Need help in correctly framing the join density function problem

Question

The problem:

$f_{XY}(x, y)= \begin{cases} \dfrac{2}{3} & \text{for $0 < x< 1, 0 < y< 3x $}\\ 0 & \text{otherwise}\\ \end{cases}$

$\text{Find } P(Y < X < 2Y)$

The solution I've worked out:

$P(Y < X < 2Y) = \int_{0}^{1} \int_{y}^{2y} \frac{2}{3} \, dx \, dy$

Step 1

$P(Y < X < 2Y) = \int_{0}^{1} \left[\dfrac{2x}{3}\right]_{y}^{2y} \, dy$

$P(Y < X < 2Y) = \int_{0}^{1} \dfrac{4y - 2y}{3} \, dy$

$P(Y < X < 2Y) = \int_{0}^{1} \dfrac{2y}{3} \, dy$

Step 2

$P(Y < X < 2Y) = \left[\dfrac{y^2}{3}\right]_{0}^{1}$

$P(Y < X < 2Y) = \dfrac{1}{3} - \dfrac{0}{3}$

Answer I got

$P(Y < X < 2Y) = \dfrac{1}{3}$

But it is wrong. The accepted answer is $0.15$ to $0.18$.

Can anyone help me understand what am I doing wrong? Am I even framing the problem correctly?

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$The integral: $$\int_0^1\int_y^{2y}(2/3)\d x\d y$$Is wrong. It is "$x$" that should run between $0$ and $1$, not $y$ (and perhaps you forgot to consider whether or not the restriction $0<y<3x$ should matter). Observe that your integral catches points where $2y$ is close to $2$, i.e. you're looking at $x$ values close to $2$ which is bad - $x$ should be restricted between $0$ and $1$.

We want $0<x<1$ and $0<y<x<2y$. There are going to be a few ways to get the area of this region, one of them being: $$\int_0^1\int_{x/2}^x(2/3)\d y\d x$$This comes out to be $1/6$, which is between $0.15$ and $0.18$ (although I don't know why you said that instead of saying $0.1666666\cdots$ or, even better, just saying $1/6$).

In my integral, note:

• "$x$" can be any value between $0$ and $1$
• $y$ is forced to be between $x/2$ and $x$
• Since $0<x/2$, we find $0<y$ and also $x/2<y$ iff. $x<2y$ (both inequalities are needed)
• We also have $y<x$

Overall we have $0<y<x<2y$ and $0<x<1$ iff. $(x,y)$ is in the region that my integral is integrating over, which is what we need. That was not true in your case.