Example 7 in Lecture 3

Suppose two discrete random variables X and Y have the following joint probability mass function:

Joint Distribution
Y\X -10% 0% 15%
2% 0 0.4 0.1
4% 0.3 0 0.2

calculate the covariance between X and Y.

cov(X,Y)=?

(Round answer to 4 decimals)

Question

Example 7 in Lecture 3

Suppose two discrete random variables X and Y have the following joint probability mass function:

Joint Distribution
Y\X	-10%	0%	15%
2%	0	0.4	0.1
4%	0.3	0	0.2

calculate the covariance between X and Y.

cov(X,Y)=?

(Round answer to 4 decimals)

BlackTom AI · Accepted Answer

The problem presents a joint pmf for discrete X and Y arranged in a table, and asks for cov(X,Y).

First, extract the joint distribution details from the table. The columns correspond to X values of -10%, 0%, and 15%, and the rows correspond to Y values of 2% and 4%. The nonzero entries are:
- P(X = -10%, Y = 4%) = 0.3
- P(X = 0%, Y = 2%) = 0.4
- P(X = 15%, Y = 2%) = 0.1
- P(X = -10%, Y = 4%) = 0.3 (already listed above)
- P(X = 15%, Y = 4%) = 0.2
- P(X = 0%, Y = 2%) = 0.4 (already listed above)
- P(X = 15%, Y = 2%) = 0.1 (already listed above)
- P(X = -10%, Y = 2%) = 0
- P(X = 0%, Y = 4%) = 0

2) Compute the marginal distributions.
- P(X = -10%) = P(X = -10%, Y = 2%) + P(X = -10%, Y = 4%) = 0 + 0.3 = 0.3
- P(X = 0%) = P(X = 0%, Y = 2%) + P(X = 0%, Y = 4%) = 0.4 + 0 = 0.4
- P(X = 15%) = P(X = 15%, Y = 2%) + P(X = 15%, Y = 4%) = 0.1 + 0.2 = 0.3
- P(Y = 2%) = P(X = -10%, Y = 2%) + P(X = 0%, Y = 2%) + P(X = 15%, Y = 2%) = 0 + 0.4 + 0.1 = 0.5
- P(Y = 4%) = P(X = -10%, Y = 4%) + P(X = 0%, Y = 4%) + P(X = 15%, Y = 4%) = 0.3 + 0 + 0.2 = 0.5

3) Compute the expected values.
- E[X] = (-10%)(0.3) + (0%)(0.4) + (15%)(0.3) = (-3%) + 0% + (4.5%) = 1.5%
  Note: using decimal form, E[X] = (-0.10)(0.3) + (0)(0.4) + (0.15)(0.3) = -0.03 + 0 + 0.045 = 0.015 in decimal units.
- E[Y] = (2%)(0.5) + (4%)(0.5) = 1% + 2% = 3%
  In decimals: E[Y] = (0.02)(0.5) + (0.04)(0.5) = 0.01 + 0.02 = 0.03

4) Compute E[XY]. Use decimal units for X and Y (i.e., X ∈ {-0.10, 0, 0.15}, Y ∈ {0.02, 0.04}). Sum x*y*p for each cell:
- For (Y = 2%, X = -10%): x*y = (-0.10)(0.02) = -0.002, but the joint probability is 0, so contribution = 0.
- For (Y = 2%, X = 0%): x*y = (0)(0.02) = 0, contribution = 0.
- For (Y = 2%, X = 15%): x*y = (0.15)(0.02) = 0.003, contribution = 0.003 × P = 0.003 × 0.1 = 0.0003
- For (Y = 4%, X = -10%): x*y = (-0.10)(0.04) = -0.004, contribution = -0.004 × 0.3 = -0.0012
- For (Y = 4%, X = 0%): x*y = 0, contribution = 0
- For (Y = 4%, X = 15%): x*y = (0.15)(0.04) = 0.006, contribution = 0.006 × 0.2 = 0.0012

Sum of contributions: E[XY] = 0 + 0 + 0.0003 + (-0.0012) + 0 + 0.0012 = 0.0003.

5) Put it together with the covariance formula: cov(X,Y) = E[XY] - E[X]E[Y].
- E[X] = 0.015, E[Y] = 0.03, so E[X]E[Y] = 0.015 × 0.03 = 0.00045.
- cov(X,Y) = 0.0003 - 0.00045 = -0.00015 in decimal units.

6) If you report the covariance in the same decimal units as X and Y, it is -0.00015. If one were to express results in percentage terms for X and Y, the units would be different (and the numerical value would differ accordingly). For the purpose of rounding,  -0.00015 rounded to four decimals is -0.0002.

7) Quick sanity check: the magnitude is small and negative, consistent with a slight negative linear relationship indicated by the table entries (e.g., higher X values tend to occur with higher Y values but there is some counterbalancing mass). The computed E[X], E[Y], and E[XY] align with the provided joint probabilities and marginals to confirm the result.

BU.510.601.W5.FA24 Practice Quiz- Requires Respondus LockDown Browser

Example 7 in Lecture 3 Suppose two discrete random variables X and Y have the following joint probability mass function: Joint Distribution Y\X -10% 0% 15% 2% 0 0.4 0.1 4% 0.3 0 0.2 calculate the covariance between X and Y. cov(X,Y)=? (Round answer to 4 decimals)

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