Chapter 4: Probability of Thresholds and Maxima

This chapter focuses on the probabilistic analysis of random processes, including threshold crossings and maxima. These concepts are crucial for understanding the reliability and safety of mechanical systems under random vibrations.

1. Summary of Gaussian Processes

Gaussian Variable

A random variable \(X\) is Gaussian if its probability density function is:

\[ P_X(x) = \frac{1}{\sqrt{2 \pi \sigma_X^2}} e^{-\frac{(x - \mu_X)^2}{2 \sigma_X^2}} \]

Gaussian Vector

A random vector \(\mathbf{X} = (X_1, \dots, X_n)^T\) is multivariate Gaussian if its joint probability density function is:

\[ P_{\mathbf{X}}(\mathbf{x}) = \frac{1}{(2\pi)^{n/2} \sqrt{\det C}} \exp \left( -\frac{1}{2} (\mathbf{x} - \boldsymbol{\mu})^T C^{-1} (\mathbf{x} - \boldsymbol{\mu}) \right) \]

where:

\(\boldsymbol{\mu} = (\mu_1, \dots, \mu_n)^T\) is the mean vector.
\(C\) is the covariance matrix with entries \(C_{ij} = \langle (X_i - \mu_i)(X_j - \mu_j) \rangle\).

Gaussian Process

A random process \(X(t)\) is Gaussian if for any times \(t_1, \dots, t_n\), the random vector \((X(t_1), \dots, X(t_n))\) is multivariate Gaussian.

Linear Transformation of a Gaussian Vector

If \(\mathbf{X} = (X_1, \dots, X_n)^T\) is a Gaussian vector and \(A\) is a deterministic matrix, then:

\[ \mathbf{Y} = A \mathbf{X} \]

is also a Gaussian vector with:

Mean: \(\boldsymbol{\mu}_Y = A \boldsymbol{\mu}_X\).
Covariance matrix: \(C_Y = A C_X A^T\).

Thus

\[ P_{\mathbf{Y}}(\mathbf{y}) = \frac{1}{(2\pi)^{n/2} \sqrt{\det A C_X A^T}} \exp \left( -\frac{1}{2} (\mathbf{y} - \boldsymbol{\mu_Y})^T (A C_X A^T)^{-1} (\mathbf{y} - \boldsymbol{\mu_Y}) \right) \]

Theorem

Any linear combination of mutually Gaussian variables is a Gaussian random variable.

Linear Time-Invariant Transformation

Let \(\mathcal{L}\) be a transformation

\[ x(t) \mapsto y(t) = \mathcal{L}[x(t)] \]

\(\mathcal{L}\) is said linear if

\[ \mathcal{L}[\alpha_1 x_1(t) + \alpha_2 x_2(t)] = \alpha_1 \mathcal{L}[x_1(t)] + \alpha_2 \mathcal{L}[x_2(t)] \]

\(\mathcal{L}\) is said time-invariant if

\[ \text{if} \: y(t)=\mathcal{L}[x(t)] \: \text{then} \: y(t-a)=\mathcal{L}[x(t-a)] \]

Example

Any mechanical system governed by a linear differential equation with constant coefficients is a linear time invariant system.

\[ a_n y^{(n)} + \dots + a_0 y = x(t) \quad x(t) \mapsto y(t) = \mathcal{L}[x(t)] \]

\(\mathcal{L}\) is linear because the differential equation is linear.
\(\mathcal{L}\) is time-invariant because the coefficients are constant.

