Let the continuous random variables X and Y have the joint density,
$
f(x, y)=
\begin{cases}
8xy,& \text{if } 0 < y<x<1\\
0, & \text{otherwise}
\end{cases}
$
Then find $P(Y\leq \frac{1}{2}, X \leq Y + \frac{1}{4})$
My approach:
$\int_0^{\frac{1}{2}}\int_y^{y+\frac{1}{4}}8xy ~dx~dy$
which leads me to $\frac{11}{96}$. My limits are correct in this case?
Best Answer
The region required is bounded by the blue,green,orange and red lines in the diagram below. As $y$ varies from $0$ to $1/2$, $x$ will vary from $y$ to $y+1/4$.
\begin{eqnarray*} \int_0^{\frac{1}{2}} \int_{y}^{y+\frac{1}{4}} 8xy dxdy =\int_0^{\frac{1}{2}} 4y \underbrace{\left[ x^2 \right]_{y}^{y+\frac{1}{4}}}_{\frac{1}{2}y+\frac{1}{16}} dy =\int_0^{\frac{1}{2}}(2y^2+ \frac{y}{4}) dy = \left[ \frac{2y^3}{3} +\frac{y^2}{8} \right]_0^{\frac{1}{2}} =\color{blue}{\frac{11}{96}}. \end{eqnarray*} So your solution is perfectly sound. $\ddot \smile$