pdf of a member of a sequence of dependent random variables

Solution 1:

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 $\log X_n$.

Solution 2:

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 $f_1(x)=1_{[0,1]}(x)$.