29  Concentration Inequalities II

Sub-Gaussian random variables

A random variable X with mean \mu is called Sub-Gaussian with parameter \sigma (X \sim SubGau (\sigma)) if

\mathbb{E}_{X} \left[ e^{s (X - \mu)} \right] \leq \exp \left[ \frac{ s^{2} \sigma^{2} }{ 2 } \right].

The definition can equivalently be expressed in terms of bounds on the tail of X. Let X \sim SubGau (\sigma) and \mu = \mathbb{E}_{X} [X]. Then for any t > 0,

\mathbb{P}_{X} \left( X - \mu \geq t \right) \leq \exp \left[ \frac{ - t^{2} }{ 2 \sigma^{2} } \right].

First use Chernoff bound to derive

\begin{aligned} \mathbb{P}_{X} (x \geq t) & \leq \inf_{s > 0} \frac{ \mathbb{E}_{X} [e^{s x}] }{ e^{s t} } \\ & \leq \inf_{s > 0} \exp \left[ \frac{ s^{2} \sigma^{2} }{ 2 } - s t \right]. \end{aligned}

Since \exp \left[ \frac{ s^{2} \sigma^{2} }{ 2 } - s t \right] is a convex function

\begin{aligned} \frac{ d }{ d s } \exp \left[ \frac{ s^{2} \sigma^{2} }{ 2 } - s t \right] & = 0 \\ (\sigma^{2} s - t) \exp \left[ \frac{ \sigma^{2} s^{2} }{ 2 } - t s \right] & = 0 \\ s & = \frac{ t }{ \sigma^{2} }. \end{aligned}

Plug s = \frac{ t }{ \sigma^{2} } back and we get derive the result

\begin{aligned} \mathbb{P}_{X} (x \geq t) & \leq \inf_{s > 0} \exp \left[ \frac{ s^{2} \sigma^{2} }{ 2 } - s t \right] \\ & = \exp \left[ \frac{ t^{2} }{ 2 \sigma^{2} } - \frac{ t^{2} }{ \sigma^{2} } \right] \\ & = \exp \left[ \frac{ - t^{2} }{ 2 \sigma^{2} } \right]. \end{aligned}

Hoeffding’s inequality

Hoeffding’s inequality provides an upper bound on the probability that the sum of bounded independent random variables deviates from its expected value by more than a certain amount.

Hoeffding’s Lemma

Let X be a bounded random variable with a \leq X \leq b and \mu = \mathbb{E}_{X} [x]. Then for all s > 0,

\mathbb{E}_{X} [e^{s (x - \mu)}] \leq \exp \left[ \frac{ s^{2} (b - a)^{2} }{ 8 } \right].

Since f (x) = e^{a x} is a convex function, the definition of the convex function states that

e^{s x} \leq \frac{ (x - a) e^{s b} }{ b - a } + \frac{ (b - x) e^{s a} }{ b - a }.

Taking the expectation on the both ends and the fact that \mathbb{E}_{X} (x) = 0,

\begin{aligned} \mathbb{E}_{X} [e^{s x}] & \leq \mathbb{E}_{X} \left[ \frac{ (x - a) e^{s b} }{ b - a } + \frac{ (b - x) e^{s a} }{ b - a } \right] \\ & = \frac{ e^{s b} \mathbb{E}_{X} [x] - a e^{s b} + b e^{s a} - e^{s a} \mathbb{E}_{X} [x] }{ b - a } \\ & = \frac{ b e^{s a} - a e^{s b} }{ b - a }. \end{aligned}

Now we will prove that

\frac{ b e^{s a} - a e^{s b} }{ b - a } \leq \exp \left[ \frac{ s^{2} (b - a)^{2} }{ 8 } \right]

which is same as proving

\log \left[ \frac{ b e^{s a} - a e^{s b} }{ b - a } \right] \leq \frac{ s^{2} (b - a)^{2} }{ 8 }.

