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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.

A best affine state estimate minimizes mean squared error among estimates $a+Cy$ formed from the available observation vector $y$. With finite second moments and invertible observation covariance matrix, it isThis follows from the orthogonal projection of centered state components onto the linear span of centered observation components. For Jointly Gaussian state and observation it equals the conditional expectation; for general laws a nonlinear estimate can have lower mean squared error.

The Bühlmann credibility premium is the best affine mean squared error estimate of a risk’s conditional claim expected value. In the Bühlmann model it is $Z\overline{X}+(1-Z)m$, where $Z=na/(na+v)$. The linear least-squares projection equations use $\mathrm{Var}(X_j)=a+v$ and $\mathrm{Cov}(X_i,X_j)=a$ for distinct years.

Given a latent risk parameter $\Theta$, observations are independent and identically distributed random variables with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. The Bühlmann credibility premium estimates $m(\Theta )$ by an affine function of past observations under mean squared error. Its population parameters are $m=\mathbb{E}m(\Theta )$, expected process variance $v=\mathbb{E}v(\Theta )$, and variance of hypothetical means $a=\mathrm{Var}(m(\Theta ))$.

For total exposure $W={\sum }_j{m}_j$ and weighted average ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$, the best affine mean squared error estimate of the conditional mean is $Z{\overline{X}}_w+(1-Z)m_0$, where $Z=Wa/(Wa+v)$. If the coefficients sum to $B$, their error is $a(1-B{)}^2+v{\sum }_j{b}_j^2/m_j$. Cauchy-Schwarz inequality minimizes the latter term at $b_j=B{m}_j/W$, and a one-dimensional quadratic minimization then gives $B=Z$.

The conditional expectation is the average of a random variable that can be predicted using specified information. It is itself a random variable: the average appropriate to one observed group may differ from that appropriate to another group. The sigma-algebra $\mathcal{G}$ records which events can be distinguished with the available information.

For a real integrable random variable $X$ on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ and a sub-sigma-algebra $\mathcal{G}$, the conditional expectation $m=\mathbb{E}[X\mid \mathcal{G}]$ is the almost surely unique integrable random variable that is measurable with respect to $\mathcal{G}$ and satisfiesIts average therefore agrees with that of $X$ on every distinguishable event. The integral characterization handles discrete observations, continuous observations, and information that is not given by a single numerical value.^[ref]

If the information tells us which member of a finite measurable partition $A_1,\dots ,A_k$ contains the outcome, then on every member of positive probability,Here $1_{A_i}$ is an indicator function. Zero-probability members do not determine a unique value, which is why uniqueness is only almost surely. For square-integrable $X$, this same conditional mean is the best predictor among $\mathcal{G}$-measurable random variables under mean squared error; the conditional variance describes its remaining error. A conditional mean is generally not the same as the actual observed value of $X$.^[ref]

The conditional variance measures the remaining squared uncertainty after specified information has been used to form a conditional mean. For a real square-integrable random variable $X$ and a sub-sigma-algebra $\mathcal{G}$, let $m=\mathbb{E}[X\mid \mathcal{G}]$. ThenIt is a $\mathcal{G}$-measurable random variable, nonnegative almost surely, rather than necessarily a single number. With the information $\mathcal{G}=\sigma (Y)$, its value is the variance in the group identified by $Y$. Its expected value is the mean squared error of predicting $X$ by its conditional mean.^[ref]

In the exact mixture distribution in law of total expectation, group $A$ contains $0,4$ around mean two, and group $B$ contains $6,10$ around mean eight. Both groups have conditional variance four because their deviations from their own means are $-2,2$ with equal probabilities. Therefore $\mathbb{E}[\mathrm{Var}(X\mid Y)]=4$. Meanwhile, the two group means have variance $\frac{1}{4}(2-13/2{)}^2+\frac{3}{4}(8-13/2{)}^2=27/4$. The law of total variance givesDirect calculation gives $\mathbb{E}[X^2]=53$ and $\mathrm{Var}(X)=53-(13/2{)}^2=43/4$, checking the decomposition.

