Finding the probability for the sum of 2 variables

Question

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.

Answer 1

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 $\mathbb{P}[X=Y]=1-\mathbb{P}[X>Y]-\mathbb{P}[X<Y]$.

Answer 2

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.