Understanding proof of McDiarmid's inequality I am working through Wasserman's lecture notes set 2 and I am unable to fill in the missing steps in the derivation of McDiarmid's inequality (p.5). Just like my previous question in the forum, I am reproducing the proof in the notes below and after the proof I will point the steps I am not able to derive.
McDiarmid's Inequality
Let $X_{1}, \ldots, X_{n}$ be independent random variables. Suppose that
\begin{align*}
\sup_{x_{1}, \ldots, x_{n}, x_{i}^{\prime}}  \left| g(x_{1}, \ldots, x_{i-1}, x_{i}, 
x_{i+1},  \ldots, x_{n}) -
g(x_{1}, \ldots, x_{i-1}, x_{i}^{\prime} , x_{i+1},  \ldots, x_{n}) \right| &\le c_{i} 
\end{align*}
for $i = 1, \ldots, n$.  Then
\begin{align*}
\mathbb{P} \left( g(X_{1}, \ldots, X_{n}) - \mathbb{E}(g(X_{1}, \ldots, X_{n})) \ge 
\epsilon \right)  & \le \exp \left( -\frac{2\epsilon^{2}}{\sum_{i=1}^{n} c_{i}^{2}} 
\right)
\end{align*}
Proof
Let $V_{i} = \mathbb{E}(g | X_{1}, \ldots, X_{i}) - \mathbb{E}(g | X_{1}, \ldots, X_{i-1})$. Then 
\begin{align*}
g(X_{1}, \ldots, X_{n}) - \mathbb{E}(g(X_{1}, \ldots, X_{n})) &= \sum_{i=1}^{n} V_{i}
\end{align*}
and $\mathbb{E}(V_{i} | X_{1}, \ldots, X_{i-1}) = 0$
Using a similar argument as in the proof of Hoeffding's Lemma we have,
\begin{align*}
\mathbb{E}(e^{t V_{i} } | X_{1}, \ldots, X_{i-1}) &\le e^{t^2c_{i}^{2}/8}
\end{align*}
Now, for any $t > 0$,
\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 e^{- t \epsilon} \mathbb{E} \left( e^{t \sum_{i=1}^{n} V_{i}} \right) \\\\
&= e^{- t \epsilon} \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 )  \\\\
&\le e^{- t \epsilon}  e^{t^2c_{n}^{2}/8} 
\mathbb{E} \left (  e^{t \sum_{i=1}^{n-1} V_{i}} \right )  \\\\
& \ldots \\\\
&\le  e^{- t \epsilon}  e^{t^2 \sum_{i=1}^{n} c_{i}^{2}/8}
\end{align*}
The result follows by taking $t = 4 \epsilon / \sum_{i=1}^{n} c_{i}^{2}$.
Questions
How to prove


*

*$\mathbb{E}(V_{i} | X_{1}, \ldots, X_{i-1}) = 0$ 

*$\mathbb{E}(e^{t V_{i} } | X_{1}, \ldots, X_{i-1}) \le e^{t^2c_{i}^{2}/8}$

*$\mathbb{E} \left( e^{t \sum_{i=1}^{n} V_{i}} \right) =
    \mathbb{E} \left( e^{t \sum_{i=1}^{n-1} V_{i}}
    \mathbb{E} \left( e^{tV_{n}} | X_{1}, \ldots, X_{n-1} \right) \right)$


Though this involves technical details, any intuition on the details will be helpful.
 A: *

*$\mathbb{E}(V_{i} | X_{1}, \ldots, X_{i-1}) = 0$
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 $t = 4 \epsilon / \sum_{i=1}^{n} c_{i}^{2}$.