For any square-integrable $\mathcal{G}$-measurable predictor $Z$, the error identity isThe cross term vanishes because $\mathbb{E}[X-m\mid \mathcal{G}]=0$. Thus using the conditional mean minimizes mean squared error, while conditional variance quantifies what this information still cannot predict. This is the orthogonal projection interpretation in L2 space.^[ref]

A credibility estimate combines a risk's individual experience with population information. In the basic equal-exposure form, the credibility factor $Z$ weights the observed sample mean, while $1-Z$ weights a prior or collective expected value. The Bühlmann credibility premium is the best affine estimate under mean squared error; an exact Bayesian posterior mean can also have this form, as in the natural conjugate credibility identity.

A distortion measure assigns a nonnegative cost to replacing a source symbol $x$ by a reconstruction $\hat{x}$. It need not be a metric: the two alphabets may differ, and neither symmetry nor the triangle inequality is required. Choosing the cost is part of the model. For numerical symbols, $d(x,\hat{x})=(x-\hat{x}{)}^2$ gives mean squared error after averaging; for exact symbol agreement, use Hamming distortion.

The Kalman filter alternates a Kalman prediction step with a Kalman measurement update. It carries a state estimate forward using a state-space model, then adjusts it by the part of the new observation that was not predicted: the innovation. The Kalman gain determines the adjustment from the uncertainty of the prediction and the measurement noise. A noisy measurement should have less influence than an equally surprising precise one.

For a precise discrete-time statement, let the state and observation be vectors in real vector spaces obeying the linear state-space modelThe matrices and control inputs are known. Assume a Gaussian random vector $x_0$ with mean $m_0$ and covariance matrix $P_0$, and mutually independent centered Gaussian noises $w_t,v_t$, independent of $x_0$, with covariance matrices $Q_t,R_t$. Suppose $R_t$ is positive-definite, so the observation innovation covariance is invertible. Then the recursions below give the exact posterior mean $m_t=\mathbb{E}[x_t\mid y_1,\dots ,y_t]$ and posterior covariance matrix $P_t$, hence the full conditional multivariate normal distribution. They also minimize mean squared error. Without the Gaussian assumptions, suitable second-order assumptions still give best affine state estimates, but do not generally give an exact posterior distribution. ^[ref].

For an exact scalar example, suppose the predicted state has normal distribution $N(10,4)$ and a measurement reports $y=14$ in the same units. With measurement noise variance $R=1$, the Kalman gain is $4/(4+1)=4/5$ and the posterior distribution isWith the same prediction and observed value but measurement noise variance $R=16$, the Kalman gain becomes $1/5$ and the posterior distribution is instead $N(10.8,3.2)$. Both updates move toward the observation; the more precise measurement moves the estimate further and reduces uncertainty more.

The useful compression is exact under the stated model: the two objects $m_t,P_t$ contain everything from the past observations needed for the next update. Their size depends on the state dimension, not on the length of the observation history. This makes causal state filtering practical when observations arrive continually. It also gives Linear-quadratic-Gaussian control an estimate of the state on which to act.

Rudolf E. Kalman's 1960 paper organized linear filtering around state transitions, error covariance and recursive computation, connecting estimation with optimal regulation. Its introduction acknowledges earlier work by Wiener, Bode, Shannon and others; the useful historical claim is this particular formulation, rather than that estimation began in 1960. ^[ref].

Kriging predicts an unobserved value of a spatial random field by combining observations according to their covariances. Nearest-neighbor interpolation uses the closest sample; kriging also asks how strongly each sample is related to the target and how much information the samples share with one another. Two nearby, highly correlated samples need not provide twice the information of one. The different forms of kriging depend on whether the mean is known, an unknown constant, or an unknown regression drift.

Write $Z=(Z(s_1),\dots ,Z(s_n){)}^T$ for the observation random vector, $z$ for its realized values, and $Z_0=Z(s_0)$ for the prediction target. Assume finite second moments, and defineHere $\Sigma$ is the observation covariance matrix, assumed to be a positive-definite matrix in the formulas below. A covariance function supplies its entries and the target covariance vector $c$. For a second-order stationary random field, the covariogram $C(h)$ depends on displacement, and the semivariogram satisfies $\gamma (h)=C(0)-C(h)$. The intrinsically stationary random field formulation permits kriging using a semivariogram even when a stationary covariogram is unavailable; the finite covariance formulation below is a convenient important case.

