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I would very much appreciate a hint for the following problem

Let $\left(X_n\right)_{n=1}^\infty$ be a sequence of random variables s.t.: $$X_1 \sim U_{[0,1]}$$ and for all $n>1$: $$X_n \sim U_{[0,X_{n-1}]}.$$ Give a general expression for $f_{X_n}$ the pdf of $X_n$.

Thanks

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Hint: Letting $V_1,V_2,\dots$ be a sequence of iid random variables distributed uniformly on $[0,1]$, show that $X_n$ has the same distribution as $V_1\cdot V_2\cdot\ldots \cdot V_n$. Next, find the distribution of $\log X_n$, which is a sum of the iid variables $\log V_i$ (what distribution does $\log V_i$ have?). The $\log$ trick is useful since pdfs of sums are easier to find than pdfs of products. Finally, use a transformation to get the pdf of $X_n$ from that of $\log X_n$.

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  • $\begingroup$ Also, a hint for the pdf of $\log V_1+\dots+\log V_n$: compute it for $n=1,2,3\dots$ until you see a pattern, then prove it by induction. $\endgroup$ – Mike Earnest Jun 6 '18 at 17:39
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I do not guarantee that this hint will lead to results.

If $F_{n}$ denotes the CDF and $f_{n}$ the PDF of $X_{n}$ then for $x\in\left[0,1\right]$ we find:

$\begin{aligned}F_{n+1}\left(x\right) & =\int_{0}^{x}P\left(X_{n+1}\leq x\mid X_{n}=y\right)f_{n}\left(y\right)dy+\int_{x}^{1}P\left(X_{n+1}\leq x\mid X_{n}=y\right)f_{n}\left(y\right)dy\\ & =\int_{0}^{x}f_{n}\left(y\right)dy+\int_{x}^{1}\frac{x}{y}f_{n}\left(y\right)dy\\ & =F_{n}\left(x\right)+x\int_{x}^{1}\frac{f_{n}\left(y\right)}{y}dy \end{aligned} $

Differentiating both sides we find:

$$f_{n+1}\left(x\right)=f_{n}\left(x\right)+\int_{x}^{1}\frac{f_{n}\left(y\right)}{y}dy-x\frac{f_{n}\left(x\right)}{x}=\int_{x}^{1}\frac{f_{n}\left(y\right)}{y}dy$$

Further we can start with $f_1(x)=1_{[0,1]}(x)$.

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  • $\begingroup$ This was the sort of direction I was taking, but I could not find a justification for the first equality which seems intuitive (looks like a variation of the law of total probability) but wasn't proven in my class. $\endgroup$ – Evan Jun 6 '18 at 17:39
  • $\begingroup$ I think it leads to $f_{n+1}\left(x\right)=\frac{1}{n!}\left(-\ln x\right)^{n}$. Sorry if it is useless for you. $\endgroup$ – drhab Jun 6 '18 at 17:54
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If $[0\le x\le1]$ is the pdf for $x$, then the cdf for $x$ is $x\,[0\le x\le1]$. Thus, the cdf for $y=\log(x)$ is $e^y\,[y\le0]$, and therefore the pdf for $y$ is $e^y\,[y\le0]$. The pdf for the sum of $n$ values of $y$ is the $n$-fold convolution of the pdf $e^y\,[y\le0]$ with itself. The Fourier Transform of this $n$-fold convolution is the $n^\text{th}$ power of the Fourier Transform of the pdf $e^y\,[y\le0]$, which is $$ \int_{-\infty}^0 e^{-2\pi iyt}e^y\,\mathrm{d}y=\frac1{1-2\pi it}\tag1 $$ Thus, the pdf for the sum of $n$ values of $y$ is $$ \begin{align} \sigma_n(y) &=\int_{-\infty}^\infty\frac{e^{2\pi iyt}}{(1-2\pi it)^n}\,\mathrm{d}t\tag{2a}\\ &=\frac{e^y}{2\pi i}\int_{1-i\infty}^{1+i\infty}\frac{e^{-yz}}{z^n}\,\mathrm{d}z\tag{2b}\\ &=e^y\frac{(-y)^{n-1}}{(n-1)!}\,[y\le0]\tag{2c} \end{align} $$ Explanation:
$\text{(2a)}$: take the inverse Fourier Transform
$\text{(2b)}$: substitute $t=\frac{1-z}{2\pi i}$
$\text{(2c)}$: if $y\gt0$, close the contour on the right half-plane, missing the singularity at $z=0$
$\phantom{\text{(2c):}}$ if $y\le0$, close the contour on the left half-plane, enclosing the singularity at $z=0$

The cdf for the sum of $n$ values of $y$ is the integral of $(2)$ $$ \Sigma_n(y)=e^y\sum_{k=0}^{n-1}\frac{(-y)^k}{k!}\,[y\le0]\tag3 $$ The cdf for the product of $n$ values of $x=e^y$ is therefore $$ \Pi_n(x)=x\sum_{k=0}^{n-1}\frac{(-\log(x))^k}{k!}\,[0\le x\le1]\tag4 $$ The pdf for the product of $n$ values of $x$ is the derivative of $(4)$ $$ \bbox[5px,border:2px solid #C0A000]{\pi_n(x)=\frac{(-\log(x))^{n-1}}{(n-1)!}\,[0\le x\le1]}\tag5 $$ The pdf of $X_n$ is given by $(5)$.

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