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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?

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  • $\begingroup$ 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. $\endgroup$
    – hardmath
    Mar 18, 2021 at 21:45

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The usual short answer to these is "draw the picture".

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.

Mathematica graphics

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 $X = 1$.

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  • $\begingroup$ My question is how you get the ranges. I see the upper bound for y being 1, but why is the lower bound -1? Shouldn't it be -infinity? Same for x. Shouldn't the upper bound be infinity and not 1? $\endgroup$
    – pasha
    Mar 18, 2021 at 22:28
  • $\begingroup$ @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 $[-1,1]\times[-1,1]$. $\endgroup$ Mar 18, 2021 at 22:31
  • $\begingroup$ Wait I misunderstood the bounds. I thought that x and y were separate bounds (not that x and y were between the [-1,1] range. That's why I was integrating x to infinity and y to -infinity $\endgroup$
    – pasha
    Mar 18, 2021 at 23:00
  • $\begingroup$ @pasha : I've made that misinterpretation of notation too. I don't ever write inequalities that way, specifically to avoid the misinterpretation you (and I, in the past) have made. $\endgroup$ Mar 18, 2021 at 23:09

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