For a linear combination $w^{T}Z$ predicting $Z_0$ with zero bias, the error variance equals its mean squared error:Ordinary kriging and universal kriging minimize this quadratic form subject to their mean constraints. Their optimality is among the specified linear predictors with zero bias under the chosen statistical model; covariance information alone does not establish optimality among arbitrary nonlinear predictors. The Gaussian random field interpretation supplies additional conclusions in simple kriging.

The reported error variance assumes a chosen covariance function, its statistical parameters, observation locations, and a mean structure. It omits uncertainty in fitted covariance parameters and does not validate the trend. A repeated-sampling mean squared error need not equal the error variance conditional on each realized data vector.^[ref]

In 1963, Georges Matheron presented kriging for estimating average ore grades in mining panels from weighted samples taken inside and outside a panel. This setting in geostatistics explains why prediction targets can be averages over regions as well as values at points.^[ref]

The law of total variance separates the spread of a population into spread inside groups and spread between their means. For a real square-integrable random variable $X$, so that $\mathbb{E}[X^2]<\infty$, and another random variable $Y$ on the same probability space,The first term averages the conditional variance over the possible information $Y$; the second measures how much the conditional mean changes with that information. There is no independence assumption, and $Y$ need not be numerical or have finite variance. Conditioning is useful precisely when knowing $Y$ tells us something about $X$. See MIT tutorial on using conditional expectation and variance.

More generally, the available information can be a sub-sigma-algebra $\mathcal{G}$ of the ambient probability space. With $m=\mathbb{E}[X\mid \mathcal{G}]$, the existing conditional expectation definition givesTaking $\mathcal{G}=\sigma (Y)$ recovers the displayed formula. The conditional mean and conditional variance are themselves random variables: before learning the group, we do not yet know which group average or group spread will apply. Their outer expectation and variance turn these varying quantities into numbers.

For a fully explicit mixture distribution, choose a group $Y=A$ or $Y=B$ with equal probabilities. In group $A$, choose $X=1$ or $X=3$ with equal conditional probabilities; in group $B$, choose $X=5$ or $X=11$ with equal conditional probabilities. Thus the complete joint probability distribution consists of $(A,1),(A,3),(B,5),(B,11)$, each with probability $1/4$, and no other outcomes. The conditional means are $2$ and $8$. Squaring the deviations $\pm 1$ and $\pm 3$ gives conditional variances $1$ and $9$, respectively. The law of total expectation gives the overall mean $\mu =(2+8)/2=5$.

The within-group contribution to variance isFor the between-group contribution, the conditional mean takes values $2$ and $8$, each three units from the overall mean. Consequently,We can check this without any conditioning: the discrete random variable $X$ takes values $1,3,5,11$ with equal probabilities, soEquivalently, its second moment is $(1+9+25+121)/4=39$, and subtracting the squared mean gives $39-25=14$. These are population variances, with the stated probabilities as weights, rather than estimates using the corrected denominator of a sample variance.

The figure makes the two sources of spread visible. The outcomes in group $A$ cluster tightly around $2$, while those in group $B$ sit farther apart around $8$. Replacing every outcome by its conditional mean removes the internal spread but leaves the separation between groups. Conversely, subtracting each group's conditional mean aligns the two groups around zero. Averaging only the conditional variances would miss nine units of variance; retaining only the conditional means would miss five. Neither summary alone describes the whole population.

Two modifications clarify what each contribution measures. Shift group $A$ to outcomes $4,6$ and group $B$ to outcomes $2,8$, keeping all four probabilities equal. Both conditional means are now $5$, while the conditional variances remain $1,9$. The between-group contribution vanishes, and total variance is just $5$. Group membership still tells us how uncertain an outcome is, even though the best point prediction stays the same. Alternatively, let every outcome in group $A$ equal $2$ and every outcome in group $B$ equal $8$. Now both conditional variances vanish, and all nine units of total variance come from group separation. Knowing the group predicts the outcome perfectly. These limiting cases distinguish differences in average outcome from differences in internal spread; the original mixture distribution contains both.

