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For standard Brownian motion, the Brownian bridge $U_t=B_t-t{B}_1$, $0\le t\le 1$, is independent of $B_1$. Every finite vector of bridge coordinates is jointly multivariate normal with $B_1$, and its cross-covariances with $B_1$ vanish. The independence of uncorrelated jointly normal variables and independence extended from generating pi-systems prove independence of the whole stochastic process.

The ordinary least squares objective has gradient $2X^{T}X\beta -2X^{T}Y$. The normal equation therefore givesThe Gram matrix $X^{T}X$ is a positive-definite matrix because $X$ has full column rank. More explicitly, for any $b$,since $X^T(Y-X\hat{\beta })=0$. This proves that the displayed solution is the unique global minimum.

The fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix isThe inverse of the symmetric matrix $X^{T}X$ is symmetric, so $H^T=H$. Multiplication gives $H^2=X(X^{T}X{)}^{-1}(X^{T}X)(X^{T}X{)}^{-1}{X}^T=H$. Thus $H$ is the orthogonal projection onto the column space of the design matrix.

The regression residual vector is $e=Y-\hat{Y}=GY$ with $G=I_n-H$. Consequently $G^T=G$, $G^2=I_n-2H+H^2=G$, and the residual sum of squares isBy fitted-residual orthogonality, $GH=0$. Using $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, the cross-covariance matrix isBoth vectors are jointly normal, so they are also independent by independence of uncorrelated jointly normal variables. The zero covariance calculation itself only needs the common error variance and lack of error correlations.

For the first wind fit, let $v_i$ be wind velocity and $Y_i$ electrical output. The simple linear regression is $Y_i=\alpha +\beta v_i+{\epsilon }_i$, with independent ${\epsilon }_i\sim N(0,{\sigma }^2)$. It has two fitted mean parameters and $25-2=23$ residual degrees of freedom. The missing analysis of variance entries are therefore velocity degrees of freedom $1$, velocity mean square $8.9296$, and residual degrees of freedom $23$. The residual mean square is $1.2816/23\simeq 0.05572$, and $8.9296/(1.2816/23)\simeq 160.25$, consistent with the printed value after rounding.

Because the model includes a regression intercept, the coefficient of determination is the explained sum of squares divided by the total centered sum of squares:A large coefficient of determination does not rule out a wrong mean function. In the PDF's first residual-versus-fitted plot, residuals are negative at both ends and positive in the middle. This curved pattern agrees with the visibly flattening output-versus-velocity relationship and motivates a polynomial regression with a quadratic term.

The second wind model is $Y_i=\alpha +{\beta }_1{v}_i+{\beta }_2{v}_i^2+{\epsilon }_i$, again with independent common-variance normal errors. Its fitted mean is $-1.155898+0.722936v_i-0.038121v_i^2$. The line marked (A) is a two-sided Student t-test of $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$, conditional on retaining the intercept and linear term. Under the null hypothesis,Its two-sided p-value is $6.59\times 10^{-8}$. This is far below $0.05$, so reject a purely linear mean in favour of the quadratic fit. Equivalently, the extra-term nested-model F-test has $F=t^2\simeq 63.15$ and null law $F_{1,22}$.

The third fit is a reciprocal-predictor regression, $Y_i=a+b/v_i+{\epsilon }_i$, with estimated mean $2.9789-6.9345/v_i$. It has two mean parameters rather than the quadratic model's three. Its residual standard error is smaller, $0.09417$ versus $0.1227$, and its coefficient of determination is larger, $0.9800$ versus $0.9676$. The corresponding residual sums of squares are approximately $23(0.09417{)}^2=0.20396$ and $22(0.1227{)}^2=0.33122$. The reciprocal fit is preferable on both these fit measures and parsimony. The two models are not nested, so an ordinary extra-term F-test between them is inappropriate. On the common normal-error likelihood, the difference ${\mathrm{AIC}}_3-{\mathrm{AIC}}_2=25\log (0.20396/0.33122)-2\simeq -14.12$ also favours the third model.

Neither lower residual sum of squares nor a higher coefficient of determination establishes adequate assumptions. The quadratic residual-versus-fitted plot removes the original pronounced curvature; the reciprocal plot also has no comparably obvious mean trend. The low-output residuals appear somewhat more spread out, so check scale-location plots and residuals against velocity for heteroscedasticity. Q-Q plots assess normality; residuals against observation order assess serial dependence; regression leverage and Cook's distance identify influential observations. Further cross-validation, replicate observations at comparable velocities, and prediction errors would help choose between their extrapolation behaviours. Both fitted shapes should be judged principally over the observed positive-velocity range.

For any finite list $t_1,\dots ,t_k\in [0,1]$, the random vector $(U_{t_1},\dots ,U_{t_k},B_1)$ has a multivariate normal distribution, since it is a linear image of a vector of Brownian motion coordinates. The Brownian covariance kernel and linearity of covariance giveThe independence of uncorrelated jointly normal variables implies that $(U_{t_1},\dots ,U_{t_k})$ is independent of $B_1$. Extending from finite-coordinate cylinder events by independence extended from generating pi-systems provesThus the Brownian bridge is independent of its endpoint. Its covariance kernel is $\mathrm{Cov}(U_s,U_t)=\min (s,t)-st$, confirming that $U$ is the standard Brownian bridge. This is the Brownian bridge independence from its endpoint property.

