# Finding the probability for the sum of 2 variables [closed]

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.

• 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. Feb 15, 2022 at 13:14
• 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. Feb 15, 2022 at 14:15
• Just use the first sentence as a suggestion, if the second is confusing - it was more of a comment. Feb 15, 2022 at 14:23
• @Henry okay thanks for the help. Feb 15, 2022 at 14:23
• No, @A.M. The joint density function contains enough information on the dependency. Feb 16, 2022 at 2:42

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.

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.