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At an interior frequency, the real and imaginary Fourier coefficients of a short-memory Gaussian time series converge jointly to independent centered normal variables, each with variance $s(\omega )/2$, where $s$ is the one-sided spectral density. Their squared sum therefore converges to $s(\omega ){\chi }_2^2/2$. Its variance tends to $s(\omega {)}^2$, so an unsmoothed periodogram is not consistent when $s(\omega )>0$. Averaging nearby ordinates can reduce this persistent variance.

Average neighboring periodogram ordinates with nonnegative weights summing to one. Let the number $L_T$ of averaged ordinates tend to infinity while their frequency span, of order $L_T/T$, tends to zero. For a smooth spectral density and suitable short-memory or fourth-order dependence conditions, the bias tends to zero and variance is of order $1/L_T$. Smooth second-order spectral density alone does not ensure this: a time-constant random scale multiplying independent white noise has a smooth unconditional spectrum but a nonergodic random limiting local power.

Add the random thermal force to the deterministic optical trap relaxation time model. In one stationary coordinate of a harmonic optical trap, the overdamped Langevin equation isHere $\xi$ is Gaussian white noise understood through its integrals, $\delta$ is the Dirac delta function, and $\zeta$ is the linear friction coefficient. The covariance coefficient has units ${\mathrm{N}}^2\mathrm{s}$; multiplying by $\delta (t-t^{\prime })$, of units ${\mathrm{s}}^{-1}$, gives squared force. This fluctuation-dissipation relation for a Langevin particle makes the stationary variance ${\sigma }^2=k_{B}T/k$, consistent with the equipartition theorem. The solution is an Ornstein-Uhlenbeck process with autocovariance functionIndeed, at positive lag the noise has zero correlation with the initial position, so $\zeta C_x^{\prime }=-k{C}_x$; its initial value is ${\sigma }^2$. This supplies both a decay time and an amplitude, making a stronger calibration test than a position histogram alone.

Specify the Fourier transform convention before using a power spectral density. For frequency $f$ in hertz, define $S_x^{(2)}(f)={\int }_{-\infty }^{\infty }{C}_x(t)e^{-2\pi ift}dt$ and the one-sided density $S_x^{(1)}(f)=2S_x^{(2)}(f)$ for $f>0$. Integration of the exponential givesThe position power spectral density has units ${\mathrm{m}}^2/Hz$; its integral is a variance, not a standard deviation. At $f=f_c$ it is half its zero-frequency plateau. Angular frequency conventions require a corresponding $2\pi$ conversion in both axis and integration measure. If $x=\beta (V-V_0)$ with position sensitivity $\beta$ in $m/V$, then $S_V^{(1)}=S_x^{(1)}/{\beta }^2$ in ${\mathrm{V}}^2/Hz$ before independent detector noise is added.

For an original invented calibration example, choose $a=0.500\mu \mathrm{m}$, dynamic viscosity $\eta =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, $T=300\mathrm{K}$, $k=0.0200pN/nm=2.00\times 10^{-5}N/m$, and $\beta =100nm/V$. The unbounded no-slip Stokes drag law givesThe corresponding voltage standard deviation is $0.14391\mathrm{V}$, and the position power spectral density plateau is $0.390369\mathrm{n}{\mathrm{m}}^2/Hz$, or $3.90369\times 10^{-5}{\mathrm{V}}^2/Hz$ in the detector. Recovering both $k=k_{B}T/{\sigma }^2$ and $k=2\pi \zeta f_c$ is an internal model check; actual agreement also tests the assumed temperature, drag and detector response.

The generator uses the exact stationary update $x_{j+1}=\rho x_j+\sigma \sqrt{1-{\rho }^2}{Z}_j$, with $\rho =e^{-\Delta t/\tau }$ and independent standard normal random variables $Z_j$. For a demeaned block of $N$ samples and its unnormalized discrete Fourier transform $X_j$, the two-sided periodogram is $\Delta t|X_j{|}^2/N$; doubling positive bins except the zero and Nyquist endpoints gives a one-sided density. Summing those bins times $\Delta f=1/(N\Delta t)$ returns the block mean square by discrete Parseval identity. Independent detector noise, finite exposure and aliasing would alter a measured spectrum and must enter a fit before interpreting $f_c$.

Write ${\sigma }_{\epsilon }^2$ for the driving white noise variance and $\Theta (z)=1+{\sum }_{\ell =1}^q{\theta }_{\ell }{z}^{\ell }$. The autoregressive polynomial factors asThe discrete Gaussian white noise values are independent. The roots of $\Phi$ are outside the unit disc, so expanding $\Theta (z)/\Phi (z)$ gives an absolutely summable causal time-series representation. Hence $x_t$ is a centered Gaussian process formed from driving white noise values at times at most $t$. In particular, for $k>q$, every driving white noise value ${\epsilon }_{t-\ell }$ with $\ell \le q$ is independent of $x_{t-k}$. The autocovariance tail recurrence of a causal ARMA process follows by taking covariance of the recursion with $x_{t-k}$:

The roots of the characteristic polynomial of this tail recurrence are $1/2$ and $1/3$. Let $k_0=\max (q-1,0)$. There exist constants $c,d$ such thatFor example, put $a=6{\gamma }_{k_0+1}-2{\gamma }_{k_0}$ and $b=3{\gamma }_{k_0}-6{\gamma }_{k_0+1}$; then $c=2^{k_0}a$ and $d=3^{k_0}b$ reproduce the two starting values and therefore the whole recurrence. Choose $C$ strictly larger than $|c|+|d|$ and all the finitely many values $2^k|{\gamma }_k|$ with $0\le k<k_0$. Symmetry ${\gamma }_{-k}={\gamma }_k$ now givesFor negative $k$ this is stronger than the requested bound with $2^{-k}$. It also proves ${\sum }_{k\ge 1}k|{\gamma }_k|<\infty$.

