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For independent standard normal random variables ${\eta }_1,\dots ,{\eta }_n$, define the displayed errors for $i\ge 2$. Each has mean zero and variance one. Nonadjacent errors depend on disjoint sets of independent variables. Adjacent covariances vanish because their product contains an independent centered factor. Thus their covariance matrix is $I_n$.

They are nevertheless dependent: the Gaussian fourth moment gives $\mathbb{E}[{ϵ}_i^2{ϵ}_{i+1}^2]=3$, and hence $\mathrm{Cov}({ϵ}_i^2,{ϵ}_{i+1}^2)=2$. Constant variance and zero pairwise covariance can conceal dependence in magnitudes. This is useful when interpreting a residual plot in linear regression.

For Brownian motion and deterministic partitions of an interval $[0,t]$ whose mesh $|\pi |$ tends to zero, the displayed limit holds in L2 space. Split the error into $U_{\pi }={\sum }_j{B}_{t_j}((\Delta B_j{)}^2-\Delta t_j)$ and the Riemann sum error $V_{\pi }={\sum }_j{B}_{t_j}\Delta t_j-{\int }_0^t{B}_{s}ds$. The summands of $U_{\pi }$ are martingale differences, hence orthogonal in L2 space, and the Gaussian fourth moment givesThe Cauchy-Schwarz inequality and the Brownian increment variance give $\mathbb{E}{V}_{\pi }^2\le (t/2){\sum }_j(\Delta t_j{)}^2\le t^2|\pi |/2$. Both vanish. This proves the limit directly without a stochastic change-of-variables formula.

For $f(x)=x$, the continuous martingale is $M_t^f=B_t$. Since $B_t$ has the centered normal distribution of variance $t$,For $f(x)=x^2$, the continuous martingale is $B_t^2-t$. The Gaussian fourth moment givesThis verifies both cases. The time integral in the printed identity is understood with its omitted $ds$.

Write $M^{(n)}=M^{f_n}$. The derivatives of the approximations areThus $|g_n|\le 1$. The Itô formula gives $M_t^{(n)}={\int }_0^t{g}_n(B_s)d{B}_s$. This use of the formula is valid despite the two second-derivative discontinuities: mollify the $C^1$ function $f_n$, whose derivative is Lipschitz. Its smooth derivatives converge boundedly, while the second derivatives are bounded by $n$ and converge except at $\pm 1/n$. For every $s>0$, the normal distribution of $B_s$ assigns those points zero probability. Dominated convergence for the drift integral and the Itô isometry for the stochastic integral pass the smooth formula to $f_n$. Equivalently, this identifies the martingale in the assumed decomposition and permits its cross-covariances to be calculated.

If $m\ge n$, then $|g_m-g_n|\le 1_{\{|x|<1/n\}}$. The Itô isometry and Doob L2 maximal inequality therefore give, for every fixed $T$,The Gaussian density is bounded by $(2\pi s{)}^{-1/2}$, soConsequently the processes are Cauchy in the complete space of continuous random paths with norm $(\mathbb{E}{\sup }_{s\le T}|Z_s{|}^2{)}^{1/2}$. They have a continuous-path limit $M$ there. Limits obtained for different integer horizons agree on overlaps; choose these versions consistently. Each $M_t$ is adapted, and passing to the $L^2$ limit in $\mathbb{E}[M_t^{(n)}\mid {\mathcal{F}}_s]=M_s^{(n)}$ shows that $M$ is a continuous martingale. Letting $m\to \infty$ in the preceding bound proves the requested convergence, with an explicit rate:The same Itô isometry identifies $M_t={\int }_0^t\mathrm{sgn}(B_s)d{B}_s$, with $\mathrm{sgn}(0)=0$.

There is a genuine factor-of-two error in the last printed conclusion. The stated $f_n$ has second derivative $n{1}_{\{|x|<1/n\}}$ almost everywhere, and the generator in the decomposition is one half of the second derivative. Hence the exact identity isBecause ${\sup }_x|f_n(x)-|x||\le 1/(2n)$ and $f_n(0)=1/(2n)$, this already proves uniform $L^2$ convergence on bounded intervals to $L_t=|B_t|-M_t$. It does not by itself prove almost-sure convergence of the full sequence, so we supply that additional step.

Take $n_k=k^2$. For each fixed integer $T$ and each $\epsilon >0$, the maximal-error bound and Markov's inequality implyThe first Borel-Cantelli lemma, applied to rational positive $\epsilon$ and integer $T$, gives uniform almost-sure convergence of this subsequence on every bounded interval. Therefore $L_{k^2}\to L$ uniformly there almost surely.

