What is the probability to form a triangle with the three pieces of the stick?

Question

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?

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I'm stuck on point 3). How would you fix it?

Thanks in advance for any help.

Answer 1

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:

$$\int_0^{\frac{1}{2}} \frac{1}{1-x}dx\int_{\frac{1}{2}}^{x+\frac{1}{2}} dy=\frac{2ln2-1}{2}\approx 0.19$$

Answer 2

Let $S$ be the region in the $xy$-plane defined by the constraints $$ \left\lbrace \begin{align*} 0 < x < 1\\[4pt] x\le y < 1\\[4pt] \end{align*} \right. $$ Then the joint density function of the random variables $X,Y$ is given by $$ f(x,y)= \begin{cases} {\Large{\frac{1}{1-x}}}&\text{if}\;\,(x,y)\in S\\[4pt] 0&\text{otherwise}\\ \end{cases} $$ Then we get \begin{align*} E[X]&=\int_0^1\int_x^1 x\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{1}{2}\\[4pt] E[Y]&=\int_0^1\int_x^1 y\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{3}{4}\\[4pt] E[XY]&=\int_0^1\int_x^1 xy\,\Bigl(\frac{1}{1-x}\Bigr)\;dy\;dx=\frac{5}{12}\\[4pt] \end{align*} hence $X,Y$ are not independent since $$ E[X]{\,\cdot\,}E[Y] = \frac{1}{2}{\,\cdot\,}\frac{3}{4} = \frac{3}{8} \ne \frac{5}{12} = E[XY] $$ For the covariance, we get $$ \text{Cov}(X,Y) = \int_0^1\int_x^1 \left( \Bigl(x-\frac{1}{2}\Bigr) \Bigl(y-\frac{3}{4}\Bigr) \right) \! \Bigl(\frac{1}{1-x}\Bigr) \;dy\;dx = \frac{1}{24} $$ The potential triangle has side lengths $a,b,c$ where $$ \left\lbrace \begin{align*} a&=x\\[4pt] b&=y-x\\[4pt] c&=1-y\\[4pt] \end{align*} \right. $$ hence noting that $a+b+c=1$, the triangle inequalities are satisfied if and only if $0 < a,b,c < {\large{\frac{1}{2}}}$.

Computing $P\bigl(a \ge {\large{\frac{1}{2}}}\bigr)$, we get $$ P\Bigl(a\ge\frac{1}{2}\Bigr) = \int_{\large{\frac{1}{2}}}^1\int_x^1 \frac{1}{1-x}\;dy\;dx=\frac{1}{2} $$ Computing $P\bigl(b\ge{\large{\frac{1}{2}}}\bigr)$, we get $$ P\Bigl(b\ge\frac{1}{2}\Bigr) = \int_0^{\large{\frac{1}{2}}}\int_{{\large{x+{\large{\frac{1}{2}}}}}}^1 \frac{1}{1-x}\;dy\;dx = \frac{1}{2}-\frac{1}{2}\ln(2) $$ Computing $P\bigl(c\ge{\large{\frac{1}{2}}}\bigr)$, we get $$ P\Bigl(c\ge\frac{1}{2}\Bigr) = \int_0^{{\large{\frac{1}{2}}}} \int _ {\large{x}} ^ {{\large{\frac{1}{2}}}} \frac{1}{1-x}\;dy\;dx = \frac{1}{2}-\frac{1}{2}\ln(2) $$ In the region $S$, we have $0 < a,b,c < 1$, hence since $a+b+c=1$, at most one of $a,b,c$ can be at least ${\large{\frac{1}{2}}}$.

It follows that the probability that $a,b,c$ qualify as the side lengths of triangle is given by \begin{align*} & 1 - \left( P\Bigl(a\ge\frac{1}{2}\Bigr) + P\Bigl(b\ge\frac{1}{2}\Bigr) + P\Bigl(c\ge\frac{1}{2}\Bigr) \right) \\[4pt] =& 1 - \left( \left( \frac{1}{2} \right) + \left( \frac{1}{2}-\frac{1}{2}\ln(2) \right) + \left( \frac{1}{2}-\frac{1}{2}\ln(2) \right) \right) \\[4pt] =& -\frac{1}{2} + \ln(2) \approx .193 \\[4pt] \end{align*}