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Let $0<\alpha <1$, $X_0=x\in [0,1]$, and let $B_{n+1}$ be a binary random variable satisfying $\mathbb{E}[B_{n+1}\mid {\mathcal{F}}_n]=X_n$. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable and $B_{n+1}$ is ${\mathcal{F}}_{n+1}$-measurable. The updateremains in $[0,1]$ and defines a bounded martingale. For example, take independent uniform random variables $U_{n+1}$ independent of the current filtration and set $B_{n+1}=1_{\{U_{n+1}\le X_n\}}$.

The binary identity $B_{n+1}^2=B_{n+1}$ givesand henceThe L2-bounded martingale convergence theorem gives a terminal random variable $X_{\infty }$ with almost sure convergence and convergence in L2. Iteration followed by the bounded convergence theorem yieldsThus $X_{\infty }$ takes only $0$ and $1$, and preservation of its expected value gives a Bernoulli distribution with parameter $x$. The terminal conditional expectation representation gives the exact mean squared errorThe finite-time paths can stay strictly inside $(0,1)$ while their limit lies entirely at the endpoints. The positive conditional variance in the interior makes permanent convergence to an interior value impossible.

Here $I_t={\sum }_{j=1}^t{1}_{\{{\xi }_j^{(1)}={\xi }_j^{(2)}\}}$ counts simultaneous collisions of two independent walks. A shared environment sign contributes $\mathbb{E}(1+\epsilon h{)}^2=1+{\epsilon }^2$; distinct signs contribute one. If the difference walk returns to zero with probability $\rho <1$, its total number of positive-time returns has law $\mathbb{P}(I_{\infty }=k)=(1-\rho ){\rho }^k$. Therefore the second moments are bounded by $(1-\rho )/(1-\rho (1+{\epsilon }^2))$ whenever $\rho (1+{\epsilon }^2)<1$. Convergence in L2 then preserves the limit's mean one.

For any $h_1,\dots ,h_m\in \mathcal{H}$ and real $u_1,\dots ,u_m$, linearity of the finite partial sums and convergence in L2 giveBy part (ii), this random variable has the normal distributionA finite family is Jointly Gaussian precisely when every real linear combination has a possibly degenerate normal distribution. Thus every finite subfamily of $(Wh{)}_{h\in \mathcal{H}}$ is Jointly Gaussian, which is the meaning of joint Gaussianity for this collection.

Enumerate the orthonormal basis consisting of $H_0$ and the Haar wavelets $H_{n,k}$ as $(e_j{)}_{j\ge 1}$, and enumerate the corresponding independent standard normal random variables as $(A_j{)}_{j\ge 1}$. Set $c_j=⟨h,e_j{⟩}_{\mathcal{H}}$. The Parseval identity for a Hilbertian basis givesThe partial sums $T_m={\sum }_{j=1}^m{c}_j{A}_j$ form a martingale for $\sigma (A_1,\dots ,A_m)$, since the next summand is independent of the past and has mean zero. Moreover,The martingale convergence theorem for a martingale bounded in L2 space proves convergence almost surely and convergence in L2. Equivalently one may apply the Kolmogorov convergence theorem to the summable variances $c_j^2$.

A finite linear combination of independent normal random variables is normal. Its characteristic function is $\mathbb{E}{e}^{iu{T}_m}=\exp (-u^2{\sum }_{j\le m}{c}_j^2/2)$. Passing to the limit identifiesFor $h=0$ this is the constant zero random variable. Each fixed $h$ has its own permissible null set; simultaneous pathwise convergence for all $h\in \mathcal{H}$ is unnecessary.

Use the same finite truncation $T_m(h)={\sum }_{j\le m}⟨h,e_j⟩A_j$ for $f$ and $g$. Since $\mathbb{E}[A_i{A}_j]={\delta }_{ij}$ by independence, zero means and unit variances,Here ${\delta }_{ij}$ is the Kronecker delta. The Cauchy-Schwarz inequality and convergence in L2 justify passing to the limit in the product's expectation. The Parseval identity for a Hilbertian basis now givesBecause each $Wh$ has mean zero, this is also its covariance identity. Thus $W$ is an isonormal Gaussian process over $\mathcal{H}$.

