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Write $C_j={\sum }_{t=1}^T{y}_t\cos ({\omega }_{j}t)$ and $S_j={\sum }_{t=1}^T{y}_t\sin ({\omega }_{j}t)$. The discrete Fourier transform coefficient is $C_j-i{S}_j$. Alternatively, by the given orthogonality relations, the harmonic obtained by linear least-squares projection has coefficients $a_j=2C_j/T$ and $b_j=2S_j/T$. Thus the source's periodogram is the squared amplitude of a fitted harmonic, scaled as power per unit angular frequency:It measures how much of the centered record varies at frequency ${\omega }_j$. For odd $T$, Fourier Parseval identity givesThe positive-frequency spacing is $2\pi /T$, so summing the power ordinates times this spacing estimates the total variance. The factor $1/(\pi T)$ corresponds to the one-sided spectral density on $(0,\pi )$; the usual two-sided periodogram is half as large.

If the observations are Gaussian white noise with common variance ${\sigma }^2$, the $2m$ harmonic sums are jointly normally distributed. Their means are zero even with a nonzero constant series mean, because each sine and cosine sums to zero. Orthogonality givesUncorrelated coordinates of a multivariate normal distribution are independent. Therefore all $C_j/\sqrt{T{\sigma }^2/2}$ and $S_j/\sqrt{T{\sigma }^2/2}$ are independent standard normals, andSince the chi-squared distribution with two degrees of freedom has mean two and variance four,Here $f_{+}$ is the one-sided white-noise spectral density. Its exact unbiasedness is accompanied by a nonconcentrating distribution: along Fourier frequencies tending to any interior frequency, $I/f_{+}$ has the same nondegenerate exponential law at every $T$. It therefore does not converge in probability to one, proving failure of statistical consistency, not just nonvanishing variance.

For a general stationary process, exact finite-sample unbiasedness is not implied. Expanding the Fourier square instead giveswhere $\gamma$ is the autocovariance function. Under suitable short-memory regularity this tends to $f_{+}(\omega )$ at frequencies ${\omega }_j\to \omega$. The periodogram is thus generally asymptotically unbiased, while the exact assertion above is for white noise.

Use local periodogram smoothing to reduce variance: average $L_T$ neighboring ordinates with weights summing to one, where $L_T\to \infty$ but $L_T/T\to 0$. At an interior frequency the shrinking frequency window makes the bias tend to zero for smooth $f_{+}$, while averaging many weakly dependent ordinates makes the variance of order $1/L_T$. For white noise with equal weights, the ordinates are exactly independent and the variance is exactly $f_{+}^2/L_T$. For a general process, impose suitable short-memory and fourth-order dependence assumptions, such as a stationary Gaussian model with sufficiently summable autocovariances. These conditions, together with smoothness, give a consistent smoothed estimator. Boundary frequencies require an appropriately one-sided or reflected window.

Smooth spectral density alone is not enough. For example let $y_t=A{ϵ}_t$, where $A>0$ is a nondegenerate random variable constant throughout the record, independent of independent unit-variance Gaussian ${ϵ}_t$, with $E[A^2]<\infty$. The unconditional autocovariance is zero off lag zero and its one-sided spectral density is the smooth constant $E[A^2]/\pi$. Nevertheless a long local average estimates the realized $A^2/\pi$, not that deterministic unconditional value. This nonergodic example explains why a dependence/ergodicity condition accompanies the smoothing argument.

For the turning point test of randomness, the null is independent identically distributed continuous observations. Let $J_t$ be one if $y_t$ is a strict local maximum or minimum among $y_{t-1},y_t,y_{t+1}$, for $2\le t\le T-1$, and set $P_T={\sum }_t{J}_t$. Three ranks are equally likely; the middle observation is an extremum in four of the six orders, giving $E[J_t]=2/3$ and $\mathrm{Var}(J_t)=2/9$. ThereforeIndicators more than two positions apart depend on disjoint observations and are independent. For neighboring indicators, both are one precisely when four observations alternate up-down-up or down-up-down. By the probability-integral transform, use four independent uniforms. The first pattern has probabilityhere $b=y_t$, $c=y_{t+1}<b$, with the two outer inequalities having probabilities $b$ and $1-c$. The reverse pattern has the same probability, so $E[J_t{J}_{t+1}]=5/12$ and $\mathrm{Cov}(J_t,J_{t+1})=5/12-4/9=-1/36$.

At separation two, the triples share only one endpoint value $c$. Conditional on it, their other observations are independent. The probability that either triple's central observation is a turning point isIt follows that $E[J_t{J}_{t+2}]={\int }_0^{1}r(c{)}^{2}dc=9/20$, so this covariance is $9/20-4/9=1/180$. Combining all terms for $T\ge 4$ givesThe standardized count is approximately standard normal for large $T$, by a central limit theorem for bounded 2-dependent indicators. A two-sided randomness test rejects for large absolute values ofToo few turns suggests persistent movement, while too many suggests alternation. Exact rank-permutation calibration is possible at small $T$. Ties invalidate the continuous-rank probabilities unless their handling is specified. The test probes this feature of randomness; nonrejection does not establish independence.