By taking p = \frac{ b }{ b - a } and 1 - p = -\frac{ a }{ b - a }

\begin{aligned} \log \left[ \frac{ b e^{s a} - a e^{s b} }{ b - a } \right] & = \log \left[ \frac{ b e^{s a} }{ b - a } - \frac{ a e^{s b} }{ b - a } \right] \\ & = \log \left[ p e^{s a} + (1 - p) e^{s b} \right] \\ & = \log \left[ e^{s a} e^{-s a} (p e^{s a} + (1 - p) e^{s b}) \right] \\ & = s a + \log \left[ (p + (1 - p) e^{s b - s a}) \right] \end{aligned}

and taking u = (b - a) s, we can get a function

\begin{aligned} \phi (u) & = s a + \log \left[ (p + (1 - p) e^{s b - s a}) \right] \\ & = (p - 1) u + \log \left[ (p + (1 - p) e^{u}) \right]. \end{aligned}

This function can be approximated using Taylor series until the second order term at point a = 0,

\phi (u) = \phi (0) + \phi' (0) x + \frac{ \phi' (0) }{ 2 } x^{2},

where

\phi' (u) = (p - 1) + \frac{ (1 - p) e^{u} }{ p + (1 - p) e^{u} } = 0,

and

\phi'' (u) = \frac{ p (1 - p) e^{u} }{ (p + (1 - p) e^{u})^{2} } = 0.

Since \phi (0) = 0, \phi' (0) = 0 and

\phi'' (0) = \frac{ p (1 - p) }{ (p + (1 - p))^{2} }

reaches its maximum of 0.25 at p = 0.5, we can derive

\log \left[ \frac{ b e^{s a} - a e^{s b} }{ b - a } \right] = \phi (u) \leq \frac{ u^{2} }{ 8 } = \frac{ s^{2} (b - a)^{2} }{ 8 }.

Therefore, we can prove the lemma

\mathbb{E}_{X} [e^{s x}] \leq \frac{ b e^{s a} - a e^{s b} }{ b - a } \leq \exp \left[ \frac{ s^{2} (b - a)^{2} }{ 8 } \right].

Equivalently, Hoeffding’s lemma states that any random variable X \in [a, b] is a subGaussian variable

X \in SubGau \left( \frac{ (b - a)^{2} }{ 4 } \right).

Hoeffding’s inequality

Let X_{1}, \dots, X_{n} be bounded independent random variables such that X_{i} \in [a_{i}, b_{i}], \forall i. Let X = \sum_{i=1}^{n} X_{i} and \mu = \mathbb{E}_{X} (x). Then for t > 0

\mathbb{P}_{X} \left( x - \mu \geq t \right) \leq \exp \left[ \frac{ - 2 t^{2} }{ \sum_{i=1}^{n} (b_{i} - a_{i})^{2} } \right].

We can first apply Chernoff’s bounding method

\begin{aligned} \mathbb{P}_{X} (x - \mu \geq t) & \leq \inf_{s > 0} \frac{ M_{X} (s) }{ e^{s t} } \\ & = \inf_{s > 0} \frac{ \prod_{i = 1}^{n} M_{X_{i}} (s) }{ e^{s t} } \\ & = \inf_{s > 0} \frac{ \prod_{i = 1}^{n} \mathbb{E}_{X} \left[ e^{s (X_{i} - \mathbb{E}_{X_{i}} [x_{i}])} \right] }{ e^{s t} }. \end{aligned}

Then by applying the Hoeffding’s Lemma,

\mathbb{E}_{X_{i}} \left[ e^{s (X_{i} - \mathbb{E}_{X_{i}} [x_{i}])} \right] \leq \exp \left[ \frac{ s^{2} (b_{i} - a_{i})^{2} }{ 8 } \right]

we can have

\begin{aligned} \mathbb{P}_{X} (x - \mu \geq t) & \leq \inf_{s > 0} \frac{ \prod_{i = 1}^{n} \mathbb{E}_{X_{i}} \left[ e^{s (X_{i} - \mathbb{E}_{X_{i}} [x_{i}])} \right] }{ e^{s t} } \\ & \leq \inf_{s > 0} \frac{ \prod_{i = 1}^{n} \exp\left[ \frac{ s^{2} (b_{i} - a_{i})^{2} }{ 8 } \right] }{e^{s t}} \\ & = \inf_{s > 0} \exp\left[ \frac{ s^{2} \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} }{ 8 } - s t \right]. \end{aligned}