For a diagonal entry, put $Z_j={\sum }_{i=1}^d(A_{ij}^2-1)$. Independence and the given moment-generating function inequality for the chi-squared distribution giveThe Chernoff bound with $a=t/4$ yieldsApply the same argument to $-Z_j$ and sum the two tails:

To obtain the off-diagonal bound, set $U=(V+W)/\sqrt{2}$ and $D=(V-W)/\sqrt{2}$. They are jointly normal with unit variances and covariance zero, so independence of uncorrelated jointly normal variables makes them independent standard normals. Since $VW=(U^2-D^2)/2$, the moment-generating function of the product of two independent standard normal random variables isFor $j\ne k$, $(A^{T}A{)}_{jk}$ is a sum of $d$ independent such products. The given inequality implies $\mathbb{E}{e}^{a(A^{T}A{)}_{jk}}\le e^{d{a}^2}$ for $|a|\le 1/2$. Taking $a=t/2$, and repeating for the negative tail, gives

Now take $t=c\sqrt{\log p/d}\in (0,1)$ and $E=A^{T}A/d-I_p$. There are $p$ diagonal entries but only $p(p-1)/2$ distinct off-diagonal entries, since $E$ is symmetric. By the union bound, the event $\Vert E{\Vert }_{\max }\le t$ has probability at leastThis is the required entrywise concentration of a Gaussian Gram matrix.

On that single event, any vector $z$ supported on at most $s$ coordinates satisfieswhere the last step is the Cauchy-Schwarz inequality on the support. If $u,v$ each have at most $s/2$ nonzero coordinates, $z=u-v$ has at most $s$. The same entrywise event applies simultaneously to all of them, so no union over vectors is needed. The sparse norm preservation from entrywise Gram control givesThe printed ratio is undefined at $u=v$. Its equivalent quadratic statement, $|\Vert A(u-v){\Vert }_2^2/d-\Vert u-v{\Vert }_2^2|\le st\Vert u-v{\Vert }_2^2$, includes that case trivially. The nontrivial concentration argument assumes $p\ge 2,c>0$; at $p=1$ the stated probability lower bound is nonpositive and hence vacuous.

The covariance matrix isIf $\mathrm{Cov}(X_1,X_2)=0$, the moment-generating function in part (ii) factors into the product of the two marginal moment-generating functions. Thus $X_1$ and $X_2$ are independent random variables; this is the independence of uncorrelated jointly normal variables.

By bilinearity of covariance,ChooseThen $(X_1,Y)$ is bivariate normal by part (i) and has zero covariance, so $Y$ is a normal random variable independent of $X_1$. Hence $X_2=a{X}_1+Y$.

Write $a=(2,1{)}^T$, so the data map is $a^{T}u$. The Gaussian likelihood and prior distribution giveCompleting the square, or using Gaussian conjugacy for a normal linear model, identifies a multivariate normal distribution with precision matrix $C_{\delta }^{-1}=I_2+{\delta }^{-2}a{a}^T$. Since $a^{T}a=5$, its posterior mean and covariance matrix areThe parameters areFor example, $(I_2+{\delta }^{-2}a{a}^T)C_{\delta }=I_2$, which checks the square completion.

The orthogonal vectorsare eigenvectors of the covariance matrix. Their respective variances areThey are independent under the posterior distribution, by independence of uncorrelated jointly normal variables. Thus uncertainty is different in the two directions even for positive noise: the data constrain $e_{\parallel }$, whereas $e_{\perp }$ spans the null space of the data map.

As $\delta \to 0$, the posterior distribution converges to the law ofThis is supported on $2u_1+u_2=m$, and its posterior mean is the minimum-norm least-squares solution. The prior distribution still determines the distribution along the unobserved null space; it has not disappeared. This example of a Gaussian posterior in the zero-noise limit distinguishes exact recovery of an observed component from recovery of the whole unknown.

For $d=1$ and $X\sim N(0,{\sigma }^2)$, the characteristic function isFor a multivariate normal distribution with covariance matrix ${\sigma }^2{I}_d$, the coordinates are uncorrelated and hence independent by independence of uncorrelated jointly normal variables. ThereforeThis is the characteristic function of an isotropic Gaussian vector.

Condition on the information available before time $n$. The current wealth is then fixed, while ${\xi }_n\sim N(b,\Sigma )$ is independent of that information. PutFor any portfolio $\theta$, use the inner product induced by the positive-definite matrix $\Sigma$ to writeThe linear image of a multivariate normal vector shows that the two portfolio returns are jointly normal, andThus independence of uncorrelated jointly normal variables makes the residual return ${\eta }^T{\xi }_n$ independent of the aligned return; it also has expected value zero.

The continuation value $V(n,\cdot )$ is concave by part (c). Conditional Jensen inequality therefore shows that adding the independent centered residual cannot improve the objective:where $m=(1+r)X_{n-1}$. Since the optimal portfolio is unique, its residual must be zero.

It remains to determine the sign. If $\lambda <0$, the portfolios $\lambda {\Sigma }^{-1}b$ and $-\lambda {\Sigma }^{-1}b$ have the same return variance, namely ${\lambda }^{2}B$, while their means are $\lambda B$ and $-\lambda B$. The latter return has the distribution of the former plus the positive constant $-2\lambda B$. Since $V(n,\cdot )$ is increasing, replacing $\lambda$ by $-\lambda$ cannot reduce the objective, contradicting uniqueness. Hence $\lambda \ge 0$.

Applying this conditional argument at every time gives nonnegative, past-measurable random variables ${\lambda }_n$ such thatThis is the Gaussian one-fund theorem.