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Let $f$ a density function such that

$f(x,y)=\left\{ \begin{array}{lcc} 2& if & x+y \leq 1, x,y\geq0 \\ \\ 0 & \text{other case} \end{array} \right.$

Find $P(X < Y)$ and $P(X+Y<1/2)$

I try this:

$P(X<Y)=\int_0^\infty\int_0^x2\,dy\,dx=\infty$ but that is a bad result. I'm stuck here can someone help me?

I have the same problem with $P(X+Y<1/2)$

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  • $\begingroup$ Check the integration area. Why do you integrate from $0$ to $\infty$? It has to be from $0$ to $1$. I hope it will help you. $\endgroup$ – Mr.M Jul 13 '18 at 15:22
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Guide:

To be found are:$$\int\int[x<y]f(x,y)dydx$$where $[x<y]$ takes value $1$ if $x<y$ and takes value $0$ otherwise.

Secondly and similarly to be found is:$$\int\int[x+y<0.5]f(x,y)dydx$$ In your effort you seem to confuse the function $f(x,y)$ with the function prescribed by $(x,y)\mapsto 2$ if $x,y>0$ and $(x,y)\mapsto 0$ otherwise.

Observe that $f(x,y)$ only takes value $2$ on the triangle that you describe.

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  • $\begingroup$ I dont understand, the integration limit wrong, when i suppose $y$ goes 0 to $\infty$? $\endgroup$ – Bvss12 Jul 13 '18 at 15:31
  • $\begingroup$ You can take the limits $0$ and $\infty$ but must take into account that for $y$ large enough the function $f(x,y)$ takes value $0$ (not value $2$). $\endgroup$ – drhab Jul 13 '18 at 15:34
  • $\begingroup$ i take limits $0$ to $1$ and $0 to $x$, is good this? $\endgroup$ – Bvss12 Jul 13 '18 at 21:35
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Note that $(X,Y)$ is uniformly distributed on the right triangle $T$ in the first quadrant with coordinates $(0,0), (1,0), (0,1)$ (which has area $1/2$). In particular $$ P((X,Y)\in A)=\frac{m(A\cap T)}{m(T)}=2m(A\cap T) $$ where $m$ denotes Lebesgue measure (area) and $A\subset \mathbb{R^2}$. In particular put $A=\{(x,y)\in\mathbb{R^2}\mid x+y< 1/2 \}$. Then $$ P((X,Y)\in A)=P(X+Y\lt1/2)=2m(A\cap T)=2\times \frac{1}{2}\times 0.5\times0.5=\frac{1}{4} $$ since $A\cap T$ is a right triangle in the first quadrant with coordinates $(0,0), (1/2, 0), (0,1/2)$.

One can reason geometrically to determine $P(X\lt Y)$ as well.

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