Question

The joint density of random variable X and Y is $f_{XY}(x,y)=\begin{cases}2x & \text{for } 0<x<1, x<y<x+1 \\ 0 & \text{otherwise}\end{cases}$ then marginal of Y is

• A. $f_{Y}(y)=\begin{cases}y^2 & 0<y<1
  y(2-y) & 1<y<2
  0 & \text{otherwise}\end{cases}$
• B. $f_{Y}(y)=\begin{cases}1 & 0<y<1
  0 & \text{otherwise}\end{cases}$
• C. $f_{Y}(y)=\begin{cases}2y & 0<y<1
  0 & \text{otherwise}\end{cases}$
• D. $f_{Y}(y)=\begin{cases}3y^2 & 0<y<1
  0 & \text{otherwise}\end{cases}$

Hint:

When the bounds of one variable depend on another, always sketch the region or use the $\max/\min$ logic to split your integral into cases. Marginal densities in such cases are often piecewise functions.

Answer 1

The Correct Option is A

To find the marginal density $f_Y(y)$, we integrate the joint density $f(x,y)$ with respect to $x$. However, the limits for $x$ depend on the value of $y$.

Step 1: \color{redIdentify the Region of Integration
The bounds are $0 < x < 1$ and $x < y < x+1$.
Rewriting the bounds for $x$ in terms of $y$:
1. $y-1 < x < y$
2. $0 < x < 1$
Combining these, we get $\max(0, y-1) < x < \min(1, y)$.

Step 2: \color{redCase 1: $0 < y < 1$
In this range, $y-1$ is negative and $y$ is less than 1.
The limits for $x$ are $0 < x < y$.
$f_Y(y) = \int_{0}^{y} 2x dx = [x^2]_{0}^{y} = y^2$.

Step 3: \color{redCase 2: $1 < y < 2$
In this range, $y-1$ is positive and $y$ is greater than 1.
The limits for $x$ are $y-1 < x < 1$.
$f_Y(y) = \int_{y-1}^{1} 2x dx = [x^2]_{y-1}^{1} = 1^2 - (y-1)^2$
$f_Y(y) = 1 - (y^2 - 2y + 1) = 2y - y^2 = y(2-y)$.

Step 4: \color{redFinal Marginal Density
Combining the cases:
$f_Y(y) = y^2$ for $0 < y < 1$ and $y(2-y)$ for $1 < y < 2$.
This matches Option (1).