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enter image description here>The continuous variables X and Y have the following joint pdf $f(x,y) = x + y, 0<x,y<1.$

Determine $P(0.5<X+Y<1.5)$.

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 $P(0.5-Y<X<1.5-Y)$, 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<X<1.5-Y) = \int_{0.5-y}^{1.5-y} (x+0.5) \,dx $, 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 $P(0.5<X+Y<1.5)$ 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 $.

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

1 Answer 1

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I know that the support of x and y is rectangular, hence they are independent

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:

$$\int_0^{0.5} \int_{0.5-y}^1 f_{X,Y}(x,y)dxdy+\int_{0.5}^1\int_0^{1.5-y} f_{X,Y}(x,y)dxdy$$

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