Q1: Calculate the covariance of the following random variables: Part A: X = 1A and Y = 1B for two indicator variables over sets A and B. For the special case of indicator functions, what is the relationship between correlated/uncorrelated and dependent/independent? Part B: X and Y, for a point (X,Y) uniformly distributed on a triangle with endpoints at (0, 0), (1, 0), and (0, 1).
The Correct Answer and Explanation is:
Q1: Covariance of Random Variables
Part A: Indicator Variables
Let X=1AX = \mathbf{1}_A, Y=1BY = \mathbf{1}_B be indicator random variables.
The covariance between XX and YY is: Cov(X,Y)=E[XY]−E[X]E[Y]\text{Cov}(X, Y) = \mathbb{E}[XY] – \mathbb{E}[X]\mathbb{E}[Y]
Since X=1AX = \mathbf{1}_A, Y=1BY = \mathbf{1}_B, and XY=1A∩BXY = \mathbf{1}_{A \cap B}, we get: E[X]=P(A),E[Y]=P(B),E[XY]=P(A∩B)\mathbb{E}[X] = \mathbb{P}(A), \quad \mathbb{E}[Y] = \mathbb{P}(B), \quad \mathbb{E}[XY] = \mathbb{P}(A \cap B) ⇒Cov(X,Y)=P(A∩B)−P(A)P(B)\Rightarrow \text{Cov}(X, Y) = \mathbb{P}(A \cap B) – \mathbb{P}(A)\mathbb{P}(B)
Special Case: Dependence vs. Correlation for Indicator Variables
- If XX and YY are independent, then P(A∩B)=P(A)P(B)⇒Cov(X,Y)=0\mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B) \Rightarrow \text{Cov}(X,Y) = 0: Uncorrelated.
- If Cov(X,Y)=0\text{Cov}(X, Y) = 0, this does not imply independence in general. However, for indicator variables, if AA and BB are independent events, then the corresponding indicators are uncorrelated.
So, for indicator functions:
- Independence ⇒ Uncorrelated
- Uncorrelated ⇏ Independence (unless under special distributional assumptions)
Part B: Uniform Distribution on a Triangle
Let (X,Y)(X, Y) be uniformly distributed on the triangle with vertices at (0,0),(1,0),(0,1)(0,0), (1,0), (0,1).
The support is the region: T={(x,y)∣x≥0,y≥0,x+y≤1}T = \{(x, y) \mid x \ge 0, y \ge 0, x + y \le 1\}
Step 1: Joint PDF
Area of triangle = 12\frac{1}{2}. So uniform PDF: f(x,y)=2for (x,y)∈Tf(x, y) = 2 \quad \text{for } (x, y) \in T
Step 2: Compute Covariance
We use: Cov(X,Y)=E[XY]−E[X]E[Y]\text{Cov}(X,Y) = \mathbb{E}[XY] – \mathbb{E}[X] \mathbb{E}[Y]
Compute expectations:
- E[X]=∫01∫01−xx⋅2 dy dx=13\mathbb{E}[X] = \int_0^1 \int_0^{1-x} x \cdot 2 \, dy\, dx = \frac{1}{3}
- E[Y]=∫01∫01−xy⋅2 dy dx=13\mathbb{E}[Y] = \int_0^1 \int_0^{1-x} y \cdot 2 \, dy\, dx = \frac{1}{3}
- E[XY]=∫01∫01−xxy⋅2 dy dx=124\mathbb{E}[XY] = \int_0^1 \int_0^{1-x} xy \cdot 2 \, dy\, dx = \frac{1}{24}
Cov(X,Y)=124−13⋅13=124−19=−572\text{Cov}(X, Y) = \frac{1}{24} – \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{24} – \frac{1}{9} = -\frac{5}{72}
Final Answers
- Part A:
Cov(X,Y)=P(A∩B)−P(A)P(B)\text{Cov}(X,Y) = \mathbb{P}(A \cap B) – \mathbb{P}(A)\mathbb{P}(B)
Independence ⇒ Uncorrelated, but not vice versa. - Part B:
Cov(X,Y)=−572\text{Cov}(X, Y) = -\frac{5}{72}
Explanation
In Part A, we examine the covariance of two indicator variables. These are binary random variables that take the value 1 if a condition or event occurs and 0 otherwise. The covariance formula simplifies nicely due to the binary nature of indicator variables: since the product XY=1XY = 1 only when both indicators are 1 (i.e., both events happen), the expected value E[XY]\mathbb{E}[XY] becomes the probability of both events occurring simultaneously: P(A∩B)\mathbb{P}(A \cap B). The covariance is then the difference between this joint probability and the product of their individual probabilities: Cov(X,Y)=P(A∩B)−P(A)P(B)\text{Cov}(X, Y) = \mathbb{P}(A \cap B) – \mathbb{P}(A)\mathbb{P}(B)
This formula highlights the connection between covariance and independence: if two indicator variables are independent, their covariance is zero. However, the converse isn’t always true—uncorrelated indicator variables aren’t necessarily independent unless additional assumptions hold.
In Part B, we consider a continuous bivariate distribution: a point uniformly distributed over a triangle. The triangle is bounded by the lines x=0x = 0, y=0y = 0, and x+y=1x + y = 1. The joint density function is constant over this region due to the uniformity. Calculating the covariance involves finding E[X]\mathbb{E}[X], E[Y]\mathbb{E}[Y], and E[XY]\mathbb{E}[XY], which are computed via double integrals over the triangular region. The results show that the covariance is negative, indicating a weak inverse relationship between XX and YY: as one increases, the other tends to decrease slightly due to the constraint x+y≤1x + y \le 1. This makes intuitive sense, as increasing one coordinate in the triangle reduces the possible range for the other.
Thus, covariance captures the directional relationship between random variables, whether they are discrete indicators or continuous variables under geometric constraints.
