Probability
Discrete Problems
Cryptography requires problems that are hard in the average case.
For our purposes, a probability space is a finite set \(\Omega = \{0,1\}^n\), and a function \(Pr:2^\Omega \rightarrow [0,1].\) such that \(Pr[F] = \Sigma_{x\in F} Pr[x]\) for all \(F \subseteq \Omega\).
(See the Wikipedia for the general definition of a probability space. Briefly, a probability space is a set \(S\), a \(\sigma\)algebra \(X\) on \(S\), and a measure \(P\) on \(X\) such that \(P(S) = 1\).)
Type 1 Random Variables
A Type 1 RV is a function \(X : \Omega \rightarrow S\), for some finite set \(S\).
A predicate is a function \(Q(x):S\rightarrow\{0,1\}\), i.e. it maps the set to a boolean. Define \(Pr[Q(x)] = \Sigma_{y \in \Omega, Q(X(y))}Pr[y]\).
Some more notation:

\(X \leftarrow \Omega\) defines a uniform random variable on \(\Omega\) (\(X \leftarrow \{0,1\}^n\)). Formally, \(Pr[A] = A/\Omega\) for any \(A \subseteq \Omega\), \(X\) is the identity function, thus \(Pr[X=a] = 1/\Omega\) for all \(a \in \Omega\).

\(A(x,r)\) denotes a randomized algorithm: \(x\) is the input, and \(r \in \{0,1\}^m = \Omega\) is the randomness.

\(X \leftarrow A(y)\) denotes a random variable \(X\) as the output of the algorithm \(A\) on input \(y\). Output of algorithm is a probability distribution.

For a predicate \(Q\), define \(Pr[Q(x)  x \leftarrow A(y)]\) as the probability that \(Q(x)\) holds for output of \(A\) on input \(y\).
Example: Let \(A(n)\) be a randomized algorithm that outputs an odd prime less than or equal to \(n\). Then
Type 2 Random Variables
A Type 2 RV is a function \(X : \Omega \rightarrow \mathbb{R}\).
We shall always use a uniform distribution on \(\Omega\), which means that for all \(r \in \mathbb{R}\),
Define
and
\(X, Y\) are independent if for all \(a,b\) we have
Large Deviation Bounds
Markoff’s Inequality:
Chebyshev’s Inequality:
Chernoff Bound: For \(i=1,...,n\), let \(X_i\) be independent random variables variables such that \(Pr[X_i = 1 ] = p\), \(Pr[X_i = 0] = 1p\), and define \(X = \Sigma^n_{i=1}X_i.\) Then