# Breaking a unit stick uniformly into two pieces

I need help with the following question:
A unit stick is broken at a point chosen at random (uniformly), what is the expected value of the longer piece?

I tried this:
X - the point chosen, $$X\sim Uni[0,1]$$
Y - the longer piece.
if $$X>\frac12$$ then $$Y=X$$, else if $$X\le\frac12$$ then $$Y=1-X$$
I wanted to calculate $$P(Y\le y)$$ but I'm having trouble with that. $$P(Y\le y)=P(\max(X,1-X)\le y) = P(1-y\le X\le y)$$
(I wanted to calculate PDF, then find density and calculate the expected value)

Can give me a hint on how to procced from here?

I have seen some similar questions asked here but I could not understand the solutions suggested

• For $\frac12\le y\le1$, $P(Y\le y)=P(X\le y)-P(X\le 1-y)=y-(1-y)=2y-1$. Commented Sep 17, 2021 at 8:43

As you always take the greater piece it's the same as taking the expected value of a uniformly distributed random variable between $$0.5$$ and $$1$$. Why?

You can write your expected value as $$E[Y] = \frac 1 2 \cdot E[Y|X\geq 0.5] + \frac 1 2 \cdot E[Y|X \leq 0.5]$$ because $$X$$ is uniform, so $$P(X\geq 0.5) = P(X\leq 0.5) = \frac 1 2$$.

However, $$E[Y|X\geq 0.5]=E[Y|X\leq 0.5]$$, because each possibility for $$Y$$ on the LHS has a correspondence on the RHS (just the shorter and longer stick switch places). So we get $$E[Y] = E[Y|X\geq 0.5] = E[X|X\geq 0.5] = E[Z] = \frac 3 4$$ where $$Z$$ is a uniformly distributed random variable on $$[0.5,1]$$, because given that $$X\geq 0.5$$ we already know that $$X$$ is the biggest stick and uniformly distributed between $$0.5$$ and $$1$$.

• Why is it not correct this way?: $E(Y) = E(Y|X>\frac12)P(X>\frac12) + E(Y|X<\frac12)P(X<\frac12) = 0.5E(X) + 0.5E(1-X)$ Commented Sep 17, 2021 at 8:39
• The first equality is correct, but the second one not. It would be correct if you write $0.5E[X|X>1/2]+0.5 E[1-X|X<1/2]$, but without the conditional expectation this doesn't work. And $E[Y]=E[Y|X\geq 1/2]$ but $0.5=E[X]\neq E[X|X\geq 1/2]=0.75$. Commented Sep 17, 2021 at 8:41
• Got it, thanks! Commented Sep 17, 2021 at 8:48
• If my answer helped you, an upvote and/or accepting the answer would be nice :) Commented Sep 17, 2021 at 9:29

I will give you two solutions, one using the method you wanted and one that I think is a little bit better and quicker.

$$\mathbb{P}[Y \leq y ] = 1 - \mathbb{P}[Y > y]$$ ( a little sketch might help now but here is how the longer piece can be greater than $$y$$. Either the first cut is less than $$1-y$$ or it is greater than $$y$$. Both have probability $$1-y$$. Hence $$\mathbb{P}[Y \leq y ] = 1 - 2(1-y) = 2y - 1$$

Note that this is only valid for $$\frac{1}{2} \leq y \leq 1$$ as the larger piece is always greater than $$\frac{1}{2}$$.

Hence the CDF of $$Y$$ is $$F_Y(y)=$$ $$\begin{cases} 0 & y< \frac{1}{2} \\ 2y-1 & \frac{1}{2}\leq y\leq 1 \end{cases}$$

Hence, by differentiation, the PDF of $$Y$$ is $$f_Y(y)=$$ $$\begin{cases} 0 & y< \frac{1}{2} \\ 2 & \frac{1}{2}\leq y\leq 1 \end{cases}$$

And lets integrate this to yield: $$\mathbb{E}[Y] = \int_{\frac{1}{2}}^1 y\cdot2 dy = \int_{\frac{1}{2}}^1 2y dy = \frac{3}{4}$$

My method

Let us take advantage of symmetry it should be clear that given $$X<\frac{1}{2}$$ and given $$X>\frac{1}{2}$$ the longest side should be the same. And so as these two cases make up every possibility (and are disjoint) we have $$\mathbb{E}[Y] = \mathbb{E}[Y | X \geq \frac{1}{2}]$$. In the case that $$X \geq \frac{1}{2}$$ the longest piece is simply $$X$$ and the expected value of a uniform variable that is greater than $$\frac{1}{2}$$ is just $$\frac{\frac{1}{2} +1}{2} = \frac{3}{4}$$