2023-Q4 - Brownian representation of a 3-point random variable
2023 Probability Prelim Exam (PDF)
Problem statement :contentReference[oaicite:0]{index=0}
Let ${B_t, t \ge 0}$ be a standard Brownian motion (SBM) equipped with its canonical filtration ${\mathcal{F}t}{t\ge0}$.
Let $X$ be a random variable which gets exactly 3 distinct and non-zero values denoted by ${x_k}_{k=1}^3$. Also let $A_k = {X = x_k}$, with $P(A_k) = p_k > 0$, $k=1,2,3$. Assume that $E(X) = 0$.
(a) Let $\mathcal{G} = \sigma{A_1, A_2 \cup A_3}$ and denote $Y = E(X \mid \mathcal{G})$.
(i) How many values does $Y$ get? What is $E(Y)$? What can you say about the event ${X = Y}$?
(ii) Use the symbols ${x_k, p_k}$, $k=1,2,3$, to present the explicit distribution of $Y$.
(b) Define a stopping time with respect to the SBM filtration, denoted by $\tau_1$, so we have $Y = B_{\tau_1}$ in distribution. Prove your answer.
(c) Find a stopping time with respect to the SBM filtration, denoted by $\tau_2$, so that
(i) $X = B_{\tau_2}$ in distribution, and
(ii) $\tau_2 \ge \tau_1$, a.s.
Hint. Differentiate between the events ${X = Y}$ and ${X \ne Y}$. Also, let $Z_t = B_{\tau_1 + t}, t \ge 0$. By the strong Markov property, $Z_t - B_{\tau_1}$, $t \ge 0$, is a SBM which is independent of $\mathcal{F}_{\tau_1}$.
Solution :contentReference[oaicite:1]{index=1}
We lightly edit your original writeup for clarity and fill in some details, keeping your structure and wording as much as possible.
(a) Conditional expectation with a coarse σ-algebra
(a)(i) Conditional expectation and basic properties
Claim.
Let $\mathcal{G} = \sigma{A_1, A_2 \cup A_3}$ and $Y = E(X \mid \mathcal{G})$. Then:
- $Y$ takes exactly two distinct values, one on $A_1$ and one on $A_2 \cup A_3$.
- $E(Y) = E(X) = 0$.
- On $A_1$ we have $X = Y$ a.s., and on $A_2 \cup A_3$ in general $X \ne Y$ except in degenerate cases.
Proof.
By definition,
- $X = x_k$ on $A_k$, with $P(A_k) = p_k > 0$, $k=1,2,3$.
- $\mathcal{G}$ has atoms $A_1$ and $A_2 \cup A_3$.
Since $E(X \mid \mathcal{G})$ is constant on each atom of $\mathcal{G}$, there exist constants $y_1,y_2$ such that \(Y(\omega) = \begin{cases} y_1, & \omega \in A_1,\\ y_2, & \omega \in A_2 \cup A_3. \end{cases}\)
On $A_1$, $X$ is deterministically $x_1$, so \(Y = E(X \mid A_1) = x_1 \quad\text{on } A_1.\) Hence $y_1 = x_1$.
On $A_2 \cup A_3$, \(Y = E(X \mid A_2 \cup A_3) = \frac{x_2 P(A_2) + x_3 P(A_3)}{P(A_2 \cup A_3)} = \frac{x_2 p_2 + x_3 p_3}{p_2 + p_3}.\) So $y_2 = \dfrac{x_2 p_2 + x_3 p_3}{p_2 + p_3}$.
Thus $Y$ takes exactly two values: $x_1$ on $A_1$ and $y_2$ on $A_2 \cup A_3$.
For the expectation: \(E(Y) = E(E(X \mid \mathcal{G})) = E(X) = 0.\)
Regarding ${X = Y}$:
- On $A_1$, $X = x_1 = Y$.
- On $A_2$ and $A_3$, $X$ takes the values $x_2$ or $x_3$, while $Y$ is the constant $y_2$ on $A_2 \cup A_3$. In general $x_2 \ne y_2$ and $x_3 \ne y_2$, so on $A_2 \cup A_3$ we typically have $X \ne Y$.