Let \(\mathcal{L}\) be a linear time-invariant transformation \[ y(t) = \mathcal{L}[x(t)] \] Substituting \[ x(t) = \int_{-\infty}^\infty x(t') \delta(t-t') \, dt' \] gives \[ y(t) = \mathcal{L} \left[ \int_{-\infty}^\infty x(t') \delta(t-t') \, dt' \right] \] By linearity \[ y(t) = \int_{-\infty}^\infty x(t') \mathcal{L}\left[ \delta(t-t') \right] \, dt' \] But in the same time \[ y(t) = \int_{-\infty}^\infty x(t') h(t-t') \, dt' \] where \(h(t)\) is the impulse response. We obtain the impulse response of a linear time-invariant system \[ h(t) = \mathcal{L}\left[ \delta(t) \right] \]

Linear Time-Invariant Transformation of a Gaussian Process

Let \(X(t)\) be a Gaussian process and \(\mathcal{L}\) a linear time-invariant transformation of impulse response \(h(t)\). We wonder if \(Y(t) = \mathcal{L}[x(t)]\) is a Gaussian process. But \[ Y(t) = \int_{-\infty}^\infty X(t') h(t-t') \, dt' \approx \sum_i X(t_i) h(t-t_i) \] where \(h(t-t_i)\) is a determistic function. Thus, \(Y(t)\) is a linear combination of the mutually Gaussian processes \(X(t_i)\) and as such, \(Y(t)\) is a Gaussian process.

Derivation as a Linear Time-Invariant Transformation

Let \(X(t)\) be a. Let show that \(\dot X(t) = \mathcal{L}[x(t)]\) is a linear time-invariant transformation. Derivation is

linear: \( \frac{d}{dt} \left( \alpha_1 X_1(t) + \alpha_2 X_2(t) \right) = \alpha_1 \frac{d}{dt} X_1(t) + \alpha_2 \frac{d}{dt} X_2(t) \)
time-ivariant: \( \frac{d}{dt} \left[ X(t-a) \right] = \dot X(t-a) \)

We also have \[ \dot X(t) = - \int_{-\infty}^\infty X(t') \delta'(t-t') \, dt' \] where \(\delta'\) denotes the derivative of the Dirac distribution. Therefore, derivation is a linear time-invariant transformation of impulse response \(h(t)=-\delta'(t)\).

2. Threshold Crossings

The rate of crossings of a level \(a\) by a random process \(X(t)\) is the average number of times per unit time that \(X(t)\) crosses the level \(a\). This is important for understanding the reliability of systems (e.g., fatigue failure).

Rate of Crossings

Figure 4: Rate of crossing a threshold for a random process \(X(t)\).

Define \(Y(t) = H[X(t) - a]\), where \(H\) is the Heaviside step function. \(Y(t)\) is true (i.e. equal to 1) when \(X(t)\) is above the threshold \(a\) and false (i.e. 0) otherwise.

The derivative of \(Y(t)\) is:

\[ \dot{Y}(t) = \dot{X}(t) \delta[X(t) - a] \]

This is a series of Dirac distributions of magnitude \(\pm 1\) depending on the sign of \(\dot X(t)\). The number of crossings of level \(a\) in the interval \([t_1, t_2]\) is obtained by counting the number of Diracs independently of their signs:

\[ N_a(t_1, t_2) = \int_{t_1}^{t_2} |\dot{X}(t)| \delta[X(t) - a] \, dt \]

The rate of crossings is:

\[ n_a(t) = \lim_{t_2 - t_1 \to \infty} \frac{N_a(t_1, t_2)}{t_2 - t_1} \]

But \(n_a(t)\) is a random process. The expectation of rate of crossings the level \(a\) is:

\[ \langle n_a(t) \rangle = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} |\dot{x}| \delta(x - a) p_{X, \dot{X}}(x, \dot{x}) \, dx \, d\dot{x} \]

Case of a Gaussian process

For a Gaussian process \(X(t)\) with mean \(\mu_X\) and standard deviation \(\sigma_X\), we know that \(\dot{X}(t)\) is also Gaussian, centred, and independent of \(X(t)\):

\[ p_{X, \dot{X}}(x, \dot{x}) = p_X(x) p_{\dot{X}}(\dot{x}) = \frac{1}{\sqrt{2 \pi \sigma_X^2}} e^{-\frac{(x - \mu_X)^2}{2 \sigma_X^2}} \cdot \frac{1}{\sqrt{2 \pi \sigma_{\dot{X}}^2}} e^{-\frac{\dot{x}^2}{2 \sigma_{\dot{X}}^2}} \]