Since \exp \left[ \frac{ s^{2} \sum_{i = 1}^{n} (a_{n} - b_{n})^{2} }{ 8 } - s t \right] is a convex function,

\begin{aligned} \frac{ d }{ d s } \exp \left[ \frac{ s^{2} \sum_{i = 1}^{n} (a_{n} - b_{n})^{2} }{ 8 } - s t \right] & = 0 \\ \left( \frac{ s \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} }{ 4 } - t \right) \exp\left[ \frac{ s^{2} \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} }{ 8 } - s t \right] & = 0 \\ \frac{ s \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} }{ 4 } - t & = 0 \\ s & = \frac{ 4 t }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } \end{aligned}

Therefore, we have proved Hoeffding’s inequality

\begin{aligned} \mathbb{P}_{X} (x - \mu \geq t) & \leq \inf_{s > 0} \exp\left[ \frac{ s^{2} }{ 8 } \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} - s t \right] \\ & = \exp \left[ \frac{ \left( \frac{ 4 t }{ \sum_{i}^{n} (b_{i} - a_{i})^{2} } \right)^{2} }{ 8 } \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} - \frac{ 4 t }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } t \right] \\ & = \exp \left[ \frac{ - 2 t^{2} }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } \right]. \end{aligned}

Another version is to bound the estimated mean.

Let X_{1}, \dots, X_{n} be bounded independent random variables such that X_{i} \in [a_{i}, b_{i}], \forall i. Let \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_{i} and \mu = \mathbb{E}_{\bar{X}} (x). Then for t > 0

\mathbb{P}_{\bar{X}} \left( x - \mu \geq t \right) \leq \exp \left[ -\frac{ 2 n^{2} t^{2} }{ \sum_{i=1}^{n} (b_{i} - a_{i})^{2} } \right]

Martingales

Martingale sequence

Given a sequence of random variables X_{1}, \dots, X_{\infty}, define another sequence of random variables Y_{1}, \dots, Y_{\infty} where Y_{n} is a measurable function of X_{1}, \dots X_{n}

Y_{n} = f (X_{1}, \dots, X_{n}).

For all Y_{n}, if \mathbb{E}_{Y_{n}} [y_{n}] is finite and

\mathbb{E}_{Y_{n + 1} \mid X_{1}, \dots, X_{n}} [Y_{n + 1} \mid X_{1}, \dots, X_{n}] = Y_{n},

then Y_{1}, \dots, Y_{\infty} is a martingale sequence with respect to X_{1}, \dots, X_{\infty}.

Martingale difference sequence

Given a sequence of random variables X_{1}, \dots, X_{\infty}, define another sequence of random variables Z_{1}, \dots, Z_{\infty} where Z_{n} is a measurable function of X_{1}, \dots X_{n}

Z_{n} = f (X_{1}, \dots, X_{n}).

For all Z_{n}, if E_{Z_{n}} [z_{n}] is finite and

\mathbb{E}_{Z_{n + 1} \mid X_{1}, \dots, X_{n}} [Z_{n + 1} \mid X_{1}, \dots, X_{n}] = 0

then Z_{1}, \dots, Z_{\infty} is a martingale difference sequence with respect to X_{1}, \dots, X_{\infty}.

A common Martingale sequence

If X_{1}, \dots, X_{\infty} is a sequence of independent random variables where \mathbb{E}_{X_{i}} [x_{i}] = 0, \forall X_{i}, then Y_{1}, \dots, X_{\infty} with Y_{n} = \sum_{i = 1}^{n} X_{i} is martingale sequence.

\begin{aligned} \mathbb{E}_{Y_{n + 1} \mid X_{1}, \dots, X_{n}} & = \mathbb{E}_{Y_{n} + X_{n + 1} \mid X_{1}, \dots, X_{n}} \\ & = \mathbb{E}_{Y_{n} \mid X_{1}, \dots, X_{n}} + \mathbb{E}_{X_{n + 1} \mid X_{1}, \dots, X_{n}} \\ & = Y_{n} + \mathbb{E}_{X_{n + 1}} \\ & = Y_{n} \end{aligned}

