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I’ve been given the joint density function: f$_X$$_,$$_Y$(x,y)=C when (X,Y) is uniform over [-1,1]$^2$. I’ve been tasked with finding P{|2X+Y|$\le$1} and P{X=Y} however I’m stuck in my question, I’ve deduced already that C=1/4 in my working, however i’m not too sure how I can apply this to find the probability tasked.

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    $\begingroup$ You want to integrate the density over the area represented by the event $|2X+Y|\le 1$ (or the event $X=Y$). You can then divide by the integrate the density over the whole area, but since with $C=\frac14$ this is $1$, it is not strictly necessary. $\endgroup$
    – Henry
    Feb 15, 2022 at 13:14
  • $\begingroup$ I'm not sure if I can understand much of this, what is 1? you can divide by the integrate the density over the whole area? This confused me more than it helped me. $\endgroup$
    – murpw2011
    Feb 15, 2022 at 14:15
  • $\begingroup$ Just use the first sentence as a suggestion, if the second is confusing - it was more of a comment. $\endgroup$
    – Henry
    Feb 15, 2022 at 14:23
  • $\begingroup$ @Henry okay thanks for the help. $\endgroup$
    – murpw2011
    Feb 15, 2022 at 14:23
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    $\begingroup$ No, @A.M. The joint density function contains enough information on the dependency. $\endgroup$ Feb 16, 2022 at 2:42

2 Answers 2

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You have $f_{X,Y}(x,y) = \frac{1}{4}$ for $(X,Y)\sim U([-1,1]\times [-1,1])$.

For both the probabilities you want to find you need to integrate twice in the area that you’re given. For the first one you need to integrate where $|2X-Y|\leq 1$ therefore you need to integrate in the area $$D=\bigg\{(x,y): \frac{-1-y}{2}\leq x \leq \frac{1-y}{2} \text{ and } y\in[-1,1]\bigg\}$$

So, integrate your $f_{X,Y}$ in that for $x$ between the above and for $y$ in $[-1,1]$.

For the second, I think an easily understandable approach is to find $\mathbb{P}[X=Y]=1-\mathbb{P}[X>Y]-\mathbb{P}[X<Y]$.

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You have a uniform distribution over a $2{\times}2$ square; specifically the $[-1,1]^2$ square. Probabilities of events within this space can be measured graphically; just compare the areas covered by the events.

Plot the lines $2X+Y=1$ and $2X+Y=-1$ within the square, and it becomes trivial to determine the probability.

The event of $\{\lvert 2X+Y\rvert\leq 1\}$ is a quadrilateral (formed of two right triangles).

The complement, $\{\lvert 2X+Y\rvert>1\}$, is two right triangles.

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