0
$\begingroup$

A stick of length 1 is broken at an arbitrary point and the leftmost piece, say "m", is selected. Stick "m" is then broken at an arbitrary point and the leftmost piece "n" is selected. What is the probability density function for the length of "n"? What is the average length of "n"?

$\endgroup$
  • 1
    $\begingroup$ This is the same as the product of two uniform i.i.d. variables. does that help? $\endgroup$ – Callus Jan 24 '14 at 5:58
  • $\begingroup$ unfortunately no!! $\endgroup$ – Alireza Majzoobi Jan 24 '14 at 6:43
  • $\begingroup$ is it correct: f(x)=1/(m-m/2)=2/m for (m/2)<L<m and 0 for others? $\endgroup$ – Alireza Majzoobi Jan 24 '14 at 6:46
1
$\begingroup$

Let's use $M$ for the length of $m$, and $N$ for the length of $n$.

Given the length $M$, the conditional expected length of $N$ is clearly $E[N\mid M]=\frac{M}{2}$ so the overall expectation of this is clearly $E[N]=E\left[\frac{M}{2}\right]=\frac14$.

The density of $M$ is uniform on $[0,1]$, taking the value $1$. Given $M=x$, the density of $N$ is uniform, taking the value $\frac{1}{x}1$, so $\Pr[N\le y\mid M=x]=\frac{y}{x}$ for $0 \le y \le x$, and is $1$ when $x \le y$. So $$\Pr[N\le y] = \int_{x=y}^1 \frac{y}{x}\,dx +\int_{x=0}^y \,dx = y(1-\log_e y)$$ ant the density for $N$ is the derivative of this, namely $$-\log_e(y).$$

You can check the expected value, as $\displaystyle \int_0^1 -y \log_e(y) \,dy = \frac14$.

$\endgroup$
0
$\begingroup$

For $y \in (0,1)$

$P(n \le y)=P(n \le y|m>y).P(m>y)+P(n \le y|m\le y).P(m \le y)$

$P(n \le y|m\le y).P(m \le y)=1.P(m \le y)=P(m \le y)=y$

$P(n \le y|m>y).P(m>y)=\int_y^1P(n \le y|x>y)P(m=x)$

$=\int_y^1 \frac{y}{x}dx=-y \ln y$

$P(n \le y)=y(1-\ln y)$ $y \in (0,1)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.