Suppose Y_{1}, \dots, Y_{n} is a martingale sequence with respect to X_{1}, \dots, X_{n}. then the sequence Z_{1}, \dots, Z_{\infty} with Z_{n} = Y_{n} - Y_{n - 1} is a martingale difference sequence.

\begin{aligned} \mathbb{E}_{Z_{n + 1} \mid X_{1}, \dots, X_{n}} & = \mathbb{E}_{Y_{n + 1} - Y_{n} \mid X_{1}, \dots, X_{n}} \\ & = \mathbb{E}_{Y_{n + 1} \mid X_{1}, \dots, X_{n}} - \mathbb{E}_{Y_{n} \mid X_{1}, \dots, X_{n}} \\ & = Y_{n} - Y_{n} \\ & = 0 \end{aligned}

Azuma Inequality

For a sequence of Martingale difference sequence of bounded random variable Z_{1}, \dots, Z_{\infty} with Z_{i} \in [a_{i}, b_{i}], \forall i, then:

\mathbb{P}_{Z_{1}, \dots, Z_{n}} \left[ \sum_{i = 1}^{n} z_{i} \geq t \right] \leq \exp \left[ \frac{ - 2 t^{2} }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } \right].

First we can apply Chernoff bound to derive

\begin{aligned} \mathbb{P}_{Z_{1}, \dots, Z_{n}} \left[ \sum_{i = 1}^{n} z_{i} \geq t \right] & \leq \inf_{s > 0} \frac{ \mathbb{E}_{Z_{1}, \dots, Z_{n}} \left[ e^{s \sum_{i = 1}^{n} z_{i}} \right] }{ e^{s t} } \\ & = \inf_{s > 0} \frac{ \mathbb{E}_{Z_{1}, \dots, Z_{n}} \left[ \prod_{i = 1}^{n} e^{s z_{i}} \right] }{ e^{s t} } \end{aligned}

Then we can use the law of total expectation to derive

\begin{aligned} \mathbb{E}_{Z_{1}, \dots, Z_{n}} \left[ \prod_{i = 1}^{n} e^{s z_{i}} \right] & = \mathbb{E}_{Z_{1}, \dots, Z_{n - 1}} \left[ \mathbb{E}_{Z_{1}, \dots, Z_{n} \mid Z_{1}, \dots Z_{n - 1}} \left[ \prod_{i = 1}^{n} e^{s z_{i}} \mid z_{1}, \dots, z_{n - 1} \right] \right] \\ & = \mathbb{E}_{Z_{1}, \dots, Z_{n - 1}} \left[ \mathbb{E}_{Z_{1}, \dots, Z_{n} \mid Z_{1}, \dots Z_{n - 1}} \left[ e^{s z_{n}} \prod_{i = 1}^{n - 1} e^{s z_{i}} \mid z_{1}, \dots, z_{n - 1} \right] \right] \\ & = \mathbb{E}_{Z_{1}, \dots, Z_{n - 1}} \left[ \prod_{i = 1}^{n - 1} e^{s z_{i}} \mathbb{E}_{Z_{n} \mid Z_{1}, \dots, Z_{n - 1}} \left[ e^{s z_{n}} \mid z_{1}, \dots, z_{n - 1} \right] \right], \end{aligned}

where the last equality is derived because \prod_{i = 1}^{n - 1} e^{s z_{i}} is a constant given z_{1}, \dots, z_{n - 1}.

We can derive a upper bound by applying the Hoeffding’s lemma

\begin{aligned} \mathbb{E}_{Z_{1}, \dots, Z_{n}} \left[ \prod_{i = 1}^{n} e^{s z_{i}} \right] & \leq \mathbb{E}_{Z_{1}, \dots, Z_{n - 1}}\ \left[ \prod_{i = 1}^{n - 1} e^{s z_{i}} \exp \left[ \frac{ s^{2} (a_{n} - b_{n})^{2} }{ 8 } \right] \right] \\ & = \exp \left[ \frac{ s^{2} (a_{n} - b_{n})^{2} }{ 8 } \right] \mathbb{E}_{Z_{1}, \dots, Z_{n - 1}}\ \left[ \prod_{i = 1}^{n - 1} e^{s z_{i}} \right] \end{aligned}

