Prove that $\mathbb{E}(T)$ is infinite for the problem A random number generator produces independent random variates $x_0, x_1, x_2, . . .$ drawn
uniformly from $[0, 1]$, stopping at the first $x_T$ that is strictly less than $x_0$. Prove that
the expected value of $T$ is infinite.
Suggest, with a brief explanation, a plausible value
of $Pr(T = ∞)$ for a real-world (pseudo-)random number generator implemented on a
computer.
I haven't come across a continuous problem like this before, I'm not sure how to prove it.
 A: Probability of lasting $T$ for a given $x_0$:
$$
x_0(1-x_0)^{T-1}\tag{1}
$$
Expected value of $T$ for a given $x_0$:
$$
\begin{align}
\sum_{T=1}^\infty Tx_0(1-x_0)^{T-1}
&=-x_0\frac{\mathrm{d}}{\mathrm{d}x_0}\sum_{T=0}^\infty(1-x_0)^T\\
&=-x_0\frac{\mathrm{d}}{\mathrm{d}x_0}\frac1{1-(1-x_0)}\\
&=-x_0\frac{\mathrm{d}}{\mathrm{d}x_0}\frac1{x_0}\\
&=\frac1{x_0}\tag{2}
\end{align}
$$
By the linearity of expectation, we can compute the the expected value of $T$ for a random $x_0\in[0,1]$:
$$
\int_0^1\frac1{x_0}\,\mathrm{d}x_0=\infty\tag{3}
$$
A: $$
\begin{align}
E(T) &= \sum_{n = 1}^\infty n\ P(T = n) \\
&= \sum_{n = 1}^\infty n\ P\left(\{X_1\geq X_0\}\cap \dots\cap\{X_{n-1} \geq X_0\}\cap\{X_n < X_0\}\right) \\
 &= \sum_{n = 1}^\infty n\ E\left(P\left(\{X_1\geq X_0\}\cap \dots\cap\{X_{n-1} \geq X_0\}\cap\{X_n < X_0\} \big|\ X_0\right)\right) \\
&= \sum_{n = 1}^\infty n\ E\left(P(X_1 \geq X_0\ |\ X_0) \cdots P(X_2 \geq X_0\ |\ X_0)P(X_n < X_0\ |\ X_0)\right) \tag{1}\label{Cynthia} \\
&= \sum_{n = 1}^\infty nE\left((1-X_0)^{n-1}X_0\right) \tag{2}\label{George} \\
&= \sum_{n = 1}^\infty n \int_0^1 (1-x)^{n-1} x\ dx\\
&= \sum_{n = 1}^\infty n B(2, n) \int_0^1 f_{\beta(2,n)}(x)\ dx \tag{3}\label{Adam} \\
&= \sum_{n = 1}^\infty n B(2, n) \\
&= \sum_{n = 1}^\infty n\cdot \frac{(n-1)!}{(n+1)!} \tag{4}\label{Ben} \\
&= \sum_{n = 1}^\infty \frac{1}{n+1} \\
&= \infty,
\end{align}
$$
where in \eqref{Adam} I used the $\beta$ distribution, and in \eqref{Ben} I represented the Beta function in terms of factorials.
Steps \eqref{Cynthia} and \eqref{George} are intuitively compelling, but formal proofs are not trivial:
Step (1) The $\sigma$-algebras $\sigma(X_0, X_1), \sigma(X_0, X_2), \dots$ are conditionally independent given $\sigma(X_0)$ (w.r.t. $P$).