The same calculation answers a prediction question. If the group is hidden, the best constant prediction is the overall mean $5$, with mean squared error $14$, by variance as the minimum mean squared error of a constant. If the group is revealed, predict $2$ for $A$ and $8$ for $B$. The errors become $-1,1,-3,3$, and their mean squared error is $5$. Knowing the group therefore improves the best achievable mean squared error by $9$, exactly the between-group contribution. This is an interpretation relative to the information available: the remaining uncertainty could shrink further if additional information were revealed.

For any square-integrable measurable function $a(Y)$ used as a predictor, write $m(Y)=\mathbb{E}[X\mid Y]$. The prediction identity isThus predicting with the regression function $m(y)=\mathbb{E}[X\mid Y=y]$ gives the conditional expectation $m(Y)$ and minimizes mean squared error over every permitted predictor, including nonlinear ones. This minimizes average squared error, not the error for each individual outcome. See Berkeley Statistics 201A notes, sections 36–37.

Why do the two pieces add cleanly? Put $R=X-m(Y)$. Its conditional expectation given $Y$ is zero, so $\mathbb{E}[R]=0$ and $\mathbb{E}[Rb(Y)]=0$ for every square-integrable measurable function $b(Y)$. In particular, $R$ and $m(Y)$ are uncorrelated random variables. Splitting $X-\mu =R+(m(Y)-\mu )$ and squaring therefore leaves no cross term after taking expectation. Geometrically, the conditional expectation is an orthogonal projection in the L2 space, and the law of total variance is its Pythagorean theorem for centered random variables.

In our example, the four products $R(m(Y)-5)$ are $3,-3,-9,9$, which average to zero. Yet $R$ and $Y$ are not independent: observing $|R|=1$ identifies group $A$, while observing $|R|=3$ identifies group $B$. This is a concrete instance of heteroscedasticity, with different conditional variances despite zero conditional means of the errors. A group can predict uncertainty even when it cannot predict the error's sign.

The law of total variance is useful whenever aggregated data conceal distinct groups. Changing mixture weights changes both contributions, so the weighting must match the population being described. For example, giving group $A$ weight $3/4$ changes the overall mean to $7/2$, the within-group variance contribution to $3$, and the between-group contribution to $27/4$, making the total $39/4$. Group differences account for spread and prediction gains; assigning them a causal explanation requires a separate causal inference argument.

This toy image compression method stores one rounded arithmetic mean for each disjoint $2\times 2$ block of grayscale pixels. The quantization rule is round to the nearest integer, with exact halves rounded upward. Each stored value is an unsigned eight-bit integer. Reconstruction fills the original block with that value, a form of nearest-neighbor interpolation. Averaging already discards information; rounding can discard more.

Here is an exact $16\times 16$ source array. Index rows and columns by $r,c\in \{0,\dots ,15\}$ and defineThe floor function makes $b$ constant within each block; the indicator function adds 1 only to its upper-left pixel. All values are integers from 20 to 215. Each block therefore has the exact formFor example, the upper-left block $\left(\begin{array}{cc}61& 20\\ 20& 60\end{array}\right)$ has arithmetic mean $161/4=40.25$ and is stored as 40. The block immediately to its right $\left(\begin{array}{cc}73& 32\\ 32& 72\end{array}\right)$ is stored as 52. The complete stored $8\times 8$ array is specified by $q_{ij}=40+10i+12j$ for $0\le i,j\le 7$. Its first row is $40,52,64,76,88,100,112,124$; its last is $110,122,134,146,158,170,182,194$.

This toy file format again has eight-bit width and height fields followed by a one-bit mode: 0 means raw eight-bit grayscale pixels, 1 means the rounded block arithmetic means. This profile accepts only width and height 16. The decoder knows the block size and reconstruction rule from the format, reads the 64 stored values in row order, and repeats each across its corresponding $2\times 2$ block. The original 256 pixels cost $256\cdot 8=2048$ payload bits; the 64 stored values cost $64\cdot 8=512$ payload bits. The reduction is $75\%$. These payload counts exclude the common 17-bit file header; complete sizes are 2065 and 529 bits, or 2072 and 536 bits after final zero padding to eight-bit units. There is no stored codebook or additional prediction rule.

The empirical mean squared error averages squared differences over the 256 original pixel positions:Every block has the same four errors, so this value is exact in squared stored shade units. If the unrounded arithmetic mean $b+1/4$ could be reconstructed, the error would instead be $410.1875$; eight-bit quantization adds $0.0625$. That hypothetical comparison does not count fractional means as free eight-bit values.

