On a stick $1$ meter long is casually marked a point $X \sim U[0,1]$. Let $X=x$, is also marked a second point $Y\sim U[x,1]$.
1) Find the density of $(X,Y)$ showing the domain.
$$\rightarrow \quad f_{XY}(x,y)=\frac{1}{1-x}\mathbb{I}_{[0,1]}(x)\mathbb{I}_{[x<y<1]}(y)$$
2) Say if $X$ and $Y$ are independent or not, and compute $\operatorname{Cov}(X,Y)$.
$$\rightarrow f_Y(y)=-\log(1-y)\mathbb{I}_{[0,1]}(y)\Rightarrow f_X(x)f_Y(y)\neq f_{XY}(x,y)\\
\Rightarrow X\text{ and }Y\text{ are not independent}$$
$$\rightarrow \operatorname{Cov}(X,Y)=-\frac{1}{6}$$
3) Now we assume to break the stick in the points $X$ and $Y$, and to form a triangle with the pieces that we have. Remembering that in a triangle the sum of the lengths of two sides must be greater than the length of the third side, what is the probability to form a triangle with the three pieces of the stick?
I'm stuck on point 3). How would you fix it?
Thanks in advance for any help.
Best Answer
If the sum of the lengths of two sides must be greater than the third side, that means that each side cannot be greater than $0.5$ so the probability is
$$\mathbb{P}[Y-X<\frac{1}{2};X<\frac{1}{2};Y>\frac{1}{2}]$$
Graphically:
In formula:
$$\int_0^{\frac{1}{2}} \frac{1}{1-x}dx\int_{\frac{1}{2}}^{x+\frac{1}{2}} dy=\frac{2ln2-1}{2}\approx 0.19$$