For $k^2\le n\le (k+1{)}^2$, the indicator windows are nested. The ratio-one interpolation for shrinking-window occupation integrals givesBoth ratios tend uniformly to one as $k\to \infty$. The two bounds converge uniformly to the continuous nonnegative process $L$, so the entire sequence does too. Thus the correct normalized conclusion isand with the factor actually printed in the question the limit isThese are the occupation approximations for Brownian local time. In the Tanaka formula convention, $L_t$ is the local time of a semimartingale at zero. It is not identically zero: $\mathbb{E}{L}_t=\mathbb{E}|B_t|=\sqrt{2t/\pi }>0$ for $t>0$, since $M$ starts at zero and is a martingale. Thus the discrepancy cannot be removed by treating the two limits as equal. The standard normalization is also given in Lalley's notes on the Itô calculus and local time.

Let ${\Delta }_j=W_{t_j}-W_{t_{j-1}}$ for $1\le j\le n$. These increments are jointly normal, because each is obtained by evaluating the isonormal Gaussian process on an interval indicator. Indicators of distinct intervals are orthogonal in $L^2([0,\infty ))$, soBy uncorrelated jointly Gaussian variables are independent, the increments on all these disjoint intervals are independent. Reversing the sign of the first increment, as in the printed list, preserves this independence.

The three requested properties have now been obtained without an existence theorem for Brownian motion. One can also obtain continuous paths: the Gaussian fourth moment gives $\mathbb{E}|W_t-W_s{|}^4=3|t-s{|}^2$. The Kolmogorov continuity theorem therefore supplies a continuous modification on every finite time interval, which can be chosen consistently on the half-line. Modification preserves every finite-dimensional distribution and hence the independent Gaussian increments. This yields Brownian motion itself.

Put ${\delta }_i=t_i-t_{i-1}$, $\Delta ={\max }_i{\delta }_i$, $v_i={\int }_{t_{i-1}}^{t_i}\sigma (s{)}^{2}ds$ and $S=\Vert {\sigma }^4{\Vert }_{\infty }$. The deterministic integrand in the Itô integral implies that the increments $Z_i=X_{t_i}-X_{t_{i-1}}$ are independent random variables with normal distributions $N(0,v_i)$. The Gaussian fourth moment givesThese identities include $v_i=0$. Since the summands of $M_n$ are centered and independent, all cross terms in its second moment vanish. Consequently,Thus $D=2$ works for every observation partition. The last step uses the total interval length, rather than assuming equally spaced observations.

Compute fitted responses ${\hat{y}}_i=\hat{\alpha }+\hat{\beta }{x}_i$ and regression residuals $e_i=y_i-{\hat{y}}_i$. A conventional residual plot shows $e_i$ against ${\hat{y}}_i$ or $x_i$; plotting against observation order additionally looks for runs. One checks for curvature, changing spread and other structure rather than random scatter around zero.

An ordinary residual plot need not reveal this failure of independence. The errors have constant mean zero, constant variance one and zero pairwise covariances, so the usual trend and changing-variance clues need not appear. A visually patternless residual plot cannot validate independence; fitted regression residuals themselves also share the fitted coefficients even when the true errors are independent.

There is detectable dependence in the magnitudes. The Gaussian fourth moment is $\mathbb{E}{\eta }_i^4=3$, giving for every adjacent pairFor $i=1$, the product is ${\eta }_1^4{\eta }_2^2$; for $i\ge 2$, it is ${\eta }_{i-1}^2{\eta }_i^4{\eta }_{i+1}^2$, with the same expected value 3. Large magnitudes tend to occur together. A lagged plot of squared regression residuals or a test directed at dependence in their squares is more informative. Thus recognition is possible with suitable diagnostics, but is not guaranteed by the standard residual plot requested here.

For the standard normal density, $P^{\prime }(x)=-xP(x)$. Using integration by parts,The boundary term vanishes because Gaussian decay dominates every power. Since the given mean is zero and variance is one, $\mathbb{E}[X^2]=1$, yielding the Gaussian fourth moment

Use the inner product $⟨f,g⟩=\mathbb{E}[f(Z)g(Z)]$ for a standard normal random variable $Z$. The degree-zero monic polynomial is $p_0=1$. Write $p_1=x+a$ and $p_2=x^2+bx+c$. The conditionsuse $\mathbb{E}Z=\mathbb{E}{Z}^3=0$ and $\mathbb{E}{Z}^2=1$. Hence these first three Probabilists' Hermite polynomials areThe Gaussian fourth moment is $\mathbb{E}{Z}^4=3$, so their squared norms are