The broad shade pattern survives, while the alternating bright and dark pixels vanish. Indeed, a second source with every block constant at $b$ produces the identical stored array $q$: exact decoding of both sources is impossible. This makes the storage benefit and the lost detail visible at once. Block averaging is useful for a coarse preview when the individual pixels are dispensable; choosing an acceptable mean squared error or visual result is part of the application, not a guarantee supplied by the encoder.

Lossy compression reduces a representation while permitting irreversible changes to the original. The encoder can map distinct inputs to the same output, so its decoder cannot recover both originals exactly. The useful question becomes how much storage is saved for an acceptable amount of distortion: mean squared error is one numerical criterion, but it does not by itself measure how noticeable a visual change is.

For example, storing the arithmetic mean of four grayscale pixels retains their average shade while discarding their arrangement. Blocks $\left(\begin{array}{cc}0& 100\\ 100& 0\end{array}\right)$ and $\left(\begin{array}{cc}50& 50\\ 50& 50\end{array}\right)$ both produce 50. The former has detail and the latter is uniform, yet the decoder receives the same value. Lossy block averaging makes this ambiguity, its bit saving and its mean squared error exact.

For the zero-truncated Poisson distribution, $21_{\{Y\text{ even}\}}$ is the unique integrable nonrandomized unbiased estimator of $p=1-e^{-\theta }$. Indeed unbiasedness gives ${\sum }_{y\ge 1}t(y){\theta }^y/y!=e^{\theta }+e^{-\theta }-2$, and power series coefficients force the parity values. Its variance is $2p-p^2$; clipping it to $1_{\{Y\text{ even}\}}$ gives mean squared error $p/2$, strictly smaller for $0<p<1$.

In the Bühlmann model, one policy has a random risk parameter $\Theta$. Conditional on $\Theta$, its annual observations $X_i$ are independent and identically distributed random variables, with mean $m(\Theta )$ and variance $v(\Theta )$. Put $m_0=\mathbb{E}[m(\Theta )]$, let $v=\mathbb{E}[v(\Theta )]$ be the expected process variance, and let $a=\mathrm{Var}(m(\Theta ))$ be the variance of hypothetical means. Assume the required second moments are finite.

The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ under mean squared error. After choosing the intercept to make the estimate unbiased, write it as $m_0+{\sum }_{i=1}^n{b}_i(X_i-m_0)$. If $B={\sum }_i{b}_i$, the law of total variance and conditional independence give its errorFor fixed $B$, the Cauchy-Schwarz inequality minimizes the last sum at $b_i=B/n$. Differentiating $a(1-B{)}^2+v{B}^2/n$ gives $B=na/(na+v)$. Thus the credibility factor and credibility premium areThe ratio form assumes $a>0$; when $a=0$ and $v>0$, the conditional mean is constant and $Z=0$. Predicting $X_{n+1}$ rather than its conditional mean adds the constant $v$ to the error and leaves the optimal affine estimate unchanged.

Here the latent intensity has a Pareto distribution with shape three and lower bound one. For $r<3$, its moments are $\mathbb{E}[{\Theta }^r]=3{\int }_1^{\infty }{\theta }^{r-4}d\theta =3/(3-r)$. ConsequentlyThe equality for $v$ uses the Poisson distribution conditional variance $v(\Theta )=\Theta$. There are two years of experience, so $Z=1/2$ and the observed sample mean is $k/2$. The Poisson credibility with a shape-three Pareto intensity is thereforeIn particular it equals $2$ when $k=5$.

For the exact Bayesian credibility estimate, the total count conditional on $\Theta =\theta$ has Poisson distribution of mean $2\theta$. Multiplication of its likelihood by the prior distribution density gives a Bayesian posterior density proportional to ${\theta }^{k-4}{e}^{-2\theta }$ on $\theta >1$. At $k=5$, its normalizing integral and first-moment integral areHence the next-year predictive expected value isThe Bühlmann credibility premium optimizes over affine estimates, whereas the Bayesian posterior mean optimizes over all square-integrable estimates. This explains why the two credibility estimates need not agree for this nonconjugate prior.

