# Lesson 29 Covariance

## Theory

The covariance measures the relationship between two random variables.

**Definition 29.1 (Covariance)**Let \(X\) and \(Y\) be random variables. Then, the

**covariance**of \(X\) and \(Y\), symbolized \(\text{Cov}[X, Y]\) is defined as \[\begin{equation} \text{Cov}[X, Y] \overset{\text{def}}{=} E[(X - E[X])(Y - E[Y])]. \tag{29.1} \end{equation}\]

The *sign* of the covariance is most meaningful:

- If \(\text{Cov}[X, Y] > 0\), then \(X\) and \(Y\) tend to move together. When \(X\) is high, \(Y\) tends to also be high.
- If \(\text{Cov}[X, Y] < 0\), then \(X\) and \(Y\) tend to move in opposite directions. When \(X\) is high, \(Y\) tends to be low.
- If \(\text{Cov}[X, Y] = 0\), then \(X\) and \(Y\) do not consistently move together. This does not mean that they are independent, just that they do not consistently move together.

**Theorem 29.1 (Covariance-Variance Relationship)**Let \(X\) be a random variable. Then: \[\begin{equation} \text{Var}[X] = \text{Cov}[X, X]. \end{equation}\]

For calculations, it is often easier to use the following “shortcut formula” for the covariance.

**Theorem 29.2 (Shortcut Formula for Covariance)**The covariance can also be computed as: \[\begin{equation} \text{Cov}[X, Y] = E[XY] - E[X]E[Y]. \tag{29.2} \end{equation}\]

*Proof.*\[\begin{align*} \text{Cov}[X, Y] &= E[(X - E[X])(Y - E[Y])] & \text{(definition of covariance)} \\ &= E[XY - X E[Y] - E[X] Y + E[X]E[Y]] & \text{(expand expression inside expectation)}\\ &= E[XY] -E[X] E[Y] - E[X] E[Y] + E[X]E[Y] & \text{(linearity of expectation)} \\ &= E[XY] - E[X]E[Y] & \text{(simplify)} \end{align*}\]

Here is an example where we use the shortcut formula.

**Example 29.1 (Roulette Covariance) **Let’s calculate the covariance between the number of bets that Xavier wins, \(X\), and
the number of bets that Yolanda wins, \(Y\).

We calculated \(E[XY] \approx 4.11\) in Lessons 25 and 27. But if we did not already know this, we would have to calculate it (usually by 2D LOTUS).

Since \(X\) and \(Y\) are binomial, we also know their expected values are \(E[X] = 3\frac{18}{38}\) and \(E[Y] = 5\frac{18}{38}\).

Therefore, the covariance is \[ \text{Cov}[X, Y] = E[XY] - E[X]E[Y] = 4.11 - 3\frac{18}{38} \cdot 5\frac{18}{38} = .744. \]

This covariance is positive, which makes sense—since the more Xavier wins, the more Yolanda wins.**Theorem 29.3 (Independence Implies Zero Covariance)**If \(X\) and \(Y\) are independent, then \(\text{Cov}[X, Y] = 0\).

*Proof.*We use the shortcut formula (29.2) and Theorem 27.1. \[\begin{align*} \text{Cov}[X, Y] &= E[XY] - E[X]E[Y] \\ &= E[X]E[Y] - E[X]E[Y] \\ &= 0 \end{align*}\]

However, the converse is not true. It is possible for the covariance to be 0, even when the random variables are not independent. An example of such a distibution can be found in the Essential Practice below.

## Essential Practice

Suppose \(X\) and \(Y\) are random variables with joint p.m.f. \[ \begin{array}{rr|ccc} y & 1 & .3 & 0 & .3 \\ & 0 & 0 & .4 & 0 \\ \hline & & 0 & 1 & 2 \\ & & & x \\ \end{array}. \] Are \(X\) and \(Y\) independent? What is \(\text{Cov}[X, Y]\)?

Two tickets are drawn from a box with \(N_1\) \(\fbox{1}\)s and \(N_0\) \(\fbox{0}\)s. Let \(X\) be the number of \(\fbox{1}\)s on the first draw and \(Y\) be the number of \(\fbox{1}\)s on the second draw. (Note that \(X\) and \(Y\) can only be 0 or 1.)

- Calculate \(\text{Cov}[X, Y]\) when the draws are made with replacement.
- Calculate \(\text{Cov}[X, Y]\) when the draws are made without replacement.

(Hint: You worked out \(E[XY]\) in Lesson 25. Use it!)

At Diablo Canyon nuclear plant, radioactive particles hit a Geiger counter according to a Poisson process with a rate of 3.5 particles per second. Let \(X\) be the number of particles detected in the first 2 seconds. Let \(Y\) be the number of particles detected in the second after that (i.e., the 3rd second). Calculate \(\text{Cov}[X, Y]\).

## Additional Practice

Consider the following three scenarios:

- A fair coin is tossed 3 times. \(X\) is the number of heads and \(Y\) is the number of tails.
- A fair coin is tossed 4 times. \(X\) is the number of heads in the first 3 tosses, \(Y\) is the number of heads in the last 3 tosses.
- A fair coin is tossed 6 times. \(X\) is the number of heads in the first 3 tosses, \(Y\) is the number of heads in the last 3 tosses.

Calculate \(\text{Cov}[X, Y]\) for each of these three scenarios. Interpret the sign of the covariance.