Stochastic Processes
A random variable maps the output of a random experiment, $\zeta$, to a real number.
A stochastic process maps the output of a random experiment, $\zeta$, to a time domain waveform.
Examples of stochastic processes.
\(X(t) = \cos(\omega_o t + \Phi)\) where \(\Phi \sim U(-\pi,\pi)\).
\(X(t) = e^{-Yt}\) where \(Y \sim U(0,b)\) and \(t \ge 0\).
First Order Moments.
\(E\{X(t)\} = E\{e^{-Yt}\} = \int_{0}^b e^{-yt} \dfrac{1}{b} dy = \dfrac{1}{bt}(1-e^{-bt})\)
\(E\{X(t)\} = E\{\cos(\omega_o t + \Phi)\} = \int_{-\pi}^{\pi} \cos(\omega_o t + \phi) \dfrac{1}{2\pi} d\phi = 0\)
Autocorrelation (Second order moments): \(R(t_1, t_2) = E\{X(t_1) X(t_2)\}\)
Auto-covariance: \(C(t_1, t_2) = R(t_1, t_2) - \mu(t_1)\mu(t_2)\)
Correlation coefficient: \(r(t_1, t_2) = \dfrac{C(t_1,t_2)}{\sigma(t_1)\sigma(t_2)}\)
Stationarity and Ergodicity
- A process is strict sense stationary (SSS) if its joint PDF and CDF are independent of a shift in the time axis.
Then obviously \(E\{X(t)\} = \mu, R(t_1, t_2) = R(t_2 - t_1) = R(\tau)\)
- A process is weak sense stationary (WSS) if
White Process
A white process means: \(C(t_1,t_2) = 0\) for \(t_1 \ne t_2\).
A stricly white process means \(C(t_1, t_2) = q(t_1) \delta(t_2 - t_1)\)
Gaussian Processes
\[f_{\mathbf{X}}(\mathbf{x}) = \dfrac{1}{\sqrt{(2\pi)^{N/2} \text{det}(\mathbf{C}_{\mathbf{X}})}} \cdot \exp [-\dfrac{1}{2}(\mathbf{x}-\mu_{X})^{\text{T}} \mathbf{C_X}^{-1} (\mathbf{x}-\mu_X) ]\]Ergodicity in the Mean
Ergodicity in the mean means
\[E\{X(t)\} = \mu_X, \quad \mu_X = \lim_{T\to\infty} \dfrac{1}{2T}\int_{-T}^{T} X(t) dt\]The ensemble average has to be a constant that doesn’t dependent on time.
Then it coincides with the time average.
Slutsky’s Theorem
- If \(E\{X(t)\} = \mu_X\) is a constant.
For finite \(T\), the average of one stochastic process relization is a random waveform
\[\mu_{X,T} = \dfrac{1}{2T} \int_{-T}^{T} X(t)dt\]Mean of the random variable
\[E\{\mu_{X,T}\} = \dfrac{1}{2T} \int_{-T}^{T} E\{X(t)\} dt = \mu_{X}\]Variance of the random varibla
\[\begin{align} \sigma_{\mu_{X,T}}^2 &= E\{(\mu_{X,T} - \mu_X)^2\}\\ &= E\bigg\{ \Big( \dfrac{1}{2T} \int_{-T}^{T} X(\alpha)d\alpha -\mu_X \Big) \Big( \dfrac{1}{2T} \int_{-T}^{T} X(\beta)d \beta -\mu_X \Big)\bigg\}\\ &= \dfrac{1}{4T^2} \int_{-T}^{T} \int_{-T}^{T} E\{[X(\alpha)-\mu_X][X(\beta)-\mu_X]\} d\alpha d\beta\\ &= \dfrac{1}{4T^2} \int_{-T}^{T} C_X(\alpha,\beta) d\alpha d\beta \end{align}\]Our stochastic process if ergodic in the mean if
\[\lim_{T \to \infty} \dfrac{1}{4T^2} \int_{-T}^{T} \int_{-T}^{T} C_X(t_1, t_2) d t_1 d t_2 = 0\]If the process is WSS, then it is ergodic in the mean if and only if
\[\lim_{T \to \infty} \dfrac{1}{2T} \int_{-T}^{T} C_X(\tau) d\tau = 0\]The proof can be get by google “Slutsky’s Theorem ergodicity” or this link.