For a weakly stationary process with finite second moments, let $\mu =\mathbb{E}{X}_t$ and $\gamma (h)=\mathrm{Cov}(X_{t+h},X_t)$, independent of $t$. If $\gamma (0)>0$, its autocorrelation function is $\rho (h)=\gamma (h)/\gamma (0)$. Given observations $x_1,\dots ,x_N$, one usual sample autocovariance function isThe correlogram plots $\hat{\rho }(h)$ against lag $h$. The partial autocorrelation function at lag $h$ is the correlation between $X_t$ and $X_{t-h}$ after linearly projecting each onto the intervening observations. Its sample version can be calculated by solving the Yule-Walker regression equationsthen taking the last coefficient ${\hat{\varphi }}_{h,h}$. Equivalently, use the corresponding residual correlation from the sample regressions, with a consistent covariance convention.

For a minimal causal autoregressive model of order $p$, the population partial autocorrelation function is zero after lag $p$, while the autocorrelation function follows an exponentially decaying or damped oscillatory tail. For a moving-average model of order $q$, its autocovariance and hence autocorrelation are zero after lag $q$, because observations farther apart share no driving-noise terms; its partial correlations usually tail off. These order identification by autocorrelation cutoffs patterns suggest AR or MA candidate orders. Sample cutoffs are approximate and should be judged with sampling uncertainty and residual diagnostics; the familiar white-noise bands near $\pm 1.96/\sqrt{N}$ are a null approximation, not universal confidence limits for a dependent series.

For the trend problem, write a symmetric estimator asReproducing arbitrary constant and quadratic trends requires $2a+2b+c=1$ and $8a+2b=0$; the linear-trend condition holds automatically by symmetry. These constraints follow by expanding $T_{t+j}=T_t+j(b_0+2c_{0}t)+c_0{j}^2$ for a trend $T_t=a_0+b_{0}t+c_0{t}^2$. Thus $b=-4a$, $c=1+6a$. Since the errors are uncorrelated and the estimator is unbiased, its mean squared error isThe strictly convex quadratic is minimized at $a=-3/35$, giving the five-point symmetric quadratic trend smootherThis weighted moving average allows negative coefficients; a convex average could not cancel the quadratic moment.

For three points, symmetry gives weights $(a,b,a)$. Unbiasedness for every quadratic requires $2a+b=1$ and $2a=0$, so the only solution is $(0,1,0)$. Thus there is no nontrivial three-point quadratic-preserving smoother. The literal claim that there is no unbiased estimator at all is false, since ${\hat{T}}_t=X_t$ is unbiased with error variance ${\sigma }^2$. This is the three-point quadratic reproduction forces the identity filter distinction.

For seasonal extraction by a centred moving average, use the additive decomposition $X_t=T_t+S_t+{\epsilon }_t$, where $S_t$ has period $m$ and its values over one period sum to zero. An average covering a full period removes the seasonal component and approximately retains a slowly changing trend. If $m$ is odd, take the centred length-$m$ average; if $m$ is even, centre it by averaging two adjacent length-$m$ averages, giving half weights at the two endpoints. Subtract this trend estimate from the data, average residuals within each seasonal phase over different cycles, and centre the resulting phase estimates to sum to zero. For a multiplicative decomposition, use ratios and normalize the phase estimates to average one.

Successive moving averages compose by convolution of their weights. Even an initially uncorrelated noise series becomes correlated: for weights $w_j$, the filtered noise has covariance ${\sigma }^2{\sum }_j{w}_j{w}_{j+h}$. Further smoothing changes this covariance again and can turn random variation into apparently persistent patterns while obscuring genuine short-scale features. Spectrally, repeated use of a transfer function $H$ multiplies the spectrum by $|H{|}^{2r}$ after $r$ applications, so repeated smoothing is not harmless repetition of the same estimate.

In simple exponential smoothing, choose $0<\alpha <1$ and update ${\ell }_t=\alpha X_t+(1-\alpha ){\ell }_{t-1}$. Iteration gives ${\ell }_t=\alpha {\sum }_{j=0}^{t-1}(1-\alpha {)}^j{X}_{t-j}+(1-\alpha {)}^t{\ell }_0$, so older observations receive exponentially decreasing weights. Forecast ${\hat{X}}_{t+h\mid t}={\ell }_t$ for every $h\ge 1$. This is appropriate for a locally stable level with no persistent trend or seasonality.