We can apply the same procedure for \mathbb{E}_{Z_{1}, \dots, Z_{n - 1}} [ \prod_{i = 1}^{n - 1} e^{s z_{i}} ]

\mathbb{E}_{Z_{1}, \dots, Z_{n - 1}} \left[ \prod_{i = 1}^{n - 1} e^{s z_{i}} \right] = \exp \left[ \frac{ s^{2} (a_{n - 1} - b_{n - 1})^{2} }{ 8 } \right] \mathbb{E}_{Z_{1}, \dots, Z_{n - 2}}\ \left[ \prod_{i = 1}^{n - 2} e^{s z_{i}} \right]

and do this iteratively to derive

\begin{aligned} \mathbb{E}_{Z_{1}, \dots, Z_{n}} \left[ \prod_{i = 1}^{n} e^{s z_{i}} \right] & = \exp \left[ \frac{ s^{2} (a_{n} - b_{n})^{2} }{ 8 } \right] \times \dots \times \exp \left[ \frac{ s^{2} (a_{1} - b_{1})^{2} }{ 8 } \right] \\ & = \exp \left[ \frac{ s^{2} \sum_{i = 1}^{n} (a_{n} - b_{n})^{2} }{ 8 } \right]. \end{aligned}

Therefore,

\begin{aligned} \mathbb{P}_{Z_{1}, \dots, Z_{n}} \left[ \sum_{i = 1}^{n} z_{i} \geq t \right] & \leq \inf_{s > 0} \frac{ \mathbb{E}_{Z_{1}, \dots, Z_{n}} \left[ e^{s \sum_{i = 1}^{n} z_{i}} \right] }{ e^{s t} } \\ & = \inf_{s > 0} \exp \left[ \frac{ s^{2} \sum_{i = 1}^{n} (a_{n} - b_{n})^{2} }{ 8 } - s t \right] \end{aligned}

Since \exp \left[ \frac{ s^{2} \sum_{i = 1}^{n} (a_{n} - b_{n})^{2} }{ 8 } - s t \right] is a convex function,

\begin{aligned} \frac{ d }{ d s } \exp \left[ \frac{ s^{2} \sum_{i = 1}^{n} (a_{n} - b_{n})^{2} }{ 8 } - s t \right] & = 0 \\ \left( \frac{ s \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} }{ 4 } - t \right) \exp\left[ \frac{ s^{2} \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} }{ 8 } - s t \right] & = 0 \\ \frac{ s \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} }{ 4 } - t & = 0 \\ s & = \frac{ 4 t }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } \end{aligned}

and therefore

\begin{aligned} \mathbb{P}_{Z_{1}, \dots, Z_{n}} \left[ \sum_{i = 1}^{n} z_{i} \geq t \right] & \leq \inf_{s > 0} \exp \left[ \frac{ s^{2} \sum_{i = 1}^{n} (a_{n} - b_{n})^{2} }{ 8 } - s t \right] \\ & = \exp \left[ \frac{ \left( \frac{ 4 t }{ \sum_{i}^{n} (b_{i} - a_{i})^{2} } \right)^{2} }{ 8 } \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} - \frac{ 4 t }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } t \right] \\ & = \exp \left[ \frac{ - 2 t^{2} }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } \right]. \end{aligned}

McDiarmid’s Inequality

Given a function f: \mathbb{R}^{n} \to \mathbb{R} with the bounded difference property, that is, the maximum change of the function output induced by replacing any x_{i} with x'_{i} is bounded by c_{i},

\lvert f (x_{1}, \dots, x_{i}, \dots, x_{n}) - f (x_{1}, \dots, x'_{i}, \dots, x_{n}) \rvert \leq c_{i},

then the function f of independent random variables X_{1}, \dots, X_{n} satisfies

\mathbb{P}_{X_{1}, \dots, X_{n}} \left[ f (X_{1}, \dots, X_{n}) - \mathbb{E}_{X_{1}, \dots, X_{n}} \left[ f (X_{1}, \dots, X_{n}) \right] \geq t \right] \leq \exp \left[ \frac{ - 2 t^{2} }{ \sum_{i=1}^{n} c_{i}^{2} } \right].

