Last time, after finishing up our discussion of data, we jumped into probability theory. Today, we're going to push our knowledge of probability theory further to talk about *random variables*. Today's material is covered in sections 3.4 and 4.3 of our textbook. (Yes, it jumps around a bit.)

As we go through this, we'll have occasion to review some of our material on probability as well.

*data*. While some of this material may seem a bit abstract, in reality these are the very tools that we use to model data. Thus, to understand the model is to understand data itself.

On page 116, the concept of a *random variable* $X$ is defined to be a random process with some numerical outcome.

- Example 0: Choose $X$ to be the sum of the scores of the next Ohio State - Michigan game
- Example 1: Flip a coin and write down
- $X=1$ if the coin lands heads or
- $X=0$ if the coin lands tails

- Example 2: Roll a standard six sided die and write down the number that comes up.
- Example 3: Roll a 10 sided die and write down
- $X=1$ if the roll comes up 1, 2, or 3,
- $X=2$ if the roll comes up 4, 5, 6, or 7, or
- $X=3$ if the roll comes up 8, 9, or 10

*discrete* random variables. In the language of probability theory that we developed last week, we would say that the sample space is the set of integers

There are certainly random variables that can, in principle, produce any real number.

- Example A: Let $X$ be the average number of points that Ohio State beats Michigan by until Michigan (someday, a long time from now) finally wins,
- Example B: Find somebody and choose $X$ to be a very precise measure of their height,
- Example C: Randomly choose a college and choose $X$ to be the average salary of all the professors.

These are all continuous random variables. In the language of probability theory that we developed last week, we would say that the sample space is the set of real numbers.

Continuous distributions are certainly very important and we will dive into them in more detail next time. Today, though, we will focus on discrete distributions.

Roughly, the *distribution* of a random variable tells you how likely that random variable is to produce certain outputs. For a discrete random variable, this boils down to listing out the probability that the random variable hits every specific value.

On page 87 of our text, the distribution of a discrete random variable is defined to be a table of all the possible outcomes together with their probabilities.

For example 3 above, we might write

X | 1 | 2 | 3 |
---|---|---|---|

p | 3/10 | 4/10 | 3/10 |

Note the all the probabilities should be non-negative and they should sum to one.

An alternative is to specify the values of a function $P$ for the possible values of $X$. For example, the previous table could be written:

- $P(X=1) = 3/10$
- $P(X=2) = 4/10$
- $P(X=3) = 3/10$

As we'll see next time, this notation will extend in a natural way to continuous distributions.

A final alternative is to represent a discrete distribution with a plot.

Pictorial representations of discrete distributions make particular sense when there are a large number of possibilities. The image below portrays a discrete distribution where the sample space is the set of integers between 1 and 100 and smaller numbers are more likely to be chosen than larger numbers.

If the sample space of a discrete distribution consists of $n$ consecutive integers, each of which has probability $1/n$, then the distribution is said to be *uniform*. For example, we might write:
$$
P(X=i) = \begin{cases} 0.01 & \text{if } 1 \leq i \leq 100 \\
0 & \mbox{otherwise.} \end{cases}
$$
This can be visualized like so:

The roll of a six-sided die or the flip of a coin can each be represented with a discrete uniform distribution.

Suppose we roll 2 six-sided die. We then form a random variable $X$ by adding the totals on the two die. We could get any number between 2 (snake eyes) and 12 (box cars) - but they are not all equally likely. According to the table on page 87 of our text, the probabilities are:

2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
---|---|---|---|---|---|---|---|---|---|---|

1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |

To understand this, note that, when we roll two die they are independent of one another. Thus the probability of getting any *particular pair* is
$$\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}.$$
Thus, the probability of getting any specific sum is the number of ways to get that sum times $/36$.

Here's an illustration of the number of ways to get certain sums of die rolls.

The binomial distribution is a discrete distribution that plays a special role in statistics for many reasons. Importantly for us, the binomial distribution allows us to see how a bell curve (in fact *the normal* curve) arises as a limit of other types of distributions.

Suppose we flip a coin 5 times and count how many heads we get. This will generate a random number $X$ between 0 and 5 but they are not all equally likely. The probabilities are:

- $P(X=0)=1/32$
- $P(X=1)=5/32$
- $P(X=2)=10/32$
- $P(X=3)=10/32$
- $P(X=4)=5/32$
- $P(X=5)=1/32$

Note that the probability of getting any *particular* sequence of 5 heads and tails is
$$\frac{1}{2^5} = \frac{1}{32}.$$
That explains the denominator of 32 in the list of probabilities.

