# One Y Variable
Alex Cardazzi

All materials can be found at
<a href="https://alexcardazzi.github.io/econ311.html"
target="_blank">alexcardazzi.github.io</a>.

## Probability

So far, we have summarized data with its mean, variance, and overall
distribution. To use these ideas to say something about a whole
population from a sample, we first need some probability.

> [!TIP]
>
> ### Probability
>
> A value between 0 and 1 inclusive that represents the likelihood a
> particular event happens.

The probability that a coin is flipped and it lands on tails is
$\frac{1}{2}$ or $0.5$ or $50\%$. What about the probabilities and
outcomes of flipping two coins and counting the number of tails? Well,
since flipping this coin is a **random variable**, it has a
distribution. That distribution is defined by all possible outcomes and
the associated probabilities:

<div class="aside">

Percentages are not *really* probabilities, but they are often
associated with them.

</div>

| Number of Tails, $y$ | Probability, $P(y)$ |
|:--------------------:|:-------------------:|
|          0           |        0.25         |
|          1           |        0.50         |
|          2           |        0.25         |

Probability Distribution of Two Coin Flips

All distributions have a mean and a variance. So, how do we find the
mean and variance of a probability distribution?

<div class="aside">

In fact, not all distributions have defined means and/or variances, but
these types of distributions are outside the scope of this course.

</div>

<div class="columns">

<div class="column" width="50%">

$$
\mu = \sum_{i = 1}^g P(y_i) \times y_i
$$

</div>

<div class="column" width="50%">

$$
\sigma^2 = \sum_{i = 1}^g P(y_i) \times (y_i - \mu)^2
$$

</div>

</div>

<div class="aside">

These formulas should look a lot like formulas for weighted averages!

</div>

To practice, let’s use the previous distribution of the number of heads
from flipping two coins.

$$
\mu = \sum_{i = 1}^g P(y_i) \times y_i = (0.25\times0) + (0.5\times1) + (0.25\times2) = 1
$$

$$
\begin{aligned}\sigma^2 &= \sum_{i = 1}^g P(y_i) \times (y_i - \mu)^2 \\
&= (0.25\times(0-1)^2) + (0.5\times(1-1)^2) + (0.25\times(2-1)^2) \\
&= (0.25\times(-1)^2) + (0.5\times(0)^2) + (0.25\times(1)^2) \\
&= 0.25 + 0 + 0.25 = 0.5\end{aligned}
$$

Flipping coins is an example of a *discrete* distribution. In other
words, individual events have non-zero probabilities of occurring.
However, when distributions are *continuous*, the probability of any
individual event occurring is precisely zero. What? How can that be
true?

Consider the lifespan of a tire on a car. Suppose the average tire lasts
about 54,000 miles. However, what is the probability that your tire
lasts **exactly** 54,000 miles? Essentially 0%.

This is because your tire could last 54,000.1 miles, 54,000.01 miles,
and so on. There are an infinite number of miles that your tire could
last, so the probability of any specific mile being the one where your
tire goes flat is zero.

When dealing with continuous probabilities, instead of asking for the
$P(y) = M$ (e.g., the probability that your tire lasts exactly $M$
miles), people ask for the $P(y) < M$, $P(y) > m$, or $m < P(y) < M$
(e.g., the probability that your tire lasts more or less than some
number of miles, or the probability that your tire lasts between two
numbers of miles.)

In the interest of being concrete, suppose tires last anywhere between
40,000 and 68,000 miles. Suppose further that the distribution is
uniform, meaning that the probability of your tire going flat is the
same from 40,000 to 68,000 miles. A question from a statistics class
might be: “what is the probability that your tire lasts more than 60,000
miles?”

To answer this, you need to know what fraction of tires last above
60,000. Well, since all tires go flat between 40 and 68, we can say the
baseline is 68-40 = 28. Then, 68-60 is the range we’re interested in.
So, the probability would be $\frac{68-60}{68-40} = 0.286$.

## The Normal Distribution

The above math highlights how you would deal with probabilities and
continuous distributions. However, not all continuous distributions are
“uniform” like we assumed above. Perhaps the most famous and widely
studied distribution is **the normal distribution**.

The normal distribution:

- is bell-shaped with it’s peak in the center
- is symmetric, meaning the mean is the same as the median, and it
  asymptotically approaches the x-axis (meaning it gets close but never
  touches it!)
- is completely characterized by its mean and standard deviation, and is
  written like $N(\mu, \sigma)$

Below is an example of what a normal distribution looks like:

<details>

<summary>

Plot
</summary>

<img src="module03-01_md_files/figure-commonmark/unnamed-chunk-2-1.svg"
style="width:90.0%" data-fig-align="center"
data-fig-alt="Example of a normal distribution"
alt="Example of a Normal Distribution" />

</details>

Normal distributions take on different shapes depending on their means
and standard deviations.

- A increasing the mean, while holding the standard deviation constant,
  results in the distribution sliding horizontally to the right along
  the x-axis.
- An increase in the standard deviation, while holding the mean
  constant, results in a shorter, wider distribution.

