We divide randomly interval $[0,1]$ into two parts, pick longer one, divide it and choose longer one. Let $X$ designate it's length. Find density.

Question

We divide randomly interval $[0,1]$ into two parts. We choose longer one and divide it randomly again. We pick longer one. Let $X$ be random variable designating length of final interval. Find its density and expected value.

EDIT

I am extremely thankful that community drew my attention to shameful inappropriateness of my question. I am giving you much needed context to understand my, otherwise ambiguous, topic. I tried to solve this problem by, first finding relevant information about random variable designating length of interval after first division. And, initially, I hoped that some kind of conditional formula could work. It didn't. I am inexperienced random variable user, so, honestly, I wanted rather to see solution , not hint.

Answer 1

The result of the first division is uniformly distributed over the interval $\left[\frac12,1\right]$. That is,

$$f_Y(y)=\begin{cases}2&\text{ if } \frac12\le y\le 1\\ 0&\text{ otherwise.}\end{cases}$$

Given that the result of the first division is $y$ then the distribution of the second division is uniform over $\left[\frac y2,y\right]$. That is the conditional probability of $X$ given $Y=y$ is

$$f_{X\mid Y=y}(x)=\begin{cases}\frac2y&\text{ if}&\frac y2\le x\le y\\ 0&\text{ otherwise.}\end{cases}$$

So, the joint distribution of the two divisions is supported by the gray area depicted below:

and

$$f_{X,Y}(y,x)=f_{X\mid Y=y}(y)f_Y(y)=4\begin{cases}\frac1y&\text{ if } \frac y2 \le x\le y\text{ and } \frac12\le y \le 1\\ 0&\text{ otherwise.}\end{cases}$$

From here, the pdf. of the second division is

$$f_X(x)=\int_{-\infty}^{\infty}f_{X,Y}(y,x)\ dy=$$ $$=4\begin{cases}\int_{\frac12}^{2x}\frac1y \ dy&\text{ if }&\frac14\le x\le \frac12\\ \int_x^1\frac1y\ dy&\text{ if }&\frac12\le x\le 1\\ 0&\text{otherwise} \end{cases}=$$ $$=4\begin{cases}\ln(4)+\ln(x)&\text{ if }&\frac14\le x\le \frac12\\ -\ln(x)&\text{ if }&\frac12\le x\le 1\\ 0&\text{otherwise.} \end{cases}$$

Here is the strange shape of this pdf.:

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As far as the expectation, we have to evaluate the following integral(s)

$$E[X]=\int_{-\infty}^{\infty}xf_X(x)\ dx=4\ln(4)\int_{\frac14}^{\frac12}x\ dx+4\int_{\frac14}^{\frac12}x\ln(x)\ dx-4\int_{\frac14}^1 x\ln(x)\ dx).$$

The key to this calculation is the simple fact that

$$\int x\ln(x)\ dx =\frac14x^2(2\ln(x)-1)+C.$$