The expectation of rate of positive crossings (\(\dot{X} > 0\)) is:

\[ \langle n_a^+ \rangle = \int_{0}^{\infty} \dot{x} p_{X \dot{X}}(a, \dot{x}) \, d\dot{x} = \frac{\sigma_{\dot{X}}}{2 \pi \sigma_X} e^{-\frac{(a - \mu_X)^2}{2 \sigma_X^2}} \]

First Rice's Formula

For a zero-mean Gaussian process (\(\mu_X = 0\)), the rate of crossings zero is:

\[ \langle n_0^+ \rangle = \frac{\sigma_{\dot{X}}}{2 \pi \sigma_X} \]

Example: Mass-Spring System

For a mass-spring system with white-noise excitation \(f(t)\) (PSD \(S_0\)):

\[ \sigma_X^2 = \langle x^2 \rangle = S_0 \int_{-\infty}^{\infty} |H(\omega)|^2 \frac{d\omega}{2\pi} = \frac{S_0}{4 \xi m^2 \omega_0^3} \]

\[ \sigma_{\dot{X}}^2 = \langle \dot{x}^2 \rangle = S_0 \int_{-\infty}^{\infty} \omega^2 |H(\omega)|^2 \frac{d\omega}{2\pi} = \frac{S_0}{4 \xi m^2 \omega_0} \]

Thus, the mean zero-crossing rate is:

\[ \langle n_0^+ \rangle = \frac{\omega_0}{2 \pi} \]

The mean zero-crossing rate is equal to the natural frequency of the resonator!

3. Occurence of Maxima

The rate of maxima beyond a level \(a\) is the average number of times per unit time that \(X(t)\) reaches a local maximum above \(a\). This is important for understanding extreme values in random vibrations.

Rate of Maxima

Figure 5: Rate of maxima above a threshold for a random process \(X(t)\).

Let \(X(t)\) be a random process. Define \(Y(t) = H[\dot X(t)]\) and \(Z(t) = H[X(t) - a]\), where \(H\) is the Heaviside step function. \(Y(t)\) is true (i.e. 1) when \(X(t)\) is increasing and false (i.e. 0) when \(X(t)\) is decreasing. A change of value in \(Y(t)\) indicates the presence of an extremum of \(X(t)\) irrespective of the level reached by \(X(t)\). \(\dot Y(t)\) is therefore a series of Dirac distributions, one for each extremum. Furthermore, \(Z(t)\) is true when \(X(t)>a\) and false otherwise. We thus cancel the extrema below the level \(a\) by multiplying \(\dot Y(t)\) and \(Z(t)\). The number of extrema beyond \(a\) in \([t_1, t_2]\) is:

\[ m_a(t_1, t_2) = \int_{t_1}^{t_2} |\ddot{X}(t)| \delta[\dot{X}(t)] H[X(t) - a] \, dt \]

The rate of extrema is:

\[ m_a(t) = \lim_{t_2 - t_1 \to \infty} \frac{m_a(t_1, t_2)}{t_2 - t_1} \]

The expectation of rate of extrema beyond \(a\) is:

\[ \langle m_a(t) \rangle = \int_{-\infty}^{\infty} \, dx \int_{-\infty}^{\infty} \, d\dot{x} \int_{-\infty}^{\infty} \:\: |\ddot{x}| \delta(\dot{x}) H(x - a) p_{X, \dot{X}, \ddot{X}}(x, \dot{x}, \ddot{x}) \, d\ddot{x} \]

This reduces to

\[ \langle m_a(t) \rangle = \int_{a}^{\infty} \, dx \int_{-\infty}^{\infty} |\ddot{x}| p_{X, \dot{X}, \ddot{X}}(x, 0, \ddot{x}) \, d\ddot{x} \]

A minimum is characterized by the condition \(\ddot{X}>0\) while a maximum verifies \( \ddot{X} < 0 \). So, if we need to count separately the minima and maxima, the reasoning is the same, excepted that the integration with respect to \(\ddot{x}\) is restricted to positive or negative values. It yields the following two formulas.

