GATE CSE 2014 | Set 1 | Question: 2

Question

Suppose you break a stick of unit length at a point chosen uniformly at random. Then the expected length of the shorter stick is ________ .

Comments

Prove that S is uniformly distributed on (0,1/2)

Let the break point be \(X\).

Because it is chosen uniformly at random, we have:

\[

X \sim \text{Uniform}(0,1)

\]

The two pieces have lengths \(X\) and \(1-X\). The shorter length is:

\[

S = \min(X, 1-X)

\]

We can find the distribution of \(S\). For \(0 \le s \le 0.5\):

\[

P(S \le s) = P(X \le s) + P(X \ge 1-s) = s + s = 2s

\]

Reasoning of  $P(S \le s) = P(X \le s) + P(X \ge 1-s)$

 

Imagine you have a stick of length 1, and you want to know the probability that the shorter piece is $\le 0.2$. There are exactly two distinct zones on the stick that will make this happen:

1) The Left Zone ($X \le 0.2$): If you break the stick very close to the left edge (between $0$ and $0.2$), the left piece is the shorter one, and its length is automatically $\le 0.2$.

2) The Right Zone ($X \ge 0.8$): If you break the stick very close to the right edge (between $0.8$ and $1.0$), the right piece is the shorter one (of length $1-X$), and its length is also automatically $\le 0.2$.

 

Because the break point $X$ is perfectly uniform, the probability of landing in the left zone is its length ($s$), and the probability of landing in the right zone is also its length ($s$). Since these two events are mutually exclusive (you can't break the stick in two places at once), you add them together:

$$P(S \le s) = s + s = 2s$$

Thus, the CDF of \(S\) is \(F_S(s) = 2s\), and its PDF is:

\[

f_S(s) = \frac{d}{ds}(2s) = 2, \quad 0 \le s \le 0.5

\]

Reasoning : This derivative tells us that the PDF is a constant value ($2$) across its entire domain $[0, 0.5]$. A constant PDF mathematically proves beyond doubt that $S$ is uniformly distributed on the interval $(0, 0.5)$.

So the shorter piece \(S\) is uniformly distributed on \(\left(0, \frac12\right)\)

We have already calculated pdf using CDF concept 

We have :

\[

f_S(s) = 2, \quad 0 \le s \le 0.5

\]

Then:

\[

E[S] = \int_0^{0.5} s \cdot 2 \, ds

= 2 \cdot \left[\frac{s^2}{2}\right]_0^{0.5}

= \left[s^2\right]_0^{0.5}

= \frac14

\]

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Some More Variation Question based on this 

Sheldon Ross Question (Q1)

Q2) A stick of unit length is broken at a point chosen uniformly at random. What is the expected length of the longer piece?

Q3) A stick of unit length is broken at two points chosen independently and uniformly at random. What is the probability that the three resulting pieces can be joined to form a triangle?

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@legend_of_cse : 

Answer 1

The length of the shorter stick can be from $0$ to $0.5$ (because if it is greater than $0.5,$ it is no longer a shorter stick).

 This random variable $L$ (length of shorter stick) follows a uniform distribution, and hence probability density function of $L$ is $\dfrac{1}{0.5-0}= 2$ for all lengths in range $0$ to $0.5$

Now expected value of $L = \int_{0}^{0.5} L*p(L) dL = \int_{0}^{0.5} L*2 dL = 2*\left[\dfrac{L^2}{2}\right]^{0.5}_0 = 0.25$

Comments

Both scenarios are identical so expectation will not change.

if you consider scenario 1 then you get 0.25, if you consider scenario 2 then you get 0.25 .

expecation = (scenario1 + scenario 2 )/2 = (0.25 +0.25) /2 = 0.25

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I think b/w 0-0.5

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Length of shorter stick will never be greater than 0.5

Answer 2

Answer = $0.25$

Let $X$ = Length of Shorter Stick

$\implies 0<X<\frac{1}{2}\implies X$ follows Uniform Distribution over $(0,\frac{1}{2})$