Below are some examples of what these changes would do to a standard
normal distribution:

<details>

<summary>

Plot
</summary>

<img src="module03-01_md_files/figure-commonmark/unnamed-chunk-3-1.svg"
style="width:90.0%" data-fig-align="center"
data-fig-alt="Examples of multiple normal distributions with different means and variances."
alt="Example of Multiple Normal Distributions" />

</details>

## Normal Distribution Application

To learn the most incredible feature of the normal distribution, you’ll
have to wait for later in the module. However, while we wait, let’s
discuss how to compare two different normal distributions.

Two particularly famous distributions that are normally distributed are
scores on the SAT and ACT. The maximum score for each is 1600 and 36,
respectively. The SAT has an average score of 1050 with a standard
deviation of 200. The ACT has an average score of 19 with a standard
deviation of 6.

So, let’s say someone scores a 1,300 on the SAT and a 29 on the ACT.
Which test did they do better on?

Well, let’s first visualize the two distributions and the two scores.

<details>

<summary>

Plot
</summary>

<img src="module03-01_md_files/figure-commonmark/unnamed-chunk-4-1.svg"
style="width:90.0%" data-fig-align="center"
data-fig-alt="SAT and ACT distributions." />

</details>

Unfortunately, this visualization is not so helpful. Since the two
distributions have different means and standard deviations, we cannot
really compare them. A first step to getting a better comparison would
be to subtract off the mean from both distributions. Then, both
distributions will have a mean of zero.

<details>

<summary>

Plot
</summary>

<img src="module03-01_md_files/figure-commonmark/unnamed-chunk-5-1.svg"
style="width:90.0%" data-fig-align="center"
data-fig-alt="SAT and ACT distributions centered at 0." />

</details>

Still, this is not helpful. Now, the variances are distorting the
comparison. To fix this, we can divide each by their respective standard
deviations. This way, both distributions have a mean of zero and a
variance of 1.

<details>

<summary>

Plot
</summary>

<img src="module03-01_md_files/figure-commonmark/unnamed-chunk-6-1.svg"
style="width:90.0%" data-fig-align="center"
data-fig-alt="SAT and ACT distributions centered at zero and scaled such that their standard deviations are both 1." />

</details>

Now, it should be quite obvious which score was “better”. The SAT score
currently has a value of 1.25 while the ACT score has a value of
1.6666667.

What do these numbers mean, though? In words, this is **the number of
standard deviations away from the mean** the original score is. This
actually has a name, and it is a $z$-score. The formula for a $z$-score
is:

$$
z = \frac{y - \mu}{\sigma}
$$

$z$-scores come from what is called a **standard normal distribution**,
which is just a normal distribution with a mean of 0 and standard
deviation (and variance) of 1, or $N(0, 1)$. We know *a lot* about the
standard normal distribution. For example, we know the probability that
a random draw from a normal distribution is greater than some number,
less than some number, or between two numbers. In most statistics
courses, this is where you’d be shown [a $z$-score
table](https://cdn.numerade.com/ask_images/9015239223ae4e049700ac6d60da49c7.jpg).
Rather than spending time on this table, here are some takeaways:

- Since the distribution is symmetric, the probability of a random draw
  being larger (or smaller) than 0 is $\frac{1}{2}$.
- There is a 15.87% chance that a random draw is larger (smaller) than
  +1 (-1).
- There is a 2.28% chance that a random draw is larger (smaller) than +2
  (-2).
  - This means that there is a 95.44% chance that the random draw is
    between -2 and +2.

Let me add some words to all of this math. Remember, if we picked a
random element out of a vector, we would expect that element to be
$\sigma$ units away from the mean. What is the $z$-score doing? It is
scaling the distance from some value $y$ to the mean $\mu$ by a measure
of the average distance from the mean, $\sigma$. Therefore, values of
$z$ that are greater in absolute value than 1 indicate that $y$ is
farther than the average element, and values less then one indicate $y$
is closer than average. Being more than two times the average distance
from the mean, for example, is pretty rare – there is only a 2.28%
chance. A first important feature of $z$-scores is that they allow for
the comparison of values from different distributions (e.g. SAT and
ACT). A second important feature is that they allow us to calculate
probabilities for every normal distribution, regardless of the mean,
variance, and/or units.

## Normal Distributions in R

R has some pre-built functions to work with normal distributions.

- `pnorm()` gives $P(Z < y)$ when `lower.tail = TRUE`. When
  `lower.tail = FALSE`, this gives $P(Z > y)$
  - `pnorm(0, lower.tail = TRUE)`: 0.5
  - `pnorm(2, lower.tail = TRUE)`: 0.9772499.
  - `pnorm(2, lower.tail = FALSE)`: 0.0227501.
- `qnorm()` is the opposite of `pnorm()` in that it converts
  probabilities into $z$-scores.
  - `qnorm(0.9772499, lower.tail = TRUE)`: 2.0000006
- `rnorm()` generates random draws from a standard normal distribution
  with a given mean and standard deviation. The default mean is 0 and
  the default standard deviation is 1. In the next lecture, you will use
  a function called `sample()` that also generates random data, but by
  drawing from values you already have.