For a local linear trend, double exponential smoothing supplies level and slope. Holt's linear trend method usesAnother standard convention is Brown's double exponential smoothing: smooth twice with the same gain, $s_t^{(1)}=\alpha X_t+(1-\alpha )s_{t-1}^{(1)}$ and $s_t^{(2)}=\alpha s_t^{(1)}+(1-\alpha )s_{t-1}^{(2)}$, then put $a_t=2s_t^{(1)}-s_t^{(2)}$, $b_t=\alpha (s_t^{(1)}-s_t^{(2)})/(1-\alpha )$ and forecast $a_t+h{b}_t$. These variants handle a nonseasonal trend; seasonal effects need separate modelling, and a long-horizon linear extrapolation assumes the current local slope persists.

The Rao-Blackwell theorem says that, for a sufficient statistic $T$ and a finite-second-moment estimator $\delta (X)$, the conditional estimator ${\delta }_T(T)=E_{\theta }[\delta (X)\mid T]$ can be chosen independently of the unknown parameter and has no greater mean squared error for estimating any specified function $g(\theta )$. Sufficiency makes the conditional distribution of the data given $T$ parameter-independent. Conditional expectation preserves the expectation, hence preserves unbiasedness. Writing $\delta -g=(\delta -{\delta }_T)+({\delta }_T-g)$, the cross term has expectation zero, because $E[\delta -{\delta }_T\mid T]=0$. ThereforeEquality holds exactly when $\delta ={\delta }_T$ almost surely. This proves the squared-loss form, including the usual unbiased-estimator variance statement.

Here $\theta >0$ is necessary for the stated uniform interval. Put $L=X_{(1)}={\min }_i{X}_i$ and $R=X_{(n)}={\max }_i{X}_i$. The joint density factors asThe factorization theorem for sufficient statistics shows that $T=(L,R)$ is sufficient. Also $E[X_1]=2\theta$, so $X_1/2$ is unbiased and has variance ${\theta }^2/12$.

For $n\ge 2$, conditional on distinct extrema $l,r$, exchangeability gives probability $1/n$ that $X_1$ is each endpoint; with probability $(n-2)/n$ it is one of the remaining observations, uniform on $(l,r)$. Its conditional mean is consequentlyThus the Rao-Blackwell estimator for a uniform scale interval iswhich is unbiased and has no greater mean squared error. For $n=1$, $L=R=X_1$ and the same formula reduces to the original estimator. As a quantitative check, the covariance of two uniform order statistics givesFor instance, after writing $X_i=\theta +2\theta U_i$, both extreme uniform variances are $n/[(n+1{)}^2(n+2)]$ and their covariance is $1/[(n+1{)}^2(n+2)]$. These formulas follow by integrating the joint extreme density $n(n-1)(r-l{)}^{n-2}$ on $0<l<r<1$; they also display strict improvement for $n>1$.

For the usual causal, stable autoregressive moving-average model with zero-mean white noise, sufficient conditions for a stationary invertible representation areThe roots of $1-\varphi z$ and $1+\theta z$ then lie outside the closed unit disc. For a minimal representation these are the usual causality and invertibility root criteria for an ARMA model. A genuinely minimal ARMA(1,1) additionally has nonzero coefficients and no cancellation $\varphi +\theta =0$; if a coefficient vanishes the order is lower. Common factors can reduce the process to white noise, so inequalities on an unreduced representation should not be asserted as necessary for every reduced process. A noncausal stationary solution of an unstable autoregressive equation is a different representation, and does not give the requested innovation-based Wold coefficients below.

Using the backshift operator $B$, write $(1-\varphi B)X_t=(1+\theta B){ϵ}_t$. The absolutely summable geometric series givesThus the Wold representation isIts coefficients are square-summable, so the series converges in mean square and defines a weakly stationary process. Invertibility makes the closed spans of past observations and past innovations equal. Since the current driving white noise is orthogonal to all past innovations, it is orthogonal to all past observations and is the linear innovation of this causal invertible process.

