Question

A joint density function of random variable $X$ and $Y$ is given by $f(x,y)=\begin{cases}kx & \text{for } 0<x<1, 0<y<1 \\ 0 & \text{otherwise}\end{cases}$ then $Cov(X,Y)$ is

• A. $-1/6$
• B. $1/6$
• C. $0$
• D. $2/3$

Hint:

Always check for factorization first! If $f(x,y) = f_X(x) f_Y(y)$, then $X$ and $Y$ are independent, and you can immediately conclude that $Cov(X,Y) = 0$ and the correlation coefficient $\rho = 0$ without performing any integration.

Answer 1

The Correct Option is C

We evaluate the covariance by checking for independence or calculating the product moment.

Step 1: \color{redFactorization of the Joint Density
The joint density is $f(x,y) = kx$ for $0 < x < 1$ and $0 < y < 1$. [cite: 4951, 4970]
Notice that the joint density can be written as a product of a function of $X$ and a function of $Y$:
$f(x,y) = g(x)h(y)$, where $g(x) = kx$ and $h(y) = 1$.
Additionally, the support of the distribution is a rectangle ($0 < x < 1$ and $0 < y < 1$).

Step 2: \color{redIdentify Independence
If the joint density function of two random variables can be factored into a function of $x$ and a function of $y$ over a rectangular support, then the random variables $X$ and $Y$ are independent.

Step 3: \color{redRelationship between Independence and Covariance
A fundamental property in statistics is that if two random variables are independent, their covariance is zero.
$Cov(X, Y) = E(XY) - E(X)E(Y) = 0$.

Step 4: \color{redConclusion
Since $X$ and $Y$ are independent in this model, $Cov(X, Y) = 0$.
Thus, the answer is 0.