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2025-Q2 – Symmetric Variables, Conditional Expectation, and Quadratic Variation

2025 Probability Prelim Exam (PDF) (link pending)

Problem 2 (verbatim)

(a)(i)

Let $X$ be a symmetric random variable (i.e., $X \stackrel{d}{=} -X$) with $P(X=0)=0$.
Prove that
\(X = \varepsilon \cdot |X| \quad \text{a.s.},\)
where $\varepsilon$ and $|X|$ are independent, and
\(P(\varepsilon = 1)=P(\varepsilon=-1)=\tfrac12.\)

(a)(ii)

Let $X,Y$ be symmetric, independent, and square-integrable random variables with
$P(X=0)=P(Y=0)=0$.
Let
\(\mathcal H = \sigma(|X|,|Y|)=\sigma(X^2,Y^2).\)
Prove that
\(E_{\mathcal H}(XY)=0.\)
Also calculate $E_{\mathcal H}[(X+Y)^2]$.

(b)

Let ${B_t : t\ge0}$ be standard Brownian motion.
Let
\(Q_n = \sum_{k=1}^n D_{k,n}^2,\qquad D_{k,n}=B_{t_{k,n}}-B_{t_{k-1,n}},\)
where
$0=t_{0,n}<t_{1,n}<\cdots<t_{n,n}=1$
and
$\Delta_{k,n}=t_{k,n}-t_{k-1,n}$.

(i) Compute
\(E(Q_n)\qquad\text{and}\qquad \mathrm{Var}(Q_n).\)

(ii) Let $\Delta_n=\max_{1\le k\le n}\Delta_{k,n}$.
Prove that if $\Delta_n\to0$ then $Q_n\to1$ in probability.

(c)

Assume that the partitions are nested:
\(\{t_{k,n}\}_{k=0}^n \subset \{t_{k,n+1}\}_{k=0}^{n+1}.\)

Let
\(\mathcal H_n = \sigma\left(\bigcup_{m=n}^\infty \{D_{k,m}^2:1\le k\le m\}\right).\)
This is a decreasing filtration.

(i) Compute $E_{\mathcal H_{n+1}}(Q_n)$ and identify the type of process ${Q_n,\mathcal H_n}$.

(ii) Prove that if $\Delta_n\to0$ then $Q_n\to1$ almost surely.

Survival Guide Solution

(a)(i) Symmetric RVs decompose into sign and magnitude

Claim

For symmetric $X$ with $P(X=0)=0$, the representation
\(X=\varepsilon |X|,\qquad P(\varepsilon=1)=P(\varepsilon=-1)=\tfrac12,\)
holds a.s., and $\varepsilon$ is independent of $|X|$.

Proof

  1. Define the sign variable.
    Let
    \(\varepsilon(\omega)=\begin{cases} +1,& X(\omega)>0 -1,& X(\omega)<0 \end{cases}.\)
    Since $P(X=0)=0$, $\varepsilon$ is well-defined a.s.
    Then trivially
    \(X=\varepsilon |X|.\)

  2. Show that $\varepsilon$ is Rademacher.
    Symmetry means
    \(P(X>0) = P(-X>0) = P(X<0).\)
    Because $P(X\ne0)=1$,
    \(P(\varepsilon=1)=P(X>0)=P(X<0)=P(\varepsilon=-1)=\tfrac12.\)

  3. Show independence of $\varepsilon$ and $|X|$.
    For any Borel set $A\subset[0,\infty)$,
    \(P(\varepsilon=1, |X|\in A) = P(X>0, |X|\in A) = P(X\in A),\) and by symmetry,
    \(P(\varepsilon=-1, |X|\in A) = P(X<0, |X|\in A) = P(X\in -A)=P(X\in A).\) Also by symmetry, \(P(\vert X\vert\in A)=P(X\in A) + P(X\in-A)=2P(X\in A) \text{ implies } P(X\in A)=\frac{1}{2}P(\vert X\vert\in A)\)

Thus
\(P(\varepsilon=\pm1, |X|\in A) = \tfrac12\, P(|X|\in A) = P(\varepsilon=\pm1)\, P(|X|\in A).\)

Conclusion

There exists an independent Rademacher $\varepsilon$ such that
\(X=\varepsilon |X| \quad \text{a.s.}.\)

Key Takeaways

  • Symmetry enables decomposition into sign and magnitude.
  • Independence follows from splitting probability into positive/negative branches.
  • “Almost sure” allows breaking ties at $X=0$ on a null set.

(a)(ii) Conditional expectations given magnitudes

Claim

For symmetric, independent $X,Y$ with $P(X=0)=P(Y=0)=0$,
\(E_{\mathcal H}(XY)=0,\)
and
\(E_{\mathcal H}[(X+Y)^2]=X^2+Y^2.\)

Proof

  1. Decompose each variable.
    From (a)(i),
    \(X = \varepsilon_X |X|,\qquad Y=\varepsilon_Y |Y|,\)
    where $\varepsilon_X,\varepsilon_Y$ are independent Rademacher variables and are independent of $\mathcal H$.

