# Joint CDF of $f(x,y)$ [closed]

I would appreciate a little help here.

Let $X$ and $Y$ be continuous random variables with a joint pdf of the form $$f(x,y)=K(x+y)$$ $$0< x <y<1$$

Find $K$. I solved it. It turns out to be $2$.

Find the marginals. Again, I found those Find the joint CDF, $F(x,y)$. This is the part I am stuck on. I used the CDF definition but I still get answers that are different from the book.

Thank you

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## closed as off-topic by Did, Davide Giraudo, anomaly, graydad, Sami Ben RomdhaneMar 19 at 17:19

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What is K(x+y) 0 ? –  ronno Nov 4 '12 at 2:17
Sorry f(x,y)=k(x+y) for 0<x<y<1 –  JouJ FarrouJ Nov 4 '12 at 2:43

The CDF will come in several parts:

For $0 \le x \le y \le 1$ you have $$F(x,y) = \int_{a=0}^x \int_{b=a}^y 2(a+b) \,db \,da = x(y^2+xy-x^2).$$

For $0 \le x \le 1 \le y$ you have $F(x,y)=F(x,1) = x+x^2-x^3.$

For $0 \le y \le 1$ and $y \le x$ you have $F(x,y)=F(y,y) = y^3.$

For $x\ge 1$ and $y \ge 1$ you have $F(x,y)=F(1,1)=1$.

For $x\le 0$ you clearly have $F(x,y)=0$. Similarly for $y\le 0$ you have $F(x,y)=0$.

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The best way to does problem is usually by drawing a picture of region of support (it should look something like a triangle with hypotenuse as $x=y$ line) Once you have that down, We know $K$ must be a constant so this integrates out to 1 so we have $$\int_{0}^{1}\int_{0}^{y}K(x+y)dxdy=K\int_{0}^{1}\int_{0}^{y}(x+y)dxdy= K\int_{0}^{1}\left.\left(\frac{1}{2}x^{2}+xy\right)\right|_{x=0}^{y}dy=K\int_{0}^{1}\frac{3}{2}y^{2}dy=K\left.\left(\frac{1}{2}y^{3}\right)\right|_{y=0}^{1}=\frac{1}{2}K$$ Thus we want $$\frac{1}{2}K=1\rightarrow K=2$$

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