So at least \(P(X = Y) \ge P(A_1) = p_1,\) and in generic non-degenerate situations $P(X=Y) = p_1$.
Conclusion.
- $Y$ has exactly two values: $x_1$ and $y_2 = \dfrac{x_2 p_2 + x_3 p_3}{p_2 + p_3}$.
- $E(Y) = 0$.
- Almost surely, $X = Y$ on $A_1$, and $X \ne Y$ on most of $A_2 \cup A_3$.
(a)(ii) Explicit distribution of $Y$
Claim.
The distribution of $Y$ is: \(P(Y = x_1) = p_1, \qquad P\!\left(Y = \frac{x_2 p_2 + x_3 p_3}{p_2 + p_3}\right) = p_2 + p_3.\)
Proof.
We already identified the values:
- On $A_1$, $Y = x_1$.
- On $A_2 \cup A_3$, $Y = \dfrac{x_2 p_2 + x_3 p_3}{p_2 + p_3}$.
Hence, \(P(Y = x_1) = P(A_1) = p_1,\) and \(P\!\left(Y = \frac{x_2 p_2 + x_3 p_3}{p_2 + p_3}\right) = P(A_2 \cup A_3) = p_2 + p_3.\)
By construction, \(E(Y) = x_1 p_1 + \left(\frac{x_2 p_2 + x_3 p_3}{p_2 + p_3}\right)(p_2 + p_3) = x_1 p_1 + x_2 p_2 + x_3 p_3 = E(X) = 0.\)
Conclusion.
$Y$ is a two-point mean-zero random variable with values $x_1$ and \(y_2 := \frac{x_2 p_2 + x_3 p_3}{p_2 + p_3},\) and probabilities $p_1$ and $p_2 + p_3$.
(b) Representing $Y$ as $B_{\tau_1}$
Claim.
There exists a stopping time $\tau_1$ for the Brownian motion such that \(B_{\tau_1} \overset{d}{=} Y.\)
Proof.
From part (a)(ii), $Y$ is mean zero and takes two values: \(Y = \begin{cases} x_1, & \text{with prob. } p_1,\\[3pt] y_2, & \text{with prob. } p_2+p_3, \end{cases}\) where $x_1\ne 0$, $y_2\ne 0$, and $x_1 \neq y_2$.
We can (and will) assume $x_1 < 0 < y_2$ by re-labeling indices if needed.
Consider the hitting time \(\tau_1 = \inf\{ t \ge 0 : B_t \in \{x_1, y_2\}\}.\)
-
Stopping time and boundedness.
$\tau_1$ is a stopping time with respect to ${\mathcal{F}t}$, and since $B_t$ started at $0$ must hit either level $x_1$ or $y_2$ almost surely, and these levels are finite, $\tau_1$ is almost surely finite, and $B{\tau_1}\in{x_1,y_2}$. -
Optional stopping and probabilities.
Since $(B_t)$ is a martingale and $B_{\tau_1}$ is bounded, the optional stopping theorem (OST) gives \(E(B_{\tau_1}) = E(B_0) = 0.\) Write \(P(B_{\tau_1} = x_1) = q_1, \qquad P(B_{\tau_1} = y_2) = q_2 = 1 - q_1.\) Then \(0 = E(B_{\tau_1}) = x_1 q_1 + y_2 (1-q_1) = y_2 + q_1(x_1 - y_2).\) Solving for $q_1$, \(q_1 = \frac{y_2}{y_2 - x_1}, \qquad q_2 = 1 - q_1 = \frac{-x_1}{y_2 - x_1}.\) -
Matching with $Y$.
Since $Y$ is also a mean-zero two-point random variable supported on ${x_1,y_2}$, its probabilities must satisfy exactly the same relations: \(x_1 p_1 + y_2 (p_2 + p_3) = 0 \quad\Longrightarrow\quad p_1 = \frac{y_2}{y_2 - x_1},\quad p_2 + p_3 = \frac{-x_1}{y_2 - x_1}.\) Hence \(P(Y = x_1) = p_1 = q_1, \quad P(Y = y_2) = p_2 + p_3 = q_2.\)
Therefore, $B_{\tau_1}$ and $Y$ have the same two-point distribution.