  2. Compute the cross term.
    \(XY = \varepsilon_X\varepsilon_Y |X||Y|.\)
    Conditional on $\mathcal H$, the magnitudes are known constants, while the signs remain independent with mean zero:
    \(E(\varepsilon_X\varepsilon_Y\mid\mathcal H)=E(\varepsilon_X)E(\varepsilon_Y)=0.\)
    Thus
    \(E_{\mathcal H}(XY)=|X||Y|\cdot 0=0.\)

  3. Compute $E_{\mathcal H}[(X+Y)^2]$.
    Expand:
    \((X+Y)^2 = X^2 + Y^2 + 2XY.\)
    Taking conditional expectation,
    \(E_{\mathcal H}(X^2)=X^2,\qquad E_{\mathcal H}(Y^2)=Y^2,\) and from step 2,
    \(E_{\mathcal H}(XY)=0.\)

Hence
\(E_{\mathcal H}[(X+Y)^2]=X^2 + Y^2.\)

Conclusion

Conditioning on magnitudes removes sign-information, killing all odd or mixed terms.

Key Takeaways

  • When conditioning on magnitudes (an even function), all odd symmetric parts vanish.
  • Independence of signs is the crucial structural simplification.
  • Linearize and expand before conditioning.

(b)(i) Expectation and variance of quadratic variation approximations

Recall
\(Q_n=\sum_{k=1}^n D_{k,n}^2,\qquad D_{k,n}\sim N(0,\Delta_{k,n}).\)

Claim

\(E(Q_n)=\sum_{k=1}^n \Delta_{k,n},\qquad \mathrm{Var}(Q_n)=2\sum_{k=1}^n \Delta_{k,n}^2.\)

Proof

  1. Expectation.
    For $D\sim N(0,\Delta)$,
    \(E(D^2)=\Delta.\)
    Since increments are independent,
    \(E(Q_n)=\sum_{k=1}^n E(D_{k,n}^2)=\sum_{k=1}^n\Delta_{k,n}=1.\)

  2. Variance.
    For $D\sim N(0,\Delta)$,
    \(\mathrm{Var}(D^2)=2\Delta^2.\)
    Independence gives
    \(\mathrm{Var}(Q_n)=\sum_{k=1}^n 2\Delta_{k,n}^2.\)

Conclusion

$E(Q_n)=1$ for any partition; variance is the sum of squared mesh lengths.

Key Takeaways

  • $D^2$ for a normal increment has variance $2\Delta^2$.
  • These sums approximate quadratic variation.
  • Brownian increments are independent and Gaussian.

(b)(ii) Convergence in probability

Claim

If $\Delta_n\to0$, then $Q_n\to1$ in probability.

Proof

From (i), \(\mathrm{Var}(Q_n)=2\sum_{k=1}^n \Delta_{k,n}^2 \le 2\Delta_n \sum_{k=1}^n \Delta_{k,n} =2\Delta_n.\)

Apply Chebyshev: \(P(|Q_n-1|>\varepsilon) \le \frac{\mathrm{Var}(Q_n)}{\varepsilon^2} \le \frac{2\Delta_n}{\varepsilon^2} \to 0.\)

Conclusion

Mesh size $\Delta_n\to0$ implies $Q_n\to1$ in probability.

Key Takeaways

  • Controlling variance is the key step for probability convergence.
  • $\sum\Delta_{k,n}=1$ is fixed, so only $\Delta_n$ matters.

(c)(i) Reverse martingale structure

Adding a point splits exactly one interval:
\((t_{k_*,n},t_{k_*+1,n}] = (t_{k_*,n+1},t_{k_*+1,n+1}] \cup (t_{k_*+1,n+1},t_{k_*+2,n+1}].\)

Thus
\(D_{k_*,n}^2 = D_{k_*,n+1}^2 + D_{k_*+1,n+1}^2,\)
and all other increments are unchanged.

Claim

\(E_{\mathcal H_{n+1}}(Q_n)=Q_{n+1}.\)

Proof

Write
\(Q_n=\sum_{k\ne k_*} D_{k,n}^2 + D_{k_*,n}^2 = \sum_{k\ne k_*} D_{k,n+1}^2 + (D_{k_*,n+1}^2 + D_{k_*+1,n+1}^2) =Q_{n+1}.\)

Since $D_{k,m}^2\in \mathcal H_m$ and $\mathcal H_{n+1}\subset\mathcal H_n$, all terms are measurable with respect to $\mathcal H_{n+1}$.
Therefore the conditional expectation is deterministic and equals $Q_{n+1}$.

Hence
\(E(Q_n\mid \mathcal H_{n+1}) = Q_{n+1}.\)

Conclusion

${Q_n,\mathcal H_n}$ is a reverse martingale.

Key Takeaways

  • Nested partitions create reverse filtrations.
  • The quadratic variation approximation decreases in terms of $\mathcal H_n$–information.
  • Reverse martingales converge almost surely (Doob’s reverse martingale theorem).

(c)(ii) Almost sure convergence of $Q_n$

Claim

If $\Delta_n\to0$ then $Q_n\to1$ a.s.

Proof

From (c)(i), $Q_n$ is a reverse martingale with
\(E(Q_n)=1,\qquad Q_n\ge0.\)

By the reverse martingale convergence theorem,
\(Q_n \to Q_\infty \quad\text{a.s.},\)
for some integrable $Q_\infty$.

But from (b)(ii),
\(Q_n \to 1 \quad\text{in probability}.\)

A reverse martingale limit is unique, so convergence in probability to $1$ forces
\(Q_\infty = 1 \quad\text{a.s.}\)

Thus $Q_n\to1$ almost surely.

Conclusion

Quadratic variation approximations converge almost surely to $1$ whenever the mesh size goes to zero.

Key Takeaways

  • Reverse martingale convergence + convergence in probability pins down the a.s. limit.
  • Quadratic variation of Brownian motion on [0,1] is literally $1$.
  • This is the foundational theorem behind stochastic calculus.
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