Example: Ergodicity in the mean
Consider a sinusoid with random phase uniformy distributed between \(\pi\) and \(-\pi\), \(X(t) = \cos(\omega t + \Phi)\). Then \(C(\tau) = \dfrac{1}{2}\cos(\omega \tau)\).
\[\lim_{T\to\infty} \dfrac{1}{2T} \int_{-T}^{T} C(\tau) d\tau = 0\]Egrodicity of the Covariance
The covariance time average
\[C_{X,T}(\tau) = \dfrac{1}{2T} \int_{-T}^{T} X(t)X(t+\tau)d t - \mu_X^2\]Spectrum of a Random Signal
Deprecated
- Partition: A partition of a set, \(S\), is a finite series of mutually exclusive subsets that completely define \(S\) such that
- The Axioms of Probability
- Conditional Probability
- Example: IC Testing Proposal
Let \(G_1\) and \(G_2\) represent chips 1 and 2 being good, respectively.
\[\begin{align} P(\{\overline{G_1} \,\overline{G_2}\}) &= 0.01 \quad P(\{G_1 \overline{G_2}\}) = 0.01\\ P(\{\overline{G_1} G_2\}) &= 0.01 \quad P(\{G_1 G_2\}) = 0.97 \end{align}\]To calculate \(P(\{\overline{G_1}\,\overline{G_2}, G_1 \overline{G_2}\} \vert \{\overline{G_1}\,\overline{G_2}, \overline{G_1}G_2\})\)
\[\begin{align} \{\overline{G_1}\,\overline{G_2}, G_1 \overline{G_2}\} \cap \{\overline{G_1}\,\overline{G_2}, \overline{G_1}G_2\} &= \{\overline{G_1}\,\overline{G_2}\}\\ P(\{\overline{G_1}\,\overline{G_2}, G_1 \overline{G_2}\} \vert \{\overline{G_1}\,\overline{G_2}, \overline{G_1}G_2\}) &= \dfrac{0.01}{0.02} = 0.5 \end{align}\]- Total Probability Theorem
If \(A_1, A_2, \dots, A_N\) is a partition of the universal set \(S\), then
\[P(B) = \sum_{i} P(B|A_i)P(A_i)\]- Bayes Theorem
using
\[P(AB) = P(BA) = P(A|B)P(B) = P(B|A)P(A)\]then
\[P(A|B) = \dfrac{P(B|A)P(A)}{P(B)} = \dfrac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\overline{A})P(\overline{A})}\]Why this can be useful? Assume the input to the system is either \(A\) or \(\overline{A}\), and the output is either \(B\) or \(\overline{B}\). The thing is the system includes some random behavior (like noise) such that \(B\) is not fully determined by \(A\). For a real application, we may have an estimation for \(P(A)\) and \(P(\overline{A})\), and we may know the noise model of the system to determine the values such as \(P(B|A)\). Then we can use the Bayes Theorem to calculate \(P(A|B)\)
- Example: Disease Testing
You’re designing a test for a disease. \(D\), that occurs in 0.5% of the population. If the disease is present, your test alerts the medical staff 97% of the time. If the disease is not present, it also give a false positive 1%. If your machine gives an alert, what’s the probability the person actually has the disease?
\[\begin{align} P(D) &= 0.005\\ P(A|D) &= 0.97\\ P(A|\overline{D}) &= 0.01\\ P(A) &= P(A|D)P(D) + P(A|\overline{D})P(\overline{D}) = 0.0148 \end{align}\]then
\[P(D|A) = \dfrac{P(A|D)P(D)}{P(A)} = 0.328\]- Independence: two events \(A\) and \(B\) are independent iff
and thus
\[P(A|B) = P(AB)/P(B) = P(A)\]For \(N\) independent events: \(P(A_1 A_2 \cdots A_N) = P(A_1)P(A_2)\cdots P(A_N)\)
Sometimes two events seems like to influence each others, but they still satisfy the definition of independence. Example: roll two six sided dice and define \(A = \{ \text{rolling a 3 on the first dice} \}\) and \(B = \{ \text{ both dice add to 7 } \}\).
\[\begin{align} A &= (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)\\ B &= (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\\ P(A) &= 1/6\\ P(B) &= 1/6\\ P(AB) &= 1/36 \end{align}\]