{i=1}^{n} V_{i} \ge \epsilon \right) \\\\
&= \mathbb{P} \left( e^{t \sum_{i=1}^{n} V_{i}} \ge e^{t \epsilon} \right) \\\\
& \le \exp(-t \epsilon) \mathbb{E} \left( e^{t \sum_{i=1}^{n} V_{i}} \right) \\\\
& \le \exp(-t \epsilon) \exp\left( \frac{1}{8} \sum_{i=1}^{n} t^{2} c_{i}^{2} \right) \\\\
& = \exp \left (-t \epsilon + \frac{1}{8} \sum_{i=1}^{n} t^{2} c_{i}^{2} \right)
\end{align*}
The third line follows from Markov's inequality. To get the final result,
we need to minimize the expression wrt $t$. This occurs at $t = 4 \epsilon / \sum_{i=1}^{n} c_{i}^{2}$.
Best Answer
I’m not sure I understood your question correctly. I’ll try to answer: try to write $$-\frac{a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta}$$ as a function of $u = t(b-a)$ : this is natural as you want a bound in $e^{u^2 \over 8}$.
Helped by the experience, you will know that it is better to chose to write it in the form $e^{g(u)}$. Then $$e^{g(u)} = -\frac{a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta}$$ leads to $$\begin{align*} g(u) &= \log\left( -\frac{a}{b-a} e^{tb} + \frac{b}{b-a} e^{ta} \right)\\ &= \log\left( e^{ta} \left( -\frac{a}{b-a} e^{t(b-a)} + \frac{b}{b-a} \right)\right)\\ &= ta + \log\left( \gamma e^u + (1-\gamma) \right)\\ &= -\gamma u + \log\left( \gamma e^u + (1-\gamma) \right),\\ \end{align*}$$ with $\gamma = - {a \over b-a}$.
Is that the kind of thing you were asking for?
Edit: a few comments on the proof
Now turn to our problem. Why is it possible to get a bound depending only on $u = t(b-a)$? Intuitively, it is just a matter of rescaling of $X$: if you have a bound $\mathbb{E}\left( e^{tX} \right) \le s(t)$ for the case $b-a=1$, then the general bound can be obtained by taking $s( t(b-a) )$. Now think to the set of centered variables with support of width 1: there isn’t so much freedom, so a bound like $s(t)$ should exist.
An other approach is to say simply that by the above lemma on $\mathbb{E}(\phi(X))$, then more generally $\mathbb{E}(\phi(tX)) \le \mathbb{E}(\phi(tX_0))$, which depends only on $u$ and $\gamma$: if you fix $u = u_0 = t_0 (b_0 - a_0)$ and $\gamma = \gamma_0 = - {a_0 \over b_0-a_0}$, and let $t, a, b$ vary, there is only one degree of freedom, and $t = {t_0 \over \alpha}$, $a = \alpha a_0$, $b = \alpha a_0$. We get $$-{a\over b-a} \phi(tb) + {b \over b-a} \phi(ta) = -{a_0\over b_0-a_0} \phi(tb_0) + {b_0 \over b_0-a_0} \phi(a_0).$$ You just have to find a bound involving only $u$.
Now we are convinced that it can be done, it must be much easier! You don’t necessarily think to $g$ to begin with. The point is that you must write everything as a function of $u$ and $\gamma$.
First note that $\gamma = -{a\over b-a}$, $1 -\gamma = {b\over b-a}$, $at = -\gamma u$ and $bt = (1-\gamma)u$. Then $$\begin{align*} \mathbb{E}(\phi(tX)) \le &-{a\over b-a} \phi(tb) + {b \over b-a} \phi(ta) \\ = & \gamma \phi((1-\gamma)u) + (1-\gamma) \phi(-\gamma u) \end{align*}$$
Now we are in the particular case $\phi=\exp$... I think you can finish.
I hope I clarified it a little bit.