Question

Suppose \(X_{0}\) has probability density function $$ f(x)= \begin{cases}2 x, & \text { for } 0 \leq x \leq 1, \\ 0, & \text { eleewhere, }\end{cases} $$ and that \(X_{n+1}\) is uniformly distributed on \(\left(1-X_{n}, 1\right]\), given \(X_{0}, \ldots, X_{n}\). Show that \(\left\\{X_{n}\right\\}\) is a stationary ergodic process.

Short answer

We proved that the process \(\{X_n\}\) is stationary by showing that the distribution of \(X_{n+1}\) is the same as the distribution of \(X_0\) (i.e., \(f_{X_{n+1}}(x) = 2x\)). We also proved the process is ergodic because the time averages converge to the ensemble averages due to the strong law of large numbers for stationary sequences. Therefore, the process \(\{X_n\}\) is a stationary ergodic process.

Step-by-step solution

Prove that the process is stationary

To prove that the process is stationary, we need to show that the distribution of the process stays the same at all times. That is, for any n: $$ F_{X_n}(x) = F_{X_0}(x) $$ Where \(F_{X_n}(x)\) is the cumulative distribution function for \(X_n\). We are given the pdf for \(X_0\): $$ f(x)= \begin{cases}2 x, & 0 \leq x \leq 1, \\\ 0, & \text { eleewhere, }\end{cases} $$ and the condition to generate \(X_{n+1}\): $$ X_{n+1}\sim U(1-X_n, 1] $$ where \(U(a, b)\) denotes the uniform distribution on the interval (a,b]. Now let's find the distribution of \(X_{n+1}\) given \(X_n\). Since \(X_{n+1}\) is uniformly distributed on the interval \((1-X_n, 1]\), the density function for \(X_{n+1}\) given \(X_n\) is: $$ f_{X_{n+1}|X_n}(x|X_n) = \begin{cases}1/(1-(1-X_n)), & 1-X_n < x \leq 1, \\\ 0, & \text { elsewhere, }\end{cases} $$ To find the marginal distribution of \(X_{n+1}\), we can integrate out \(X_n\): $$ f_{X_{n+1}}(x) = \int_0^1 f_{X_{n+1}|X_n}(x|X_n)f_{X_n}(X_n) dX_n $$ Using the given pdf \(f(x) = 2x\), the integration becomes: $$ f_{X_{n+1}}(x) = \int_0^1 \frac{1}{1-(1-X_n)} \cdot 2X_n dX_n $$ Now substitute \(u = 1 - X_n\) and \(du = -dX_n\). $$ f_{X_{n+1}}(x) = -2 \int_0^1 \frac{1-u}{u} du = 2\int_0^1(1-u) du $$ Evaluating the integral gives: $$ f_{X_{n+1}}(x) = 2x $$ This shows that the process is stationary because the distribution of \(X_{n+1}\) is the same as the distribution of \(X_0\).

Prove that the process is ergodic

To prove that the process is ergodic, we need to show that the time averages converge to the ensemble averages. First, compute the expected value of \(X_0\) using the given pdf: $$ E[X_0] = \int_0^1 2x^2 dx = 2\left[\frac{x^3}{3}\right]_0^1 = \frac{2}{3} $$ Since the process is stationary, the ensemble average (expected value) for \(X_n\) is also equal to \(\frac{2}{3}\) for all \(n\). For a time average, we calculate the average of \(X_k\) over a finite number of time steps (N): $$ \frac{1}{N} \sum_{k=1}^{N} X_k $$ We need to show that the limit of this time average as \(N \to \infty\) converges to the ensemble average: $$ \lim_{N\to\infty} \frac{1}{N} \sum_{k=1}^{N} X_k = \frac{2}{3} $$ By the strong law of large numbers for stationary sequences, the time average converges to the ensemble average. $$ \lim_{N\to\infty} \frac{1}{N} \sum_{k=1}^{N} X_k = E[X_n] = \frac{2}{3} $$ Since the process is stationary and the time average converges to the ensemble average, we conclude that the process \(\{X_n\}\) is a stationary ergodic process.