1
$\begingroup$

A stick of length $1$ is broken in a random place that is $U(0, 1)$. Let $X$ be the length of the longer piece and let $Y$ be the length of the shorter piece.

$a)$ Find the pdf and expected value of $X$.

$b)$ Find the pdf and expected value of $Y$.

$U \sim U(0,1)$

I got to the point where I have an equation for the cdf of $X$:

$$\frac{1}{2}[P(1-U \leq X \:\:|\:\: U < \frac{1}{2}) + P(U \leq X \:\:|\:\: U > \frac{1}{2})]$$

but I don't know how to get the pdf from that at all.

Any help on this question would be appreciated.

$\endgroup$
2
$\begingroup$

$U \sim U(0,1)$

Define $X = \text{max}\{U, 1-U\}$

so that

$$X = \begin{cases} 1-U, & 0 < U < 1/2 \\ U, & 1/2 \leq U < 1 \end{cases} $$

It follows that

$$X = \frac{1 + |2U - 1|}{2}$$


$$F_X(x) = P (X \leq x), \:\:\:\: 1/2 < x < 1$$

$$=P\left(\frac{1 + |2U - 1|}{2} \leq x \right)$$

$$=P(|2U - 1| \leq 2x-1)$$

$$=P(-2x+1 \leq 2U - 1 \leq 2x-1)$$

$$=P(1-x \leq U \leq x)$$

$$=P(1-x < U \leq x)$$

$$=F_U(x) - F_U(1-x)$$

Differentiating with respect to $x$:

$$f_X(x) = f_U(x) + f_U(1-x), \:\:\:\:\: \text{for}\:\:\: 1/2 < x < 1$$

It follows that

$$f_X(x) = \begin{cases} 2, & \text{if}\:\:\:1/2 < x < 1, \\ 0, & \text{otherwise} \end{cases} $$

and since we have for a uniform variable

$$f_Z(z) = \begin{cases} \frac{1}{b-a}, & \text{if}\:\:\:a < z < b, \\ 0, & \text{otherwise} \end{cases} $$

$$E[Z] = \frac{b+a}{2} $$

It follows that

$$E[X] = \frac{1 + 1/2}{2} = \frac{3}{4} $$

$\endgroup$
0
$\begingroup$

The CDF of $X$ is that for $u\in(1/2;1)$

$$\mathsf P(X\leq u) ~=~ \tfrac 12 \mathsf P(1-U\leq u\mid U\leq 1/2)+\tfrac 1 2\mathsf P(U\leq u\mid U> 1/2)$$

If $U$ is uniformly distributed on $(0;1)$, then on condition that $U\leq \tfrac 12$, $U$ is uniformly distributed on $(0;1/2)$.

$$\begin{align}\mathsf P(U\leq u\mid U\leq \tfrac 12) ~ = & ~ \dfrac{\mathsf P(U\leq u)\mathbf 1_{0\leq u\leq 1/2}}{\mathsf P(U\leq 1/2)}\\ ~ = & ~ 2u~\mathbf 1_{u\in(0;1/2)}\end{align}$$

And similarly

$$\begin{align}\mathsf P(U\leq u\mid U\geq \tfrac 12) ~ = & ~ \dfrac{\mathsf P(1/2\leq U\leq u)\mathbf 1_{1/2\leq u\leq 1}}{\mathsf P(U\geq 1/2)}\\ ~ = & ~ (2u-1)~\mathbf 1_{u\in(1/2;1)}\end{align}$$

Continue from here.


PS: To simplify maters:

$$\begin{align}f_X(x) ~=& ~ \frac{\operatorname d ~}{\operatorname d x}\mathsf P(X\leq x) \\ = & ~ \left\lvert\frac{\operatorname d ~}{\operatorname d x}\tfrac 12 \mathsf P(1-U\leq x\mid U\leq 1/2)\right\rvert+\left\lvert\tfrac 1 2\mathsf P(U\leq x\mid U> 1/2)\right\rvert \\ = & ~ (f_U(1-x)+f_U(x))~\mathbf 1_{x\in(1/2;1)} \\ = & ~ 2~\mathbf 1_{x\in(1/2;1)} \end{align}$$

$\endgroup$
  • $\begingroup$ Sorry I really don't understand a lot of the notation here, especially in the PS... $\endgroup$ – Sam N. Mar 15 '16 at 5:39
0
$\begingroup$

To try it another way:

Let $U$ measure the break point.   This is uniformly distributed over the length of the stick.$$U\sim\mathcal U(0~;1)$$

With probability $1/2$ the stick is broken close to one end ($0$) and with probability $1/2$ the stick is broken close to the other ($1$).

Either way, the break point will be uniformly distributed over whichever half it occurs on.   The conditional distributions are thus:

$$U\mid_{(U<1/2)}~\sim~\mathcal U(0~;1/2)$$ $$U\mid_{(U\geq 1/2)}~\sim~\mathcal U[1/2~;1)$$

When $U<1/2$ the length of the longer piece, $X$, equals $1-U$, otherwise it is $U$.   So:

$$X\mid_{(U<1/2)}~\sim~\mathcal U(1/2~;1)$$ $$X\mid_{(U\geq 1/2)}~\sim~\mathcal U[1/2~;1)$$

Since there is equal probability of the break point being on either half, then the length of the longer piece is uniformly distributed over the interval $1/2$ to $1$:

$$X~\sim~\mathcal U(1/2~;1)$$

Given this, you can find the CDF, pdf, and expected value of $X$.

Those for $Y$ are found similarly.

As a check, $\mathsf E(X)+\mathsf E(Y) = 1$ (why?).

$\endgroup$

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.