Previously… (1/2)

[PSDR] Definition 3.1

Let \(S\) be the sample space of an experiment. A random variable is a function from \(S\) to the real line. Random variables are usually denoted by a capital letter. Many times we will abbreviate the words random variable with rv.

Discrete versus Continuous Random Variables

Discrete Random Variable: A random variable that can take on only a finite or countably infinite set of outcomes. We can say that a discrete R.V. has distinct values that can be counted.
Example: Coin Flips, Dice Rolls, Outcomes of a Test (positive/negative), gender (M/F), etc.
Continuous Random Variable: A random variable that can take on any value along a continuum (but may be reported “discretely”)
Example: Height (really continuous, but we usually just report to the nearest inch/centimeter), temperatures, etc.

Previously… (2/2)

Discrete versus Continuous Probability Functions

Probability Functions = Probability Distributions Table, Graph, or Formula that describes values a random variable can take on, and its corresponding probability (discrete random variable) or density (continuous random variable).
We call a Probability Function as Probability Mass Function (PMF) for discrete random variables and Probability Density Function (PDF) for continuous random variables.

Discrete Random Variables and Probability Mass Functions

[PSDR] Definition 3.3

A discrete random variable is a random variable that takes integer values. A discrete random variable is characterized by its probability mass function (PMF). The PMF \(p\) of a random variable \(X\) is given by \[p(x) = P (X=x).\]

[PSDR] Theorem 3.1

Let \(p\) be the probability mass function of \(X\).

\(p(x) \ge 0 \text{ for all } x\).
\(\sum_x p(x) = 1\).

The Bernoulli Probability Mass Function

Let \(X\) be a Bernoulli random variable with PMF given as

\[P(X=x; p) = p^x (1-p)^{1-x}.\]

The Bernoulli PMF models the probability of a success-failure trial with \(p\) as the probability of success.

Applying the properties of the PMF:

\(P(X=x; p) = p^x (1-p)^{1-x} \ge 0 \text{ for all } x \text{ and } 0 < p < 1\).
\[\begin{align*} \sum_{x=0}^{1} P(X=x; p) = & \sum_{x=0}^{1} p^x (1-p)^{1-x} \\ = & p^0 (1-p)^{1-0} + p^1 (1-p)^{1-1} \\ = & (1-p) + p \\ = & 1 \end{align*}\]

The Binomial Probability Mass Function (1/2)

Let \(X\) be a Binomial random variable with PMF given as

\[P(X = x; n, p) = \binom{n}{x} p^x (1-p)^{(n-x)} \hspace{5px} \text{ for } x = 0,1,2,\cdots, n\] where \(\binom{n}{x} = \frac{n!}{x!(n-x)!}\).

The parameters of this function is \(p\) and \(n\). The binomial distribution assumes that \(p\) is fixed for all \(n\) trials.

\(p\) is the probability of success.
\(n\) is the number of Bernoulli trials.

The Binomial Probability Mass Function (2/2)

Applying the properties of the PMF:

\(P(X=x; p; n) = \binom{n}{x} p^x (1-p)^{(n-x)} \ge 0 \text{ for all } x \text{ and } 0 < p < 1 \text{ and } n \ge 1\).
\[\begin{align*} \sum_{x=0}^{n} P(X=x; p) = & \sum_{x=0}^{n} \binom{n}{x} p^x (1-p)^{(n-x)} \\ = & 1 \end{align*}\]

In your mini-assignment today, you are going to prove that the Binomial PMF satisfies the second property.

Independent Discrete Random Variables

Independent random variables and independent events share many similarities.

It’s important to keep in mind that two events \(A\) and \(B\) are independent if \(P(A,B) = P(A \cap B)=P(A)P(B)\). The definition for independent discrete random variables is as follows.

Take \(X\) and \(Y\), two discrete random variables.

If \[P(X=x,Y=y)=P(X=x)P(Y=y),\] then \(X\) and \(Y\) are said to be independent for any \(x\) and \(y\).

For \(n\) discrete random variables \(X_1,X_2,X_3,\cdots,X_n\). We say that \(X_1,X_2,X_3,\cdots,X_n\) are independent if \[P(X_1=x_1,X_2=x_2,\cdots,X_n=x_n)=P(X_1=x_1)P(X_2=x_2)\cdots P(X_n=x_n)\] for all \(x_1,x_2,\cdots,x_n\).

Example Problem 1

I toss a coin twice and define \(X\) to be the number of heads I observe. Then, I toss the coin two more times and define \(Y\) to be the number of heads that I observe this time.

Find \(P((X<2) \cap (Y>1))\).

Since \(X\) and \(Y\) are the result of different independent coin tosses, the two random variables \(X\) and \(Y\) are independent. Also, note that both random variables have the Bernoulli PMF. We can write

\[\begin{align*} P((X<2) \cap (Y>1)) = & P(X<2)P(Y>1) \text{ (because X and Y are independent) } \\ = & \left(P_X(0) + P_X(1)\right)P_Y(2) \\ = & \left(\frac{1}{4}+\frac{1}{2}\right)\frac{1}{4} \\ = & \frac{3}{16} \\ \end{align*}\]

The Sum of Two Independent Discrete Random Variables

This is also known as the convolution of two random variables.

The general formula for the distribution of the sum \(Z=X+Y\) of two independent discrete random variables is

\[P(Z=z)=\sum _{x \in S} P_X(x)P_Y(z-x)\]

Where \(P_X\) and \(P_Y\) the probability mass functions of discrete random variables \(X\) and \(Y\) respectively.

Example Problem 2

Suppose you have \(X\) and \(Y\) discrete random variable with Bernoulli PMFs with the same probability of success \(p\). What is the PMF of \(Z = X + Y\).

Again, the Bernoulli PMFs is given as

\[P_X(x) = p^x (1-p)^{1-x} \text{ and }\]

\[P_Y(x) = p^y (1-p)^{1-y}.\]

So, \[\begin{align*} P_Z(z) = & \sum _{x = 0}^{1} P_X(x)P_Y(z-x) \\ = & \sum _{x = 0}^{1} p^x (1-p)^{1-x} p^{z-x} (1-p)^{1-(z-x)} \\ \end{align*}\]

Discrete Random Variables