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Given a stick of length $l$. We cut the stick twice. Let $X$ be the random variable defined by the length of the stick after the first cut, and $Y$ be the random variable defined by the length of the stick after the second cut. What is the probability for $Y>\frac{l}{4}$?

First, the PDF of $X$ is uniform, thus $f_{X}(x)=\frac{1}{l}$ (since $0\le X\le l$). $Y$ is uniform in $[0,X]$, then $f_{Y|X}(y|x)=\frac{1}{x}$. Then the probability of $Y>\frac{l}{4}$ is $$\int_{l/4}^{x}\frac{1}{x}dy=1-\frac{l}{4x}$$

On the other hand, we have $$f_{X,Y}=f_Xf_{Y|X}=\frac{1}{lx}$$ We write the probability of $Y>\frac{l}{4}$ as follow: $$P\left(\frac{l}{4}<Y<x|\frac{l}{4}\le X \le l\right)=\frac{P(l/4<Y<x,l/4\le X \le l)}{P(l/4\le X\le l)}=\frac{\int_{l/4}^{x}\int_{l/4}^{l} \frac{1}{lx}dxdy}{\frac{1}{l}}$$

which leads to a different result from $1-\frac{l}{4x}$.

My question is which of the solution is correct, and what's wrong with the incorrect solution?

Thanks in advance.

Tien Kha Pham
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    Once you start cutting, what is "the stick"? – Ted Shifrin May 23 '15 at 17:01
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    The second approach seems absurd - to begin with, assuming that $l=L$, what is $x$ and why $P\left(\frac{l}{4}<Y<x\mid \frac{l}{4}\le X \le L\right)$ is related to $P\left(Y>\frac{l}{4}\right)$? – Did May 23 '15 at 17:05
  • @TedShifrin I don't really understand what you mean. At the beginning, the stick is a straight piece of wood of length $l$... – Tien Kha Pham May 23 '15 at 17:06
  • @Did typo, I will fix it. I mean $P(l/4<Y<x|l/4\le X=x\le l)$. Since $Y>l/4$, $X$ must be also greater than $l/4$. – Tien Kha Pham May 23 '15 at 17:08
  • Still absurd. Please explain. – Did May 23 '15 at 17:11
  • @Did Assume $X=x$. If $x<l/4$, then the probability is $0$. Thus we only need to consider the case $x\ge l/4$. Then we need to find the probabilty for which $l/4<Y<x$, given $l/4\le x\le l$. That is a conditional probability, I think. – Tien Kha Pham May 23 '15 at 17:18
  • There will be two, and then three, pieces. Which of the three lengths are you talking about? – Ted Shifrin May 23 '15 at 17:22
  • @TedShifrin after each cut, you can throw away the piece on the right and just consider the length of the piece on the left (so at the beginning, lay the stick parallel to your face) – Tien Kha Pham May 23 '15 at 17:26

2 Answers2

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Both solutions are incorrect. Without loss of generality we might assume $l=1$. The first solution computes $P(Y > 1/4 | X = x)$ and is furthermore not valid if $x < 1/4$.

The second solution has a very confusing notation (the $x$ should be $X$?) but I think you try to compute $P(Y > 1/4 | X > 1/4)$, however the integration order is swapped. You could integrate with respect to x for the outer integral with limits $1/4$ to $1$ and y for the inner integral with limits from $1/4$ to $x$. Or you could use the order you have but then outer integral (for y) should have limits from $1/4$ to 1 and inner integral (for x) should have limits from $y$ to $1$.

The probability in the denominator should be $P(X > 1/4) = 3/4$ but is irrelevant since they didn't ask for a conditional probability.

dioid
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1

Let the random variables $U$ and $V$ be uniformly distributed in $[0,1]$. We have to compute the probability $P$ that $UV\geq{1\over4}$. This probability is equal to the area above the hyperbola $uv={1\over4}$ in the square $[0,1]^2$, and is given by $$P=\int_{1/4}^1\left(1-{1\over 4 u}\right)\>du=\left(u-{1\over4}\log u\right)\Biggr|_{1/4}^1={3\over4}-\log\sqrt{2}\doteq0.4034\ .$$

Christian Blatter
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