Let X_{1}, \dots, X_{n} be a sequence of independent random variable, and Y_{1}, \dots, Y_{n} be a Martingale sequence with respect to X_{1}, \dots, X_{n} where

Y_{i} = g (X_{1}, \dots, X_{i}) = \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, X_{n}) \mid X_{1}, \dots, X_{i} \right].

Note that we have

\begin{aligned} Y_{0} & = \mathbb{E}_{X_{1}, \dots, X_{n}} \left[ f (X_{1}, \dots, X_{n}) \right], \\ Y_{n} & = \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{n}} \left[ f (X_{1}, \dots, X_{n}) \mid X_{1}, \dots, X_{n} \right] = f (X_{1}, \dots, X_{n}). \end{aligned}

Let Z_{1}, \dots, Z_{n} be an independent random variable with Z_{i} = Y_{i} - Y_{i - 1}, which is proved to be a Martingale difference sequence.

Note that

\sum_{i = 1}^{n} Z_{i} = Y_{n} - Y_{0} = f (X_{1}, \dots, X_{n}) - \mathbb{E}_{X_{1}, \dots, X_{n}} \left[ X_{1}, \dots, X_{n} \right]

For each Z_{i}, its upper bound U_{i} and lower bound L_{i} can be written as

\begin{aligned} U_{i} & = \sup_{x_{i} \in X_{i}} \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, x_{i}, \dots X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] - \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i - 1}} \left[ f (X_{1}, \dots, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] \\ L_{i} & = \inf_{x_{i} \in X_{i}} \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, x_{i}, \dots, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] - \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i - 1}} \left[ f (X_{1}, \dots, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] \end{aligned}

The difference between U_{i} and L_{i} is less than c_{i} because of the bounded property property of f,

\begin{aligned} U_{i} - L_{i} & = \sup_{x_{i} \in X_{i}} \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, x_{i}, \dots X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] - \inf_{x_{i} \in X_{i}} \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, x_{i}, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] \\ & = \sup_{x_{i}, x'_{i} \in X_{i}} \left[ \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, x_{i}, \dots, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] - \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, x'_{i}, \dots, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] \right] \\ & = \sup_{x_{i}, x'_{i} \in X_{i}} \left[ \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ f (X_{1}, \dots, x_{i}, \dots, X_{n}) - f (X_{1}, \dots, x'_{i}, \dots, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] \right] \\ & = \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ \sup_{x_{i}, x'_{i} \in X_{i}} f (X_{1}, \dots, x_{i}, \dots, X_{n}) - f (X_{1}, \dots, x'_{i}, \dots, X_{n}) \mid X_{1}, \dots, X_{i - 1} \right] \\ & \leq \mathbb{E}_{X_{1}, \dots, X_{n} \mid X_{1}, \dots, X_{i}} \left[ c_{i} \mid X_{1}, \dots, X_{i - 1} \right] \\ & \leq c_{i}. \end{aligned}

Therefore, we have a Martingale difference sequence of bounded random variable Z_{1}, \dots, Z_{n}, where Z_{i} \in [a_{i}, b_{i}], \forall i and b_{i} - a_{i} \leq c_{i}. We can apply Azuma inequality to obtain McDiarmid’s inequality.

\begin{aligned} \mathbb{P}_{Z_{1}, \dots, Z_{n}} \left[ \sum_{i = 1}^{n} z_{i} \geq t \right] & = \mathbb{P}_{Z_{1}, \dots, Z_{n}} \left[ \mathbb{E}_{X_{1}, \dots, X_{n}} \left[ X_{1}, \dots, X_{n} \right] \right] \\ & \leq \exp \left[ \frac{ - 2 t^{2} }{ \sum_{i = 1}^{n} (b_{i} - a_{i})^{2} } \right] \\ & \leq \exp \left[ \frac{ - 2 t^{2} }{ \sum_{i = 1}^{n} c_{i}^{2} } \right]. \end{aligned}