Proof Accoding to the penultimate paragraph on p. 109 of [1], it suffices to show that, for every $n \in \{1, 2, \dots\}$,
 $$
  \sigma(X_0, X_1, \dots, X_n) \perp_{\sigma(X_0)} \sigma(X_0, X_{n+1}). $$
 According to [1] Corollary 6.7 on p. 110, this is tantamount to showing that, for every $n \in \{1, 2, \dots\}$,
 $$
  \sigma(X_0, X_1, \dots, X_n) \perp_{\sigma(X_0)} \sigma(X_{n+1}). \tag{5}\label{Dan}
 $$
Let $n \in \{1, 2, \dots\}$. According to [1] Proposition 6.6 ("conditional independence, Doob") on p. 110, in order to show \eqref{Dan} it suffices to show that, for every $H \in \sigma(X_{n+1})$,
 $$
 P\left(H\ |\ \sigma(X_0, X_1, \dots, X_n)\right) \overset{P-\mathrm{a.s.}}{=} P(H\ |\ \sigma(X_0)).
 $$
 But, since $\sigma(X_0), \sigma(X_1), \dots, \sigma(X_{n+1})$ are $P$-independent, then, by [1] Corollary 3.7 ("grouping") on p. 51,
 $$
 \sigma(X_0, X_1, \dots, X_n) \perp \sigma(X_{n+1}).
 $$
Therefore, by the comment in the beginning of p. 105 of [1],
 $$
 P\left(H\ |\ \sigma(X_0, X_1, \dots, X_n)\right) \overset{P-\mathrm{a.s.}}{=} P(H) \overset{P-\mathrm{a.s.}}{=} P(H\ |\ \sigma(X_0)).
 $$
Q.E.D.
Step (2) For every $n \in \{1, 2, \dots\}$,
 $$
 \begin{align}
 P\left(X_n < X_0\ |\ \sigma(X_0)\right) &\overset{P-\mathrm{a.s.}}{=} X_0, \\
  P\left(X_n \geq X_0\ |\ \sigma(X_0)\right) &\overset{P-\mathrm{a.s.}}{=} 1 - X_0.
 \end{align}
 $$
Proof Let $n \in \{1, 2, \dots\}$. Define $B := \{(x,y) \in \mathbb{R}^2 \ |:\ x > y\}$. Observe that $B$ is a Borel set. Then
 $$
 P\left(X_n < X_0\ |\ \sigma(X_0)\right) = E\left(\mathbb{1}_{B}(X_0, X_n)\ |\ \sigma(X_0)\right).
 $$
According to [1] Theorem 6.3 ("Conditional Distribution") on p. 107, there exists a regular conditional distribution $P\left(X_n \in B\ |\ X_0 = x\right)$. In fact, since $X_0$ and $X_n$ are $P$-independent, then, by the comment in the beginning of [1] p. 107, we may assume that
 $$
 P\left(X_n \in B\ |\ X_0 = x\right) = P_{X_n}(B).
 $$
 Therefore, by [1] Theorem 6.4 ("Disintegration") on p. 108, we have
 $$
 \begin{align}
 E\left(\mathbb{1}_{B}(X_0, X_n)\ |\ \sigma(X_0)\right) &\overset{P-\mathrm{a.s.}}{=} \left(x \mapsto \int \mathbb{1}_B(x, y)\ P_{X_n}(dy)\right) \circ X_0 \\
 &= \left(x \mapsto P_{X_n}\left((-\infty,x)\right) \right)\circ X_0 \\
 &= \left(x \mapsto P_{U(0,1)}\left((-\infty,x)\right) \right)\circ X_0 \\
 &= \left(x \mapsto \begin{cases}0 & , x \leq 0 \\ x &, 0 < x \leq 1 \\ 1 &, x > 1\end{cases}\right) \circ X_0 \\
 &\overset{P-\mathrm{a.s.}}{=} X_0.
 \end{align}
 $$
This implies, in turn,
 $$
 \begin{align}
 P\left(X_n \geq X_0\ |\ \sigma(X_0)\right) &= E\left(1-\mathbb{1}_{B}(X_0, X_n)\ |\ \sigma(X_0)\right) \\
 &\overset{P-\mathrm{a.s.}}{=} 1 - E\left(\mathbb{1}_{B}(X_0, X_n)\ |\ \sigma(X_0)\right) \\
 &\overset{P-\mathrm{a.s.}}{=} 1 - X_0.
 \end{align}
 $$
Q.E.D.

References
[1] Kallenberg, Olav (2001). Foundations of Modern Probability (2nd ed.). York, PA, USA: Springer. ISBN 0-387-95313-2.
