Q: Probability of $P(x+y<1/2)$ It's a basic problem but I'm stuck at it practicing for an upcoming exam. I'm supposed to get the answer $\frac{1}{96}$ but i can only seem to muster $\frac{1}{48}$.
The problem is find $P(X+Y \le \frac{1}{2})$ with Probability Density Function $P_{XY} (x,y)=8xy$ for $0<y<x<1$. The two variables are not independent.
My failed solution is to evaluate
$\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}-x}8xy dydx$
Please tell what i'm doing wrong. Thanks

A: Since $x + y < (1/2)$, and $y > 0$, you can not have $x \geq (1/2).$
Further, consider the expression
$$\int_{0}^{1/2} ~\int_0^{1/2 - x} 8xy ~dydx. \tag1 $$
Given the constraint that $y < x$, the expression in (1) above has a flaw.  That is, it allows specific instances when $y \geq x.$
For example, if $x = (1/10)$, then the expression in (1) above allows $y$ to range from $0$ to $4/10$.  This violates the constraint that $y$ must be less than $x$.
If $(1/4) < x < (1/2)$ then you can have $0 < y < (1/2) - x.$
However, when $0 < x < (1/4)$, then you must have that $0 < y < x.$
So, the correction to the expression in (1) above is
$$\int_{0}^{1/4} ~\int_0^x 8xy ~dydx + \int_{1/4}^{1/2} ~\int_0^{1/2 - x} 8xy ~dydx. \tag2 $$.
Let $~\displaystyle I = \int_0^x 8xy ~dy, ~~J = \int_0^{1/2 - x} 8xy ~dy.$
Then,

*

*$\displaystyle ~I = 4xy^2 \bigg|_{y = 0}^{y=x} = 4x^3.$


*$\displaystyle ~J = 4xy^2 \bigg|_{y = 0}^{y=[(1/2) - x]} = x - 4x^2 + 4x^3.$
So, the expression in (2) above becomes
$$\int_{0}^{1/4} I ~dx + \int_{1/4}^{1/2} J ~dx$$
$$=\int_{0}^{1/4} 4x^3 ~dx + \int_{1/4}^{1/2} [x - 4x^2 + 4x^3] ~dx $$
$$= x^4 \bigg|_{x = 0}^{x=1/4} + 
\left[\frac{x^2}{2} - \frac{4}{3}x^3 + x^4\right] \bigg|_{x = 1/4}^{x=1/2} = \frac{1}{96}.$$

A: If $0 < x \leq 0.25$, then $x+y \leq 0.5$ by default since $y < x$. If $x > 0.25$, then the only way that $x + y \leq 0.5$ is if $y \leq 0.5 -x$ (also note that since $y > 0$ we require that $x < 0.5$). Put it all together and
$$ P(X+Y \leq 0.5) = \int_0^{0.25} \int_0^x 8xy dy dx + \int_{0.25}^{0.5} \int_0^{0.5-x} 8xy dy dx = \frac{1}{96}$$