For the Fourier calculation, take $D=2m+1$ positive and $T=DN$ positive. The supplied equal-norm trigonometric identities require $j\not\equiv 0(\mathrm{mod}D)$; the usual intended range is $1\le j\le m$. Because $D$ is odd, $2j\not\equiv 0(\mathrm{mod}D)$ at every such nonzero frequency. The geometric series of $e^{i\omega t}$ and $e^{2i\omega t}$ over $T$ terms vanish. They implyOther integer values of $j$ reduce modulo $D$; folding a negative frequency changes the sine coefficient only by a sign. The permitted zero frequency is handled below.

Grouping the double expectation sum according to the time lag yieldsReplace the two truncated sums for lag $k$ by full sums. Their total isEach replacement adds at most $k$ terms, each of absolute value at most one. This gives the finite-record Fourier covariance bound:Both coefficients have expected value zero, so this expectation is their covariance.

For each finite $T$, $(A,B)$ has a multivariate normal distribution because it is a linear transformation of the Gaussian observation vector. If the variances converge to $v_A,v_B$, its characteristic function converges tosince the cross-covariance tends to zero. This is the characteristic function of two independent centered normal variables; zero limiting variances are allowed.

We can calculate the variance limits rather than merely assume their existence. Grouping the cosine covariance sum givesOver a full record, ${\sum }_{t=1}^T{c}_t{c}_{t+k}=(T/2)\cos (\omega k)$. Truncation changes this by at most $k$. The same statements hold for the sine products. Consequentlyand the identical main term and bound apply to $\mathbb{E}{B}^2$. Absolute summability now gives

Specify the spectral convention to identify this limit. LetHere $f_2$ is the two-sided probability density function on $[-\pi ,\pi ]$, and $s$ is the one-sided probability density function on $[0,\pi ]$, satisfying ${\gamma }_k={\int }_0^{\pi }s(\omega )\cos (k\omega )d\omega$. Thus each variance limit is $f_2(\omega )=s(\omega )/2$. For this particular model, the filter representation also gives

For the specified squared-sum estimator, an exact expectation calculation uses $c_t{c}_u+s_t{s}_u=\cos (\omega (t-u))$:The missing tail and the $k/T$ terms vanish; indeed the bias is $O(T^{-1})$ by the exponential covariance bound. Therefore $I(\omega )$ is asymptotically unbiased for the one-sided probability density function $s(\omega )$. Under the common two-sided convention its limit in expectation is $2f_2(\omega )$, so $I(\omega )/2$ is the appropriately normalized estimator.

At an interior frequency with $s(\omega )>0$, the joint Gaussian limit gives the exponential limit of a one-sided periodogram:Moreover, for a centered Gaussian pair,The nondegenerate exponential limiting law rules out convergence in probability to $s(\omega )$. Thus the unsmoothed periodogram is not consistent at frequencies with positive spectrum, despite its asymptotic unbiasedness. If $s(\omega )=0$, the nonnegative estimator has expectation tending to zero and is consistent there by Markov's inequality. Smoothing an increasing number of nearby ordinates while shrinking the frequency bandwidth can remove the positive-spectrum variance obstruction.

For $j\equiv 0(\mathrm{mod}D)$, the sine coefficient is identically zero and the cosine coefficient is the normalized sum of the observations. The corrected limits areThe joint limit is $(N(0,s(0)),0)$, independent in the degenerate sense. The exact expectation formula for $I$ still applies and tends to $s(0)$, but now $I(0)/s(0)\stackrel{d}{\to }{\chi }_1^2$ when $s(0)>0$, with limiting variance $2s(0{)}^2$. The estimator is again inconsistent. This covers the zero frequency allowed by the printed inequality on $j$.

The figure uses the allowed cancellation $\Theta =\Phi$, so the observations reduce to white noise. Orthogonal Fourier projections give the exponential law exactly at each plotted record length; increasing the record does not narrow the individual periodogram ordinate.

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.

With angular frequency $\lambda$, the periodogram isAt Fourier frequencies $2\pi j/T$ it is the squared modulus of the normalized discrete Fourier transform. If a nonzero mean is unknown, subtract the sample mean first.

For a zero-mean stationary process, expanding the square givesUnder absolute summability of the covariances this converges to the spectral density of a stationary process. Thus the periodogram is generally biased at finite $T$, but asymptotically unbiased under this short-memory time series condition.

It is nevertheless not a pointwise estimator with statistical consistency unless it is smoothed. For Gaussian white noise, at a nonzero Fourier frequency other than the Nyquist frequency, the real and imaginary Fourier components are independent normal variables. Exactly,where $f={\sigma }^2/(2\pi )$. The variance does not decrease with $T$. Under usual short-memory time series assumptions the same exponential limit is asymptotic for general processes. Averaging nearby frequencies or using a lag-window estimator reduces variance; a frequency bandwidth tending to zero while $T$ times that bandwidth tends to infinity can give statistical consistency. The raw plot remains useful for detecting strong periodic peaks, but increasing the record length alone does not remove its pointwise noise.