Nested Uniform Random Variables

Expectation · Hard · Free problem

Define a chain of uniform random variables as follows: $X_1 \sim U(0,1)$, and for $k \geq 2$, $X_k \sim U(X_{k-1}, 1)$. So each variable is uniform on the interval from the previous value up to 1.

  1. Find the probability density function of $X_n$.
  1. Evaluate $\lim_{n \to \infty} E\!\left[\prod_{i=1}^{n} X_i\right]$.

Hints

  1. Try computing the PDF of $X_2$ by integrating out $X_1$, then look for a pattern involving powers of $-\ln(1-t)$.
  2. Substitute $u = -\ln(1-t)$ to transform $X_n$ into a familiar distribution. What family has density $u^{n-1} e^{-u} / (n-1)!$?
  3. For the product limit, analyze $\sum E[\ln X_k]$. Use the Gamma connection: if $-\ln(1-X_k) \sim \text{Gamma}(k,1)$, what is $E[e^{-G}]$ for $G \sim \text{Gamma}(k,1)$?

Worked Solution

How to Think About It: Each $X_k$ is uniform between the previous value $X_{k-1}$ and

$, so the chain drifts upward toward
$. For part (1), a slick substitution $Y_k = -\ln(1-X_k)$ turns the nested-uniform structure into a sum of independent exponentials -- a Gamma. For part (2), the product $\prod X_i$ is a tug-of-war: each factor approaches
$, but we keep multiplying more of them. The clean way to resolve it is not to chase $\sum E[\ln X_k]$ (that converges to $-\pi^2/6$, and Jensen makes it the wrong quantity anyway), but to set up a one-step recursion for the expected product and find its fixed point.

Quick Estimate: $E[X_1] = 1/2$. $E[X_1 X_2] = E[X_1\cdot\frac{X_1+1}{2}] = \frac12(\frac13+\frac12) = \frac{5}{12}\approx 0.417$. $E[X_1X_2X_3]=\frac{7}{18}\approx0.389$. The values are falling but clearly leveling off toward something near $0.37$, not toward $0$.

Approach: Part (1): iterate the conditional density and recognize a Gamma via $u=-\ln(1-t)$. Part (2): define $G_m(x)=E[\text{product of the next }m\text{ factors}\mid\text{current value }x]$, derive its integral recursion, and solve for the fixed point.

Formal Solution:

*Part (1): PDF of $X_n$.* The conditional density of $X_k$ given $X_{k-1}=x$ is $f(t\mid x)=\frac{1}{1-x}$ for $t\in(x,1)$. Starting from $f_{X_1}(t)=1$: $f_{X_2}(t)=\int_0^t \frac{1}{1-x}\,dx = -\ln(1-t),$ $f_{X_3}(t)=\int_0^t \frac{1}{1-x}\bigl(-\ln(1-x)\bigr)dx = \frac{(-\ln(1-t))^2}{2}.$ By induction, $\boxed{f_{X_n}(t)=\frac{(-\ln(1-t))^{\,n-1}}{(n-1)!}},\qquad t\in(0,1).$ Substituting $u=-\ln(1-t)$ gives the density of $U=-\ln(1-X_n)$ as $\frac{u^{n-1}}{(n-1)!}e^{-u}$, i.e. $U\sim\text{Gamma}(n,1)$. So $X_n = 1-e^{-U}$ with $U\sim\text{Gamma}(n,1)$; as $n\to\infty$, $U\to\infty$ and $X_n\to1$ almost surely.

*Part (2): Limit of $E[\prod_{i=1}^n X_i]$.* Let $P_n = E[\prod_{i=1}^n X_i]$. Define, for a chain started at a given value $x$, $G_m(x) = E\Bigl[\prod_{j=1}^{m} Y_j \;\Big|\; Y_1\sim U(x,1),\ Y_j\sim U(Y_{j-1},1)\Bigr], \qquad G_0(x)\equiv1.$ Conditioning on the first draw $Y_1\sim U(x,1)$, $G_m(x) = E_{Y_1}\bigl[Y_1\, G_{m-1}(Y_1)\bigr] = \frac{1}{1-x}\int_x^1 y\,G_{m-1}(y)\,dy.$ Because $X_1\sim U(0,1)$, we have $P_n = G_n(0)$. One checks $G_1(0)=\frac12$, $G_2(0)=\frac{5}{12}$, $G_3(0)=\frac{7}{18}$ -- matching the direct values.

The sequence $G_m(x)$ is decreasing in $m$ and bounded below by $0$, so $G(x)=\lim_{m\to\infty}G_m(x)$ exists and satisfies the fixed-point equation $(1-x)\,G(x) = \int_x^1 y\,G(y)\,dy.$ Differentiating both sides in $x$: $\;-G(x) + (1-x)G'(x) = -x\,G(x)$, which simplifies to $(1-x)G'(x) = (1-x)G(x)$, i.e. $G'(x) = G(x) \;\Longrightarrow\; G(x) = C\,e^{x}.$ To pin $C$, use the boundary at $x=1$: by L'Hopital, $\lim_{x\to1}\frac{1}{1-x}\int_x^1 yG(y)\,dy = x\,G(x)\big|_{x=1}=G(1)$, so the operator fixes the value at $x=1$; since every $G_m(1)=1$ (indeed $G_0(1)=1$), we get $G(1)=1$. Thus $Ce^{1}=1$, so $C=e^{-1}$ and $G(x)=e^{\,x-1}.$ One verifies directly that $(1-x)e^{x-1}=\int_x^1 y\,e^{y-1}\,dy$. Therefore $\lim_{n\to\infty} P_n = G(0) = e^{-1} = \boxed{\tfrac1e}.$

(Sanity check: the exact values $\tfrac12,\tfrac{5}{12},\tfrac{7}{18},\tfrac{1631}{4320},\dots$ decrease monotonically to $e^{-1}\approx0.3679$, confirmed by simulation. Note the earlier-tempting route via $\sum_k E[\ln X_k]$ fails twice over: that sum converges to $-\pi^2/6$, and even so $E[\prod X_i]\ne \exp\!\sum E[\ln X_i]$ by Jensen's inequality.)

Answer: (1) $f_{X_n}(t)=\dfrac{(-\ln(1-t))^{n-1}}{(n-1)!}$ on $(0,1)$, equivalently $-\ln(1-X_n)\sim\text{Gamma}(n,1)$. (2) $\displaystyle\lim_{n\to\infty}E\!\left[\prod_{i=1}^n X_i\right]=\frac1e$.

Intuition

The nested uniform construction is a beautiful example of how iterated random operations can produce clean distributional results. The key trick is the substitution $u = -\ln(1-t)$, which linearizes the nested uniform structure: each step of "draw uniformly between the previous value and 1" becomes "add an independent $\text{Exp}(1)$ to the running total." Sums of independent exponentials are Gamma, and suddenly the whole chain has a transparent structure. This log-transform trick appears constantly in order statistics, record values, and Poisson process theory.

The product limit in part (2) tests whether you can distinguish between "each factor approaches 1" and "the product actually converges." Because $X_k$ approaches 1 exponentially fast (specifically

- X_k \approx e^{-G}$ with $G$ of order $k$), the deviations $\ln X_k \approx -e^{-G_k}$ shrink geometrically and their sum converges. This is the same phenomenon as $\prod (1 - a_k)$ converging to a positive limit when $\sum a_k < \infty$. In practice, this kind of reasoning shows up whenever you need to assess whether accumulated small multiplicative effects (transaction costs, survival probabilities, compounding errors) eventually kill a quantity or leave a positive residual.

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