# Discrete Random Variables

In this section, we are going to cover some important Discrete Random Variables. Note that we will be writing X ~ F to denote that X has distribution F. Some discrete random variables that we are going to cover are:

• The Point Mass Distribution
• The Discrete Uniform Distribution
• The Bernoulli Distribution
• The Binomial Distribution
• The Geometric Distribution
• The Poisson Distribution

### The Point Mass Distribution

X has a point mass distribution at a, written X ~ δa, if P(X = a) = 1 in which case The probability function is f(x) = 1 for x = a and 0 otherwise.

### The Discrete Uniform Distribution

Let k > 1 be a given integer, We say that X has a uniform distribution on {1, …., k} if X has probability mass function given by ### The Bernoulli Distribution

Let X represent a binary coin flip. Then P(X = 1) = p and P(X = 0) = 1 – p for some p ∈ [0, 1].
X has a Bernoulli distribution written X ~ Bernoulli(p) with probability function as

f(x) = px(1 – p)1 – x                  for x ∈ {0, 1}

Note that Bernoulli distribution has only two possible outcomes, either a success or a failure( You can consider head or success and tail as failure). The probability function can also be expressed as,
f(x) = p*x + (1 – p)(1 – x)

To get intuitions, lets put values of x and check the outcome.
For discrete P(X = 1) = fX(1) = p1(1 – p)0 = p
P(X = 0) = fX(0) = p0(1 – p)1 = 1 – p
We can see that we get the desired value P(X=1) = p and P(X = 0) = 1 -p

### The Binomial Distribution

Suppose we have a coin that falls head with a probability p. For n flips, let the random variable X determine the number of heads. Then, this X is considered to follow a binomial distribution.

X ~ Binomial(n, p) with mass function
f(x)  =  nCx px(1 – p)n-x          for x = 0, . . . , n
0                                 otherwise

Binomial distribution for n=1 is Bernoulli distribution. Bernoulli’s experiment when conducted repeatedly with replacement results in Binomial distribution.
Here n and p are parameters whereas x is a value of X.

Let us understand the probability mass function of the binomial distribution. As stated, consider X as the number of heads/success (outcome with probability p) for n flips/outcomes.
As we know for discrete distributions P(X = x) = fX(x)
Therefore we need to calculate the probability of x heads in total n outcomes. Since the events are independent, we simply need to multiply after choosing x outcomes from the total(n). This yields to the mathematical formula
nCx px(1 – p)n-x
x out of n (nCx) outcomes are chosen and the probability of them being head is derived (px). Rest of them (n – x) must be tail ((1 – p)n-x).

Here are Probability mass function and Cumulative distribution function for Binomial Distribution.

### The Geometric Distribution

X has a geometric distribution with parameter p ∈ (0, 1), written X ~ Geom(p), if
P(X = k) = p (1 – p)k – 1,        k ≥ 1

Think of X as the number of flips needed until the first head when flipping a coin. So if k flips are needed to get the first head, definitely the previous k – 1 flips are tail. Therefore, the probability of k – 1 tail is (1 – p)k – 1 along with last head p.

Notice the behavior of PMF according to the probability of head. For p = 0.2 i.e. low probability of getting head, the chance that we might need 1 toss to get head is less whereas that of p=0.8 is high but as we go further the nature of the curves changes. The probability of getting head on the fourth toss is higher for p=0.2 and lower for p=0.8. Note an interesting property that the chance of getting a head on second toss (and tail on first) is the same for p=0.2 and p=0.8. You can calculate the same from the formula.

### The Poisson Distribution

X has a Poisson distribution with parameter λ, written X ~ Poisson(λ) if Here λ is the event rate/rate parameter. Poisson is often used for count of rare events like traffic accidents and radioactive decay.

Let us suppose that the average number of goals in a FIFA match is 2.5. Therefore λ = 2.5 and according to Poisson distribution,
P( k goals in a match) = fX(k) = e λ * λx/x!