# Understanding the bounds of integration.

This is our homework question:

5. Let $$X$$ and $$Y$$ be random variables with joint pdf $$f(x,y) = \begin{cases} 1/4 &-1\le x,y \le 1 \\ 0 &\text{otherwise} \end{cases}$$ Find $$P(2X - Y \gt 0)$$.

The solution given to us is the following:

$$\displaystyle P(2X - Y \gt 0) = \int_{-1}^1 \int_{\frac{y}{2}}^1 \frac{1}{4} dxdy = \frac{1}{2}$$.

I don't quite get the bounds of integration used. I get that we'd do $$P(2X>Y)$$ leading to $$P(X>Y/2)$$. But why is upper bounded by 1? Don't the constraints mean it should go to infinity? Why is y bound between -1 and 1?

• I've replaced your images with $\LaTeX$ syntax, which you've probably seen before. I didn't remove the image URLs in case you wanted to reuse them or revise my Edit (or rollback). The last few statements are not quite perfect, but I wasn't confident in my ability to reword them for you, especially since these are the heart of your Question. Mar 18, 2021 at 21:45

Here, $$X \in [-1,1]$$ and $$Y \in [-1,1]$$ and we want the subset where $$2X-Y > 0$$. This region is shown in blue below.
Partitioning along the $$Y$$-axis (into horizontal strips), $$Y$$ ranges from $$-1$$ to $$1$$, the left bound is $$X = Y/2$$, and the right bound is $$X = 1$$.
• @pasha : What is $f(x,y)$ for $y> 1$? Is there any point in integrating all those zeroes? There is no benefit in extending the region of integration beyond where the integrand is nonzero. We're told in the definition of $f$ that the only interesting $(x,y)$ pairs are on $[-1,1]\times[-1,1]$. Mar 18, 2021 at 22:31