We know, In uniform distribution, the mean (first moment) of the distribution is:

${\displaystyle E(X)={\frac {1}{2}}(b+a).}$   Link : Continuous uniform distribution - Wikipedia

Hence, the expected length of the shorter stick is = $Mean$ = $E[X] = \frac{b+a}{2} = \frac{0+\frac{1}{2}}{2} = \frac{1}{4} = 0.25$

Comments

Beautiful
Short & simple approach

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crazy soln! wow

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smart work

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Let $X =\text{Lenght of shortest stick}$ Then how do we know that distribution of $X$ will be uniform, i mean somehow you have prove that either PDF of $X$ is constant over an interval (or) the cdf is linear right.

Answer 3

Expectation in case of a Uniform Random Variable = $\frac{b-a}{2}$

https://www.ucd.ie/msc/t4media/Uniform%20Distribution.pdf

As shorter length stick is specified in question, take a=0 and b=0.5

$\frac{0.5}{2}$ = ¼ = 0.25

Comments

@ashley they have given it wrong, right one is a+b/2 not b-a/2 you can try to calculate what they have they did a mistake in calcuation so be careful

Answer 4

Firstly lets calculate probability density function of chosing a point on a unit length stick -- the point has a constant probability between 0 and 1 (uniform distribution) and 0 otherwise.

so

$$
\int_1^0 c \, dx
$$
 

which makes c=1

 

Now lets say i break a stick at x, then stick is divided into two parts of length x and 1-x.

but i want the shorter stick, upto 0 to 0.5 the shorter length is x and from 0.5 to 1 the shorter length is 1-x.

so integration of cx from 0 to 0.5 + c(1-x) from 0.5 to 1

$$
\int_0^{0.5} c x \, dx + \int_{0.5}^1 c(1 - x) \, dx
$$
 

substitute c=1, and this uses lotus (law of unconscious statistician) and ans = 1/4 or 0.25

Answer 5

Ans

Answer 6

The expected length is 0.25 (or 1/4).

Here is the step-by-step reasoning:

1. Define the Break Point: Let $X$ be the random variable representing the point where the stick is broken. Since the stick has unit length 1 and the point is chosen uniformly, $X$ is a uniform random variable on the interval $[0, 1]$.

2. Define the Stick Lengths: When the stick is broken at point $X$, the two pieces have lengths:

   • Piece 1: $X$

   • Piece 2: $1 - X$

3. Define the Shorter Stick: We are interested in the length of the shorter stick. Let's call this new random variable $L$.

    

   $$L = \min(X, 1 - X)$$

4. Calculate the Expected Value: The expected value of $L$ is found by integrating $L(x)$ over the range of $X$, which is from 0 to 1.

    

   $$E[L] = \int_{0}^{1} \min(x, 1-x) \,dx$$

5. Solve the Integral: To solve this, we must determine where $\min(x, 1-x)$ changes. The two terms are equal when $x = 1 - x$, which means $2x = 1$, or $x = 0.5$.

   • If $0 \le x \le 0.5$, the shorter piece is $x$.

   • If $0.5 \le x \le 1$, the shorter piece is $1-x$.

   We split the integral at this point:

    

   $$E[L] = \int_{0}^{0.5} x \,dx + \int_{0.5}^{1} (1-x) \,dx$$

   • First part:

      

     $$\int_{0}^{0.5} x \,dx = \left[ \frac{x^2}{2} \right]_{0}^{0.5} = \frac{(0.5)^2}{2} - 0 = \frac{0.25}{2} = 0.125$$

   • Second part:

      

     $$\int_{0.5}^{1} (1-x) \,dx = \left[ x - \frac{x^2}{2} \right]_{0.5}^{1} = \left(1 - \frac{1}{2}\right) - \left(0.5 - \frac{(0.5)^2}{2}\right) = 0.5 - (0.5 - 0.125) = 0.125$$

   • Total Expected Value:

      

     $$E[L] = 0.125 + 0.125 = 0.25$$