We have a stick of a unit length. Two points on the stick will be selected randomly (uniformly along the length of the stick) and independently. Then we break the stick at these two points so that we get three pieces of the stick. What is the probability that these three pieces form a triangle?

Solution.

Let us call the left end of the stick the origin. Let $X$ be the length from the origin to the first selected point. Let $Y$ be the length from the origin to the second selected point.

Note that when $X=Y$, we have only two pieces and they cannot form a triangle, so we assume $X \neq Y$.
We first assume that $X < Y$. Then after breaking the stick, we have three pieces of length $X$, $Y-X$, and $1-Y$, respectively.

Now, let us consider a condition so that three segments of length $a$, $b$, and $c$ form a triangle in general. It is necessary and sufficient that the sum of the lengths of any two segments is bigger than the length of the rest to form a triangle.
Namely, we must satisfy all of the following inequalities.
\begin{align*}
a + b &> c\\
b + c &> a\\
c + a &> b.
\end{align*}

Applying this condition to the current situation, we must have
\begin{align*}
X + (Y-X) &> 1-Y\\
(Y-X) + (1-Y) & > X\\
(1- Y) + X &> Y – X.
\end{align*}
Simplifying these, we obtain
\begin{align*}
Y &> \frac{1}{2}\\
X &< \frac{1}{2}\\ Y &< X + \frac{1}{2}. \end{align*} The region in the unit square satisfying these inequality is depicted in the figure below (orange region).

Note that the joint density function of $X$ and $Y$ are uniformly distributed on the unit square. Thus, the probability that three pieces form a triangle under the assumption $X < Y$ is given by the area of the orange triangular region, which is $1/8$. Note that by symmetry, the case $X > Y$ gives the same probability. Hence, the desired probability is
\begin{align*}
&P(\text{form a triangle}) \\
&= P(\text{form a triangle} \mid X < Y) + P(\text{form a triangle} \mid X > Y)\\
&\frac{1}{8}+\frac{1}{8} = \frac{1}{4}.
\end{align*}

Alternatively, we can find the probability for the case $X > Y$ as follows. By symmetry argument, we simply need to exchange $X$ and $Y$ and obtain the conditions
\begin{align*}
X &> \frac{1}{2}\\
Y &< \frac{1}{2}\\ X &< Y + \frac{1}{2}. \end{align*} The blue region below shows the area surrounded by these inequalities. The area of this blue region is $1/8$. Thus, the total area is $1/8 + 1/8 = 1/4$ as before.

The post Probability that Three Pieces Form a Triangle first appeared on Problems in Mathematics.

Let $X, Y$ be discrete random variables. Prove the linearity of expectations described as
\[E(X+Y) = E(X) + E(Y).\]

Solution.

The joint probability mass function of the discrete random variables $X$ and $Y$ is defined by
\[p(x, y) = P(X=x, Y=y).\] Note that the probability mass function of $X$ can be obtained from $p(x, y)$ by
\begin{align*}
p_X(x) &= P(X=x)\\
& = \sum_{y} p(x, y).
\end{align*}
Similarly, the probability mass function of $Y$ is expressed as
\[p_Y(y) = \sum_{x} p(x, y).\]

Using these equalities and the definition of expectations, we can compute $E(X+Y)$ as follows.

\begin{align*}
E(X+Y) &= \sum_{x} \sum_{y} (x+y) p(x, y)\\
&= \sum_{x} \sum_{y}\left(x p(x, y) + y p(x, y)\right)\\
&= \sum_{x} \sum_{y} x p(x, y) + \sum_{x} \sum_{y} y p(x, y)\\
&= \sum_x x \sum_y p(x,y) + \sum_y y \sum_x p(x,y)\\
&= \sum x p_X(x) + \sum_y y p_Y(y)\\
&= E(X) + E(Y)
\end{align*}

This proves the desired linearity of expectations $E(X+Y) = E(X) +E(Y)$.

The post Linearity of Expectations E(X+Y) = E(X) + E(Y) first appeared on Problems in Mathematics.