# Finding P(a< u(X,Y) <b) given a rectangular support

>The continuous variables X and Y have the following joint pdf $$f(x,y) = x + y, 0

Determine $$P(0.5.

I know that the support of x and y is rectangular, hence they are independent. I'm not sure if this point would be crucial to solving the problem.

I can rewrite the probability as $$P(0.5-Y, and I can calculate this given that I can calculate the marginal pdf of X, which is (X+0.5). So this means I solve $$P(0.5-Y, and I got 1.5-y as my answer. I'm not quite sure how to get a numerical answer for this.

I did think of calculating $$P(0.5 using the joint pdf but I wasn't too sure what the limits would be for x and y.

Edit: Based on comment, I guess I have to use the joint pdf to determine the probability. Now I have added an image, and the shaded part represents what needs to be calculated. I'm not too sure what the limits are, what I'm thinking of is $$\int_{0}^{1}\ \int_{0.5-y}^{1.5-y} (x+y) \,dx dy$$.

• You're right, I think you will need to use the joint PDF in order to evaluate $P(.5 <X+Y<1.5).$ Jun 3, 2020 at 5:18

This is not enough for independence. You'd need $$f_{X,Y}(x,y)=f_X(x)f_Y(y)$$ and $$x+y$$ doesn't factorise as such. $$X$$ and $$Y$$ are dependent. And, the calculations you tried to do with marginals are not correct.
You'll use the joint distribution as @BruceET commented. The integral should be divided into two summands because when you fix $$y$$ on the outer integral, the entry and exit points also differ based on $$y<0.5$$ or $$y\geq 0.5$$. So, it'll look like as follows:
$$\int_0^{0.5} \int_{0.5-y}^1 f_{X,Y}(x,y)dxdy+\int_{0.5}^1\int_0^{1.5-y} f_{X,Y}(x,y)dxdy$$