The numerator in the table on the previous slide is the number of ways to get the sum. For example, there are 10 ways to get 2 heads in 5 flips:

The interactive tool below allows you to play with the parameters defining a binomial distribution.

The mean and standard deviation that we learned for data can be extended to random variables using the idea of a *weighted average*.

The expectation of a discrete random variable is $$ E(X) = \sum x_i P(X=x_i) = \sum x_i p_i.$$ We might think of this as a weighted mean.

Recall our weighted die roll with probability distribution

X | 1 | 2 | 3 |
---|---|---|---|

p | 3/10 | 4/10 | 3/10 |

The expected value of a roll of this die is $$1\frac{3}{10} + 2\frac{4}{10} + 3\frac{3}{10} = 2.$$

I've got a weighted coin that comes up heads $75\%$ of the time, in which case I write down a one. If it comes up tails, I write down a zero. The expectation associated with one flip is $$E(X) = 1\times \frac{3}{4} + 0 \times \frac{1}{4} = \frac{3}{4}.$$

The variance of a discrete random variable $X$ is $$\sigma^2(X) = \sum (x_i - \mu)^2 p_i.$$ We might think of this as a weighted average of the squared difference of the possible values from the mean. The standard deviation is the square root of the variance.

The variance of our weighted die roll in example above is $$(1-2)^2\frac{3}{10} + (2-2)^2\frac{4}{10} + (3-2)^2\frac{3}{10} = \frac{6}{10}.$$

The variance of our weighted coin flip is $$\sigma^2(X) = (1-3/4)^2\frac{3}{4} + (0-3/4)^2 \frac{1}{4}=\frac{3}{16}.$$

One nice thing about expectation and variance is that they are *additive*. That is, if
$X_1$ and $X_2$ are both random variables, then
$$E(X_1 + X_2) = E(X_1) + E(X_2)$$
and
$$\sigma^2(X_1 + X_2) = \sigma^2(X_1) + \sigma^2(X_2).$$

Suppose I flip my weighted coin that comes up heads 75\% of the time 100 times and let $X_i$ denote the value of my $i^{\text{th}}$ flip. Thus, $$X_1 + X_2 + \cdots X_{100}$$ represents the total number of heads that I get and, by the additivity of expectation, we get $$E(X_1 + X_2 + \cdots X_{100}) = 100 \times \frac{3}{4} = 75.$$

Similarly, for the variance we get $$\sigma^2(X_1 + X_2 + \cdots X_{100}) = 100 \times \frac{3}{16} = \frac{75}{4}.$$ Of course, this means that the standard deviation is $\sqrt{75/4}$.

*Note*: The standard deviation of one flip is $\sqrt{3/16} \approx 0.433013$ and the
standard deviation of 100 flips is $\sqrt{75/4} \approx 4.33013$. The second is 10 times
larger in magnitude but, *relative to the total number of flips* it's $0.0433013$, which
is tens times smaller.

Suppose that $X$ is a random variable with

$$P(X=1)=p \text{ and } P(X=0)=1-p.$$Then, $$E(X) = p\times1+(1-p)\times0 = p$$ and $$\sigma^2(X) = p(1-p)^2 + (1-p)(0-p)^2 = p(1-p).$$ The standard deviation is then $$\sigma(X) = \sqrt{p(1-p)}.$$ These are forumulae that we will definitley use!

Now suppose that $S$ represents the sum of $n$ independent runs of $X$. Then, since mean and variance are additive, we have

$$E(S) = np,$$$$\sigma^2(S) = np(1-p),$$and $$\sigma(S) = \sqrt{np(1-p)}.$$

Judging from our class survey, around $10.5%$ of the population has green eyes. Suppose we now draw a random sample of 150 people. Assuming our estimate to be a good one,

- How many people from the sample do we expect to have green eyes?
- What is the standard deviation associated with that process?

We expect about $150\times0.105 = 15.75$ people to have green eyes.

The standard deviation would be $$ \sigma \approx \sqrt{150\times0.105\times(1-0.105)} \approx 3.75. $$