Let (X, Y) have joint density f(x, y) = cx for 0 < x ≤ 1 and 0 ≤ y ≤ x, and 0 otherwise. Find c, both marginal densities, and decide whether X and Y are independent.

Step-by-step solution

1. 1

   Normalise over the triangular support. Integrating $y$ first is easiest because $x$ is constant in the inner integral:

   $\int_{0}^{1}\!\!\int_{0}^{x}cx\,dy\,dx=\int_{0}^{1}cx\cdot x\,dx=c\int_{0}^{1}x^{2}dx=\frac{c}{3}=1$

   $c=3$

2. 2

   Find the marginal density of X. Fix $x\in(0,1]$ and integrate out $y$ over the slice $0\le y\le x$:

   $f_X(x)=\int_{0}^{x}3x\,dy=3x\cdot x=3x^{2},\qquad 0<x\le 1$

3. 3

   Find the marginal density of Y. Fix $y\in[0,1]$; the support requires $x\ge y$, so $x$ ranges from $y$ to $1$:

   $f_Y(y)=\int_{y}^{1}3x\,dx=\left[\frac{3x^{2}}{2}\right]_{y}^{1}=\frac32\left(1-y^{2}\right),\qquad 0\le y\le 1$

   Getting the limits of this one right — $x$ from $y$ to $1$, not $0$ to $1$ — is the crux.

4. 4

   Test independence by multiplying the marginals.

   $f_X(x)f_Y(y)=3x^{2}\cdot\frac32\left(1-y^{2}\right)=\frac92x^{2}\left(1-y^{2}\right)$

   which is not equal to $f(x,y)=3x$.

5. 5

   Give the decisive structural reason. Even before comparing formulas, the support itself settles it: the region $0\le y\le x$ is not a rectangle, so the range of $Y$ depends on the value of $X$. Independence would require the support to be a product set.

   $\boxed{c=3;\ f_X(x)=3x^{2},\ f_Y(y)=\tfrac32\left(1-y^{2}\right);\ X\text{ and }Y\text{ are NOT independent}}$

6. 6

   Check that both marginals integrate to 1. $\int_{0}^{1}3x^{2}dx=1$ ✓ and $\int_{0}^{1}\tfrac32\left(1-y^{2}\right)dy=\tfrac32\left(1-\tfrac13\right)=1$ ✓, confirming the value of $c$ and both limit choices.