Consider a coin toss. The outcome of the coin toss is a random variable \(y\) that takes on two possible distinct outcomes: Heads or Tails.
An example to a Bernoulli experiment: Imagine you are bout to toss a coin 10 times in a row and you wonder how likely it is that you will end up getting Heads 5 times out of those 10 tosses.
This is a typical Bernoulli experiment as it consists of:
\(n=10\) Bernoulli trials that are independent of each other
We are interested in the likelihood of observing \(k=5\) successes (H) that occur.
Note that the probability that we get Heads for each coin toss is \(p=0.5\) (if it is a fair coin).
\[ f(y) = \left\{ \begin{array}{cc} p & \mathrm{if\ } y=H \\ 1-p & \mathrm{if\ } y=T \\ \end{array} \right. \]
In this case, the number of successes \(k\) in Bernoulli experiment follows a binomial distribution. We denote this as: \[ k \sim B(n,p) \]
If \(k \sim B(n,p)\), the probability of observing \(k\) successes in the experiment \(B(n,p)\) is given by the following function:
\[ P(k)= \left( \begin{array}{cc} n\\ k \end{array} \right) p^k (1-p)^k = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} \]
If a random variable \(x\) follows a normal distribution then its probability distribution function takes the following form:
\[ f(x)=\frac{1}{\sqrt{2 \pi \sigma^2}} e^{\frac{-(x-\mu)^2}{2 \sigma^2} } \]
A normal distribution is characterized bu its mean \(\mu\) and its standard deviation \(\sigma\) and is expressed by
\[ x \sim N(\mu,\sigma^2) \]
When we have \(\mu=0\) and \(\sigma=1\), it is usually referred to as the standard normal distribution.
Standard normal variabes are often denoted by \(Z\).
The standard normal PDF is denoted by \(\phi\) and the standard normal CDF is denoted by \(\phi\). Hence,
\[ \phi(c)=\Phi'(c), \;\; \Phi(c)=P(Z \leq c), \;\; Z \sim N(0,1) \]