For any real orthonormal basis $(G_i)$, the argument of (b)(ii) gives almost sure convergence and convergence in L2 of $\tilde{W}h$. Finite linear combinations have normal distributions, and the Parseval identity for a Hilbertian basis givesThus $W$ and $\tilde{W}$ are centered Gaussian processes with the same covariance function. More explicitly, their finite-dimensional characteristic functions are bothThis proves the basis invariance of an isonormal Gaussian process: equality of all finite-dimensional distributions, and hence equality in law of the indexed collections.

For the specified cosine orthonormal basis, put ${\lambda }_i=\pi (i-\frac{1}{2})$. Its coefficient against an interval indicator isConsequently the Brownian half-integer sine expansion, a Karhunen–Loève expansion, isFor every fixed $t$, the sum converges almost surely and in L2 space because the coefficient squares sum to $\Vert 1_{[0,t]}{\Vert }^2=t$. Its continuous modification is a standard Brownian motion by (b)(v). The expansion also holds in $L^2(\Omega \times [0,1])$: the expectation of the squared tail is the integral of the coefficient-square tail, which tends to zero by the dominated convergence theorem, with bound $t$.

To realize this expansion for the original $B$, choose $b_i=W{G}_i$. These are Jointly Gaussian with covariance matrix equal to the identity, hence are independent standard normal random variables. The finite orthogonal projections of $1_{[0,t]}$ onto $G_1,\dots ,G_N$ converge in $\mathcal{H}$, and the identity $\mathbb{E}|Wh{|}^2=\Vert h{\Vert }^2$ transfers that convergence to the displayed expansion for $B_t$. With an independently chosen sequence $(b_i)$ the resulting Brownian motion has the same law; a pathwise identification uses the coupling just given.

Fix $t>0$, a deterministic partition of an interval $\pi$, and write $\eta (s)=t_j$ on $[t_j,t_{j+1})$. Its mesh is $|\pi |={\max }_j(t_{j+1}-t_j)$. The Brownian increment variance givesThus the left-endpoint sums ${\sum }_j{B}_{t_j}\Delta B_j$ converge in L2 space to the Itô integral. Although these step coefficients are unbounded, clipping them at finite levels first makes them bounded simple predictable processes; the Itô isometry and their finite second moments remove the clipping.

The elementary identity $B_{t_{j+1}}^2-B_{t_j}^2=2B_{t_j}\Delta B_j+(\Delta B_j{)}^2$ givesThe squared Brownian increments are independent, with means $\Delta t_j$ and variances $2(\Delta t_j{)}^2$, since a centered normal random variable of variance $v$ has fourth moment $3v^2$. ThereforePassing to the convergence in L2 limits yieldsThis holds for each fixed $t$; continuous sample paths of both sides extend equality from rational times to all $t$ on one event of probability one.

For the finite probability distribution, put ${\xi }_j=(X_j+1)/2$. These are independent fair zero-one variables, andThe binary integer ${\sum }_{j=1}^n{2}^{n-j}{\xi }_j$ is uniform on $\{0,\dots ,2^n-1\}$. HenceThese are the midpoints of $2^n$ equal subintervals of $[-1,1]$. For every bounded continuous $f$, the Riemann sums giveThus $\boxed{Y_n\Rightarrow \mathrm{Unif}[-1,1]}$ in the sense of weak convergence of probability measures.

For the almost-sure and $L^2$ assertions, the Dyadic Rademacher series $Y={\sum }_{j\ge 1}{2}^{-j}{X}_j$ converges absolutely on every sign sequence, since $|X_j|=1$. Its tail satisfies $|Y-Y_n|\le 2^{-n}$, proving almost sure convergence and convergence in L2. More precisely, zero means and independence eliminate the cross terms:This identity follows first for finite tails and then by dominated convergence. Bounded continuous test functions and bounded convergence theorem identify the probability distribution of $Y$ with the weak limit above, namely the uniform distribution on $[-1,1]$.

