Probability that maximum and minimum of two segments is less than a given value

Question

I'm working on probability theory and recently I've got stuck with the following task:

Random point $A$ divides the segment $[0, 1]$ on two segments. Let $M$ be the size of biggest segment and $m$ be the size of least one. Calculate the probabilites $P(M \le x)$ and $P(m \le x)$ for every $x \in [0; 1]$.

This is my solution:

Suppose that $X$ is the length of the second segment(from $0$ to $A$). And $Y$ be the length of the second(from $A$ to $1$).

The first slippery moment: $P(X \le x) = x$ and $P(Y \le x) = 1-x$

By symmetry $P(X \le Y) = P(Y \le X)$

Than $P(m \le x) = P(X \le Y)P(X \le x)+ P(Y \le X)P(Y \le x) = \frac{1}{2}x+ \frac{1}{2}(1-x)=\frac{1}{2}$

$P(M \le x) = P(X \ge Y)P(X \le x)+ P(Y \ge X)P(Y \le x) = \frac{1}{2}x+ \frac{1}{2}(1-x)=\frac{1}{2}$

I'm very unsure about this solution. Am I right?

Answer 1

the two laws are both uniform:

$m \in [0;\frac{1}{2}]$

$M \in [\frac{1}{2};1]$

Thus the two CDF's are the following

$\mathbb{P}[M \leq x]=\frac{x-\frac{1}{2}}{\frac{1}{2}}$

$\mathbb{P}[m \leq x]=\frac{x}{\frac{1}{2}}$

Written in a better way, we have

$\mathbb{P}[M \leq x]=2(x-\frac{1}{2})\mathbb{1}_{[\frac{1}{2};1]}+\mathbb{1}_{[1;+\infty]}$

$\mathbb{P}[m \leq x]=2x \mathbb{1}_{[0;\frac{1}{2}]}+\mathbb{1}_{[\frac{1}{2};+\infty]}$

Answer 2

It appears to me that you've made an error: $$(m≤x)=P(X≤Y)P(X≤x)+P(Y≤X)P(Y≤x)$$ This is incorrect. Intuitively, why? First, it is correct that this probability is the summation of the two equiprobable cases $Y>=X$ and $Y<=X$ . But, when X<=Y, then we must multiply it by the CONDITIONAL probability $P(X<=x)$ given that $X<=0.5$. This information makes a difference. If $x=0.6$, then $P(X<=x)=1$ given that$ X<=Y $(or that X<=0.5). Your error is from the use of the absolute probability $P(X<=x)$ rather than the conditional one, which is $P(X<=x)_{|(X<=Y)}=x$ if $x<=0.5$, and 1 otherwise.

My solution would be that $M$ is uniformly distributed over $[0.5,1]$ and $m$ over $[0,0.5]$, and similar to the analysis above, I solve for $P(m<=x)$:

For a randomly selected $x$ in $[0,1]$, it is equivalent to being distributed over $[0,0.5]$ with a 0.5 chance, and over $[0.5,1]$ with a 0.5 chance. In the first case, $x$ has the same distribution as $m$, and hence $P(m<=x)=0.5$ for that case. In the second case, $x>=m$ with certainty, so $P(m<=x)=1$. As mentioned, these two cases are equiprobable, so we sum the conditional probabilities multiplied by 0.5 each: $$P(m<=x)=0.5*0.5+0.5*1=0.75$$

$P(M<=x)$ follows using the same technique. The critical difference between my approach and yours is that I used conditional probabilities whereas you neglected them in your definition, rendering your result invalid.

A hint that points to the invalidity of your result is that $m<M$ in all cases, so it doesn't make sense that they are have the same probability of being greater than x.