The expectation of rate of minima beyond \(a\) is:

\[ \langle m_a^-(t) \rangle = \int_{a}^{\infty} \, dx \int_{0}^{\infty} \ddot{x} p_{X, \dot{X}, \ddot{X}}(x, 0, \ddot{x}) \, d\ddot{x} \]

The expectation of rate of maxima beyond \(a\) is:

\[ \langle m_a^+(t) \rangle = -\int_{a}^{\infty} \, dx \int_{-\infty}^{0} \ddot{x} p_{X, \dot{X}, \ddot{X}}(x, 0, \ddot{x}) \, d\ddot{x} \]

In particular, all maxima irrespective of the level \(a\) are obtained when \(a\to -\infty\)

\[ \langle m_a^+(t) \rangle = -\int_{-\infty}^{\infty} \, dx \int_{-\infty}^{0} \ddot{x} p_{X, \dot{X}, \ddot{X}}(x, 0, \ddot{x}) \, d\ddot{x} \]

By permuting the two integrals and introducing the marginal probability density \(p_{\dot{X}, \ddot{X}}(\dot{x}, \ddot{x})\), it yields the expectation of rate of maxima

\[ \langle m^+(t) \rangle = -\int_{-\infty}^{\infty} \ddot{x} p_{ \dot{X}, \ddot{X}}(0, \ddot{x}) \, dx \, d\ddot{x} \]

Case of a Gaussian process

When \(X(t)\) is a stationary Gaussian process two times derivable, we know that \(X(t)\), \(\dot{X}(t)\), and \(\ddot{X}(t)\) are mutually stationary and mutually Gaussian. We also know that by stationarity \(\langle \dot{X} \rangle = \langle \ddot{X} \rangle = 0\) and \( \langle X \dot{X} \rangle = \langle \dot{X} \ddot{X} \rangle = 0\).

To find the joint probability density \(p_{\dot X, \ddot X}(y,z)\), we must calculate the covariance matrix

\[ \mathbf C = \left( \begin{array}{cc} \langle \dot X^2 \rangle & \langle \dot X \ddot X \rangle \\ \langle \dot X \ddot X \rangle & \langle \ddot X^2 \rangle \end{array} \right) = \left( \begin{array}{cc} \sigma_{\dot X}^2 & 0 \\ 0 & \sigma_{\ddot X}^2 \end{array} \right) \]

Since \(p_{\dot X, \ddot X}(y,z)\) is Gaussian

\[ p_{\dot X,\ddot X}(y,z) = \frac{1}{2\pi \sigma_{\dot X} \sigma_{\ddot X}} e^{- \frac{1}{2} \left[ \frac{y^2}{\sigma_{\dot X}^2} + \frac{z^2}{\sigma_{\ddot X}^2} \right] } \nonumber \]

The rate of maxima is then

\[ \langle m^+ \rangle = - \int_{-\infty}^0 z p_{\dot X,\ddot X}(0,z) \, dz = \frac{-1}{2\pi \sigma_{\dot X} \sigma_{\ddot X}} \int_{-\infty}^0 z e^{\left(- \frac{z^2}{2 \sigma_{\ddot X}^2} \right) } \, dz \]

After simplifying, we get

Second Rice's Formula

For a Gaussian process, the rate of maxima is:

\[ \langle m^+ \rangle = \frac{\sigma_{\ddot X}}{2\pi \sigma_{\dot X} } \]

The second Rice formula is similar to the first Rice formula by considering that a maximum of \(X(t)\) is a zero-crossing of \(\dot{X}(t)\).