The spectral density of an ARMA process follows from the squared frequency response of the linear filter of a stationary time series:This displayed density uses the usual two-sided convention. For the one-sided spectral density used in Question 5, multiply it by two and restrict to $0\le \omega \le \pi$.

Let ${\psi }_0=1$ and ${\psi }_j=(\varphi +\theta ){\varphi }^{j-1}$ for $j\ge 1$. Orthogonality of the driving white noise gives $\gamma (h)={\sigma }^2{\sum }_{j\ge 0}{\psi }_j{\psi }_{j+h}$ for $h\ge 0$. HenceFor $h\ge 1$, separate the $j=0$ term and sum the remaining geometric series:Negative lags satisfy $\gamma (-h)=\gamma (h)$. With $\varphi =0$, interpret the lag-one factor as one and later powers as zero, recovering the MA(1) covariance.

To express prediction entirely through observations, invert the moving-average factor:The best linear predictor from the whole past is thereforeIndeed $X_t-{\hat{X}}_t={ϵ}_t$ is orthogonal to the closed linear span of the past. For any other past-measurable linear predictor $L$, the mean squared error decomposes as $E[(X_t-L{)}^2]={\sigma }^2+E[({\hat{X}}_t-L{)}^2]$, establishing optimality and error variance ${\sigma }^2$. The result only uses second-order properties; identifying this linear predictor with a full conditional expectation requires stronger assumptions such as independent innovations or joint Gaussianity. In the cancellation case $\theta =-\varphi$ within the stable range, $X_t={ϵ}_t$, the off-zero covariances and predictor vanish, and the spectrum is constant.

Let $\Phi (z)=1-{\varphi }_{1}z-\cdots -{\varphi }_p{z}^p$. The causality root criterion for an autoregressive model,is a sufficient condition for a weakly stationary process solution, and gives the causal solution $X_t={\sum }_{j\ge 0}{\psi }_j{\epsilon }_{t-j}$. Indeed $1/\Phi$ is analytic in a disk of radius greater than one, so its Taylor coefficients ${\psi }_j$ decay geometrically. The series therefore has mean-square convergence and defines a linear filter of a stationary time series; multiplying by $\Phi$ verifies the equation.

For the specified order-two model, put $\phi =(1+\sqrt{5})/2$. Since $\phi -{\phi }^{-1}=1$,When $\alpha \ne 0$, its roots are $1/(\alpha \phi )$ and $-\phi /\alpha$. Both have modulus greater than one exactly when $|\alpha |<1/\phi$. The case $\alpha =0$ is white noise and satisfies the same inequality. ConsequentlyThis is the causal stationarity condition selected above. Without causality, a two-sided autoregressive model can also have a noncausal stationary autoregression when its polynomial has no unit-circle zeros. Thus the inequality is not a necessary condition for all conceivable weak stationarity solutions; the causal interpretation is the one required by the subsequent innovations and forecasting questions.

To find the Wold representation, expand $1/\Phi (z)={\sum }_{j\ge 0}{\psi }_j{z}^j$. Matching coefficients gives ${\psi }_0=1$, ${\psi }_1=\alpha$ and ${\psi }_j=\alpha {\psi }_{j-1}+{\alpha }^2{\psi }_{j-2}$. HenceHere $F_1=F_2=1$ are the Fibonacci numbers. The coefficient formula also follows by partial fractions in the displayed factorization. Absolute summability follows from $|\alpha \phi |<1$ and $|\alpha /\phi |<1$. The past observation span equals the past noise span, because the causal series gives one inclusion and ${\epsilon }_t=X_t-\alpha X_{t-1}-{\alpha }^2{X}_{t-2}$ gives the reverse inclusion. Thus these noises are the linear innovations, and there is no additional deterministic component. This is the Fibonacci-coefficient autoregression.

The best linear prediction from an infinite past is the orthogonal projection onto the closed span of $X_T,X_{T-1},\dots$. Future white noise is orthogonal to that span. Substitution in the recursion therefore givesThe respective errors are ${\epsilon }_{T+1}$ and $\alpha {\epsilon }_{T+1}+{\epsilon }_{T+2}$, so their mean squared errors are ${\sigma }^2$ and $(1+{\alpha }^2){\sigma }^2$. No normal distribution assumption is needed for these linear forecasts.