For the requested product identity, each Rademacher random variable has characteristic function $\mathbb{E}{e}^{it{X}_j}=\cos t$. The characteristic function of a sum of independent variables therefore yieldsBy part (a), or directly by bounded convergence theorem, this converges to the characteristic function of a uniform distribution, $\frac{1}{2}{\int }_{-1}^1{e}^{itx}dx=\sin t/t$. Consequently the dyadic cosine product isBoth sides are one at zero. At zeros of the sinc function the identity is understood as the limit of the finite products, so no logarithm or division by a vanishing factor is needed.

The real L2 space on a probability space consists of equivalence classes of measurable functions $X:\Omega \to \mathbb{R}$ satisfying $\mathbb{E}[X^2]<\infty$, where two random variables represent the same element if they are equal almost surely. Its norm isPassing to equivalence classes makes this a genuine norm: $\Vert X{\Vert }_2=0$ exactly when $X=0$ almost surely. The Minkowski inequality gives the triangle inequality.

Let $(X_n)$ be a Cauchy sequence in this norm. Choose an increasing sequence of indices $n_k$ such thatThis is possible by choosing $n_k$ beyond a Cauchy threshold for tolerance $2^{-k}$. Set $D_k=X_{n_{k+1}}-X_{n_k}$. The Cauchy-Schwarz inequality on a probability space gives $\mathbb{E}|D_k|\le \Vert D_k{\Vert }_2$. By Tonelli theorem,Thus the series of absolute differences is finite almost surely. Define $X=X_{n_1}+{\sum }_{k\ge 1}{D}_k$ on this probability-one event, and give it the value zero on its complement. It is a measurable function, and $X_{n_k}\to X$ almost surely.

For $l>k$, the Minkowski inequality givesApply the Fatou lemma to the squared difference as $l\to \infty$. It follows thatIn particular $X-X_{n_1}\in L^2$, and therefore $X\in L^2$ by the Minkowski inequality. This proves convergence in L2 of the subsequence. To recover the entire sequence, for a prescribed $\epsilon >0$ choose a Cauchy threshold $N$ for $\epsilon /2$, then choose $k$ with $n_k\ge N$ and $\Vert X-X_{n_k}{\Vert }_2<\epsilon /2$. For every $n\ge N$,Hence $\boxed{X_n\to X\text{ in }{L}^2}$, proving completeness of Lp spaces for $p=2$ without assuming the desired completeness in the argument.

Set $M_n={\sum }_{j=0}^n{X}_j$ and ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$. The independence and zero expected values give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$, so $(M_n)$ is a martingale. The variance additivity for independent random variables givesThe Cauchy-Schwarz inequality implies ${\sup }_n\mathbb{E}|M_n|\le \sqrt{V}$. Applying the martingale convergence theorem from (b),In fact the L2 martingale convergence theorem also gives convergence in L2. Independently, for $n>m$ the same variance additivity for independent random variables yields $\mathbb{E}|M_n-M_m{|}^2={\sum }_{j=m+1}^n\mathbb{E}[X_j^2]\to 0$, confirming that the partial sums form a Cauchy sequence in $L^2$.

Suppose first thatThis is convergence in L2, hence convergence in probability, which implies convergence in distribution. By part (i), the respective laws are ${\mu }_n$ and $\nu$, so ${\mu }_n$ converges weakly to $\nu$. Part (a)(i) also gives

Conversely, suppose that ${\mu }_n$ converges weakly to $\nu$ and the displayed second moments converge. By the stated quantile coupling for convergence in distribution,Part (i) identifies their second moments with those of the measures, so part (a)(ii) applies and gives convergence in $L^2$. ThereforeThis is the one-dimensional second-Wasserstein convergence criterion realized by the common-quantile coupling.

Let $S_N$ be the Fourier partial sum through frequency $N$. By orthogonality and the coefficient formulas,Expanding the nonnegative squared L2 norm of the error gives the exact finite identityFor $f$ in the L2 space, completeness of the Fourier basis makes the left-hand side tend to zero. Dividing the resulting equality by $2L$ proves Parseval's identity:This argument uses convergence in L2 rather than an unqualified exchange of an infinite sum with the integral of its square. The function in part (c) is continuous and piecewise polynomial, so these hypotheses hold.

