Dividing the ruler into $3$ equal parts and throwing out the middle one

Question

John decided to understand the theory of probability and got ruler $100$ cm long. At the first stage, he said that the segment $[0,100]$ is equal to $1$. Then he divided the ruler into $3$ equal parts and threw out the middle one, and set the probability of $\frac{1}{2}$ to the other $2$ intervals. He continues to split remaining parts step by step, throwing out the middle ones, and for the remaining ones we estimate the probability to be half as small as at the previous step. What variance of a random variable will John get in the limit?

As far as I understand, a random variable, which is mentioned in the text is number that is written on the ruler.

We need to find $\operatorname{Var}[X] = E[X^2] - (E[X])^2$

$E[X]$ does not change during steps and equals $50$. However, calculating limit of $(E[X^2])$ seems tricky to me.

Answer 1

Here is a possible approach.

Suppose John has repeated the throw-out-the-middle third step many times. Let $X_1$ be a number randomly chosen from the remaining pieces of the first one-third of the ruler and let $X_2$ be a number randomly chosen from the remaining pieces of the last one-third of the ruler. Let $X = X_1$ with probability $\frac12$, $X = X_2$ with probability $\frac12.$

Assuming that $X_1$ and $X_2$ are distributed according to the description of what happens to the first and last third of the ruler in the question, $X$ is also distributed as described in the question.

The distribution of $X_1$ is approximately the same as the distribution of $\frac13 X$ (and exactly the same in the limit), while the distribution of $X_2$ is approximately the same as the distribution of $\frac13(200 + X).$

Glossing over the question of whether $E[X]$ and $E[X^2]$ exist, if we assume that they do then you can use these relationships to express $E[X_1],$ $E[X_1^2],$ $E[X_2],$ and $E[X_2^2]$ in terms of $E[X]$ and $E[X^2]$ using the factor $\frac13$ and you can express $E[X]$ and $E[X^2]$ in terms of the other expectations using the construction of $X$ from $X_1$ and $X_2.$

The expectations of the non-squared variables will work out to the obvious values. The interesting part will be the expectations of the squared variables.

Answer 2

$X$ is uniformly distributed on the set of numbers between $0$ and $1$ which only have $0$s and $2$s in their ternary expansion. In other words, $$ X=\sum_{n=1}^{+\infty}\frac{1}{3^n}X_n $$ where $(X_n)_{n\in\mathbb N^*}$ are i.i.d. and uniformly distributed on $\{0,2\}$. As $\mathbb E[X_n]=\operatorname{Var}(X_n)=1$ we deduce that $$ \mathbb E[X]=\sum_{n=1}^{+\infty}\frac{1}{3^n}=\frac12 $$ and $$ \operatorname{Var}(X)=\sum_{n=1}^{+\infty}\frac{1}{3^{2n}}=\frac18\cdot $$

Now if the unit is $100$cm then multiply $\mathbb E[X]$ by $100$ and $\operatorname{Var}(X)$ by $100^2$.

Answer 3

Variance of the Cantor Distribution

The question seems to be about an analog of the Cantor distribution, but supported on the interval $[0,100]$ rather than the usual $[0,1]$. The usual approach to computing the variance of this distribution is to invoke the Law of Total Variance: given "sufficiently nice" random variables $X$ and $Y$, $$\DeclareMathOperator{Var}{Var} \Var[X] = E\bigl[ \Var[X\mid Y] \bigr] + \Var\bigl[ E[X\mid Y]\bigr].\label{eq1}\tag{1} $$ Take $X$ to be the random variable of interest, i.e. the variable described in the question. Then define $Y$ in terms of $X$: $$ Y = \begin{cases} 0 & \text{if $X \in [0,\frac{1}{3}]$, and} \\ 1 & \text{if $X \in [\frac{2}{3}, 1]$.} \end{cases} $$ Essentially, $Y$ is the random variable which tells us if $X$ is in the left part of the Cantor set, or the right part. To compute $E[\Var[X\mid Y]]$, note that $$ \Var[X\mid Y=0] = \Var[\tfrac{1}{3} X] = \frac{1}{9} \Var[X], $$ since $X\mid Y=0$ is a scaled copy of $X$ (it is $X$, scaled by a factor of $1/3$). Similarly, $$ \Var[X\mid Y=1] = \Var[\tfrac{1}{3} X + \tfrac{2}{3}] = \frac{1}{9} \Var[X]. $$ Averaging these gives \begin{align} E[\Var[X\mid Y]] &= \frac{1}{2} \left( \Var[X\mid Y=0] + \Var[X\mid Y=1] \right)\\ &= \frac{1}{9} \Var[X]. \label{eq2}\tag{2} \end{align} For the $\Var\bigl[ E[X\mid Y]\bigr]$ term, observe that $E[X \mid Y=0]$ is the expected value of $X$ given that $X$ is in $[0,1/3]$—by symmetry, this is $1/6$. Similarly, $E[X\mid Y=1]$ is the expected value of $X$, given that $X$ is in $[2/3,1]$—again, using symmetry, this is $5/6$. This means that $E[X\mid Y]$ is a random variable with the distribution $$ E[X\mid Y] = \begin{cases} \frac{1}{6} & \text{with probability $1/2$, and} \\ \frac{5}{6} & \text{with probability $1/2$.} \end{cases} $$ Hence $$ \Var\bigl[ E[X\mid Y] \bigr] = \frac{1}{2}\left( \left(\frac{1}{6} - \frac{1}{2} \right)^2 + \left( \frac{5}{6} - \frac{1}{2} \right)^2 \right) = \frac{1}{9}. \label{eq3}\tag{3}$$ Plugging (\ref{eq2}) and (\ref{eq3}) into (\ref{eq1}) gives $$ \Var[X] = \frac{1}{9} \Var[X] + \frac{1}{9} \iff \frac{8}{9}\Var[X] = \frac{1}{9} \iff \Var[X] = \frac{1}{8}. $$

Variance of the Distribution Described

The distribution described in the question here is a scaled copy of the Cantor distribution. Specifically, the random variable here is $100$ times the Cantor random variable. Recall that for any real $a$ and $b$, $$ \Var[aZ + b] = a^2 \Var[Z].$$ This fact was actually used above, but it is worth specifically calling out here. If $X$ is a "Cantor distributed" random variable, then the variable of interest is $100X$. Hence $$\Var[100X] = 100^2 \Var[X] = \frac{100^3}{8}. $$