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The question is devoted to some basic probability theory.

Imagine the following procedure: we throw a point $X_1$ into the segment $[0,1]$ (we mean that $X_1$ is uniformly distributed). After that we choose the longer of two parts $[0,X_1]$ or $[X_1,1]$ and also "break" it randomly into two parts: thus, $X_2$ is distributed uniformly in $[0,X_1]$ or $[X_1,1]$.

How can I get the distribution of $X_2$? Can I use the formula $$f_{X_2}(x)=\int\limits_{\mathbb{R}}f_{X_2|X_1}(x\,|\,y)\cdot f_{X_1}(y)\,dy,\,\,\,$$ where $f_{X_2|X_1}(x\,|\,y)$ is the conditional pdf of $X_2$ given $X_1=y$? If so, why?

Thanks in advance!

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  • $\begingroup$ Is it obvious, or it is derived from the formula, that I mentioned above? $\endgroup$ – Mikitka Dec 1 '15 at 15:58
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Let $X_1\sim U(0,1)$, then $X_2$ has conditional density $$f_{X_2\mid X_1}(x_2\mid x_1) = \frac1{1-x_1}\mathsf 1_{\left(0,\frac12\right)}(x_1)\mathsf 1_{(x_1,1)}(x_2) + \frac1{x_1}\mathsf 1_{\left(\frac12,1\right)}(x_1)\mathsf 1_{(0,x_1)}(x_2). $$ Note that $$\mathsf 1_{\left(0,\frac12\right)}(x_1)\mathsf 1_{(x_1,1)}(x_2)= \mathsf 1_{\left(0,\frac12\right)}(x_2)\mathsf 1_{(0,x_2)}(x_1) + \mathsf 1_{\left(\frac12,1\right)}(x_2)\mathsf 1_{\left(0,\frac12\right)}(x_1)$$ and $$\mathsf 1_{\left(\frac12,1\right)}(x_1)\mathsf 1_{(0,x_1)}(x_2) = \mathsf 1_{\left(0,\frac12\right)}(x_2)\mathsf 1_{(x_2,1)}(x_1) + \mathsf 1_{\left(\frac12,1\right)}(x_2)\mathsf 1_{\left(\frac12,1\right)}(x_1), $$ so \begin{align} f_{X_2\mid X_1}(x_2\mid x_1) =& \left(\frac1{1-x_1}\mathsf 1_{(0,x_2)}(x_1) + \frac1{x_1}\mathsf 1_{\left(\frac12,1\right)}(x_1) \right)\mathsf 1_{\left(0,\frac12\right)}(x_2)\\ +& \left(\frac1{1-x_1}\mathsf 1_{\left(0,\frac12\right)}(x_1) + \frac1{x_1}\mathsf 1_{(x_2,1)}(x_1) \right)\mathsf 1_{\left(\frac12,1\right)}(x_2). \end{align} The marginal density of $X_2$ is obtained by computing \begin{align} f_{X_2}(x_2) &= \int_{\mathbb R} f_{X_2\mid X_1}(x_1\mid x_2)f_{X_1}(x_1)\ \mathsf dx_1\\ &= \left[\int_{\mathbb R} \left(\frac1{1-x_1}\mathsf 1_{(0,x_2)}(x_1) + \frac1{x_1}\mathsf 1_{\left(\frac12,1\right)}(x_1) \right)\ \mathsf dx_1\right] \mathsf 1_{\left(0,\frac12\right)}(x_2)\\ &\quad+\left[\int_{\mathbb R}\left(\frac1{1-x_1}\mathsf 1_{\left(0,\frac12\right)}(x_1) + \frac1{x_1}\mathsf 1_{(x_2,1)}(x_1) \right)\ \mathsf dx_1\right]\mathsf 1_{\left(\frac12,1\right)}(x_2)\\ &=\left[\int_0^{x_2}\frac1{1-x_1}\ \mathsf dx_1+\int_{\frac12}^1\frac1{x_1}\ \mathsf dx_1 \right]\mathsf 1_{\left(0,\frac12\right)}(x_2)\\ &\quad+\left[\int_0^{\frac12}\frac1{1-x_1}\ \mathsf dx_1+\int_{x_2}^1\frac1{x_1}\ \mathsf dx_1 \right]\mathsf 1_{\left(\frac12,1\right)}(x_2)\\ &= \log\left(\frac2{1-x_2}\right)\mathsf 1_{\left(0,\frac12\right)}(x_2) + \log\left(\frac2{x_2}\right)\mathsf 1_{\left(\frac12,1\right)}(x_2). \end{align}

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