Take the inner product of the Fourier series with $\sin (n\pi x/L)$. Fourier orthogonality eliminates the constant, all cosine terms, and all sine terms except the $n$th. The remaining squared norm issoFor a square-integrable function, the calculation is justified by taking inner products of finite Fourier partial sums and passing to their mean-square limit.

Write $A=g(X)$ and $B=g(Y)$. The two conditional expectation identities supply a $\sigma (Y)$-measurable version of $B$ and a $\sigma (X)$-measurable version of $A$. Work with those versions, which do not change any almost sure equality. Both lie in L2 space, and the Cauchy-Schwarz inequality makes their product an integrable random variable.

Testing the first conditional expectation identity against $B$, and the second against $A$, givesFor completeness, these tests with unbounded multipliers are valid by first using the bounded truncations $B_m=(-m)\vee (B\wedge m)$ and $A_m=(-m)\vee (A\wedge m)$, then passing to the limit with the Cauchy-Schwarz inequality and convergence in L2. Consequently,A nonnegative random variable with zero expected value vanishes almost surely. Therefore $g(X)=g(Y)$ almost surely. This is mutual conditional expectation rigidity in L2; it also follows from the equality case for conditional second moments.

For a counterexample to equality of the original random variables, let $X$ give probability $1/4$ to each of $-2,-1,1,2$, set $Y=-X$, and use the nonconstant function $g(t)=t^2$. Then $g(X)=g(Y)=X^2$ takes both values $1$ and $4$. It is measurable with respect to both generated sigma-algebras, so both required conditional expectation identities hold. Symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$, while neither random variable can be zero. Thus

The L2-bounded martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because $0<X_n<1$. Thus there is $X_{\infty }\in [0,1]$ almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound $|X_n-X_{\infty }{|}^2\le 1$ gives convergence in L2 by the dominated convergence theorem.^[ref]

The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ givesTaking expected values and iterating provesBy bounded convergence theorem, $\mathbb{E}[X_{\infty }(1-X_{\infty })]=0$. The random variable inside this expected value is nonnegative, so it is zero almost surely; on $[0,1]$ this forces $X_{\infty }\in \{0,1\}$. Again by bounded convergence theorem, $\mathbb{E}{X}_{\infty }={\lim }_n\mathbb{E}{X}_n=x$. HenceThis is the Bernoulli polarization martingale with update size $1/2$. It also has an exact mean squared error: the terminal representation $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ and the equality case for conditional second moments give

The floor function gives the deterministic boundfor every outcome, including $X=1$. Thus all three convergence assertions hold: there is pointwise, and hence almost sure convergence, as well as convergence in probability and convergence in L2. In detail, for $\epsilon >0$ the probability $\mathbb{P}(|X-X_n|>\epsilon )$ is zero as soon as $2^{-n}\le \epsilon$; alsoThe uniform distribution even gives the exact second moment of the error by integrating over the dyadic partition:

The uniformly integrable martingale convergence theorem states that a uniformly integrable martingale $(U_n)$ has an integrable terminal random variable $U_{\infty }$ such that $U_n\to U_{\infty }$ almost surely and in the L1 norm, andThe last identity follows by passing to the limit of a sequence in $\mathbb{E}[U_m{1}_A]=\mathbb{E}[U_n{1}_A]$ for $A\in {\mathcal{F}}_n$ and $m\ge n$, using convergence in L1.

Part (b)(iii) verifies uniform integrability for $(Z_n)$, soMoreover $|Z_{\infty }|\le L$ almost surely. In this particular problem the common bound also yields convergence in Lp for every finite $p\ge 1$, including convergence in L2: the dominated convergence theorem applies to $|Z_n-Z_{\infty }{|}^p\le (2L{)}^p$. General uniform integrability alone does not assert these higher-power conclusions.