Conclusion.
The stopping time \(\tau_1 = \inf\{ t \ge 0 : B_t \in\{x_1,y_2\}\}\) satisfies $B_{\tau_1} \overset{d}{=} Y$.
(c) Representing $X$ as $B_{\tau_2}$ with $\tau_2 \ge \tau_1$
(c)(i) Constructing the 3-point stopping time
Claim.
There exists a stopping time $\tau_2$ such that \(B_{\tau_2} \overset{d}{=} X \quad\text{and}\quad \tau_2 \ge \tau_1 \text{ a.s.}\)
Proof.
We distinguish the events ${X = Y}$ and ${X \ne Y}$.
We know:
- $Y = x_1$ on $A_1$.
- $Y = y_2 = \dfrac{x_2 p_2 + x_3 p_3}{p_2 + p_3}$ on $A_2 \cup A_3$.
- $X$ takes values $x_1, x_2, x_3$ with probabilities $p_1,p_2,p_3$.
We already have $\tau_1$ such that $B_{\tau_1} \overset{d}{=} Y$.
-
On the event ${X = Y}$.
This is exactly $A_1$, since there $X = x_1 = Y$.
On this event, we can set \(\tau_2 = \tau_1.\) Then $B_{\tau_2} = B_{\tau_1} = x_1$ matches $X$ on $A_1$. -
On the event ${X \ne Y}$, i.e., $A_2 \cup A_3$.
On $A_2 \cup A_3$, $Y = y_2$ is some value between $x_2$ and $x_3$, and we want to further “split” $y_2$ into the values $x_2$ and $x_3$ with probabilities \(P(X = x_2 \mid A_2 \cup A_3) = \frac{p_2}{p_2 + p_3}, \qquad P(X = x_3 \mid A_2 \cup A_3) = \frac{p_3}{p_2 + p_3}.\)After time $\tau_1$, define the shifted process \(Z_t = B_{\tau_1 + t}, \quad t \ge 0.\) By the strong Markov property, $(Z_t - B_{\tau_1}){t\ge0}$ is a SBM independent of $\mathcal{F}{\tau_1}$, so $Z_t$ is a Brownian motion started at $B_{\tau_1}=Y$.
Define \(T = \inf\{t \ge 0 : Z_t \in \{x_2, x_3\}\},\) and set \(\tau_2 = \tau_1 + T.\)
Then \(B_{\tau_2} = Z_T \in \{x_2,x_3\}.\)
The hitting probabilities of $x_2$ and $x_3$ by a SBM starting at $y_2$ (with $x_2 < y_2 < x_3$) are \(P(Z_T = x_3) = \frac{y_2 - x_2}{x_3 - x_2}, \qquad P(Z_T = x_2) = \frac{x_3 - y_2}{x_3 - x_2}.\)
Now check these match the conditional probabilities:
-
Using $y_2 = \dfrac{x_2 p_2 + x_3 p_3}{p_2 + p_3}$, \(y_2 - x_2 = \frac{x_2 p_2 + x_3 p_3}{p_2 + p_3} - x_2 = \frac{(x_3 - x_2)p_3}{p_2 + p_3},\) therefore \(\frac{y_2 - x_2}{x_3 - x_2} = \frac{p_3}{p_2 + p_3} = P(X = x_3 \mid A_2 \cup A_3).\)
-
Similarly, \(x_3 - y_2 = (x_3 - x_2) \frac{p_2}{p_2 + p_3},\) and \(\frac{x_3 - y_2}{x_3 - x_2} = \frac{p_2}{p_2 + p_3} = P(X = x_2 \mid A_2 \cup A_3).\)
Thus, conditional on $Y = y_2$, $B_{\tau_2}$ has exactly the same conditional distribution as $X$ on $A_2 \cup A_3$.
-
Putting these pieces together, we have:
- On $A_1$, $B_{\tau_2} = x_1$ with prob. $p_1$.
- On $A_2 \cup A_3$, $B_{\tau_2}$ takes values $x_2$ and $x_3$ with conditional probabilities $p_2/(p_2+p_3)$ and $p_3/(p_2+p_3)$, hence overall probabilities $p_2$ and $p_3$.
So $B_{\tau_2}$ and $X$ have the same three-point distribution.
Conclusion.
The construction above yields a stopping time $\tau_2$ such that $B_{\tau_2} \overset{d}{=} X$.
(c)(ii) Verifying $\tau_2 \ge \tau_1$ and stopping-time property
Claim.
The random time $\tau_2$ defined by
\(\tau_2 =
\begin{cases}
\tau_1, & \text{on } \{X = Y\} = A_1,\\[3pt]
\tau_1 + T, & \text{on } \{X \ne Y\} = A_2 \cup A_3,
\end{cases}\)
is a stopping time with respect to $(\mathcal{F}_t)$ and satisfies $\tau_2 \ge \tau_1$ a.s.
Proof.
- Inequality $\tau_2 \ge \tau_1$.
By construction:- On $A_1$, $\tau_2 = \tau_1$.
- On $A_2 \cup A_3$, $\tau_2 = \tau_1 + T$ with $T\ge 0$. So always $\tau_2 \ge \tau_1$.
-
Stopping-time property.
- $\tau_1$ is a stopping time by definition as a hitting time of a closed set.
- $T$ is a hitting time for the process $Z_t = B_{\tau_1 + t}$, hence a stopping time with respect to the shifted filtration. By the strong Markov property and standard theory, $\tau_1 + T$ is then a stopping time in the original filtration.
- Since $\tau_2$ is defined piecewise using events in $\mathcal{F}_{\tau_1}$ and stopping times $\tau_1$ and $\tau_1+T$, it is again a stopping time.
Formally, for each $t$, \(\{\tau_2 \le t\} = \big(\{X=Y\}\cap\{\tau_1\le t\}\big) \;\cup\; \big(\{X\ne Y\}\cap\{\tau_1+T\le t\}\big),\) which lies in $\mathcal{F}_t$ since both $\tau_1$ and $\tau_1+T$ are stopping times and ${X=Y}$, ${X\ne Y}$ are measurable.
Conclusion.
$\tau_2$ is a stopping time, and $\tau_2 \ge \tau_1$ a.s., with $B_{\tau_2}$ distributed exactly as $X$.
Key Takeaways
-
Coarse σ-algebras compress information.
Conditioning $X$ on $\mathcal{G} = \sigma{A_1, A_2\cup A_3}$ collapses three distinct values into a two-point random variable $Y$. -
Two-point mean-zero distributions and Brownian motion.
Any two-point mean-zero variable ${a,b}$ with $a<0<b$ can be realized as $B_\tau$ where $\tau$ is the first hitting time of ${a,b}$ for SBM. The probabilities are determined by the harmonic property: \(P(B_\tau = b) = \frac{|a|}{|a|+|b|}, \quad P(B_\tau = a) = \frac{|b|}{|a|+|b|}.\) -
Strong Markov property for “multi-step” constructions.
We first reach the “coarse” random value $Y$ by $\tau_1$, then, conditional on $Y$, we restart a fresh Brownian motion and run it until it hits the finer targets $x_2,x_3$. This is a standard Skorokhod-embedding style idea in miniature. -
Matching conditional probabilities via hitting probabilities.
The conditional mean $y_2 = E(X \mid A_2 \cup A_3)$ sits between $x_2$ and $x_3$ in such a way that Brownian hitting probabilities from $y_2$ to $x_2$ and $x_3$ reproduce precisely \(P(X=x_2\mid A_2\cup A_3),\quad P(X=x_3\mid A_2\cup A_3).\) -
Piecewise stopping times.
You can glue stopping times together along measurable events, and you can add a stopping time to another by using the strong Markov property: if $T$ is a stopping time for the shifted process after $\tau_1$, then $\tau_1+T$ is a stopping time for the original filtration.
This problem is a clean template for representing discrete random variables as values of Brownian motion at (carefully chosen) stopping times.
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