# Tag Info

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White Gaussian noise in the continuous-time case is not what is called a second-order process (meaning $E[X^2(t)]$ is finite) and so, yes, the variance is infinite. Fortunately, we can never observe a white noise process (whether Gaussian or not) in nature; it is only observable through some kind of device, e.g. a (BIBO-stable) linear filter with transfer ...

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Math tools We can do the calculation using some basic elements of probability theory and Fourier analysis. There are three elements (we denote the probability density of a random variable $X$ at value $x$ as $P_X(x)$): Given a random variable $X$ with distribution $P_X(x)$, the distribution of the scaled variable $Y = aX$ is $P_Y(y) = (1/a)P_X(y/a)$. The ...

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Just suppose I give you a series of numbers, and I tell you they were picked randomly. And you know I am not trying to deceive you. Numbers are: $3$, $1$, $4$, $1$, $5$, $3$, $2$, $3$, $4$, $3$. I now propose you to predict the next one, or at least, to be as close as possible. Which number would you pick? [Think] [Compute] I bet most of the readers ...

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A random process is a collection of random variables, one random variable for each time instant. It is best to write the random process as $$\{X(t)\colon -\infty < t < \infty\} \tag{1}$$ where the $\{$ and $\}$ indicate that a set (or collection) of objects is being defined, and the interior says that a typical member of this set is denoted by $X(t)$ ...

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when we observe the Random Process at a specific time $t_k$, that is the value at $X(s_1,t_k), X(s_2,t_k),\ldots,X(s_n,t_k)$, if we denote them by $(a_1,a_2,\ldots, a_n)$. Now the mapping between the outcomes $(s_1,s_2,s_3,\ldots,s_n)$ and its probabilities $(a_1,a_2,\ldots,a_n)$ are collectively called as Random Variable. Not true. Random variable ...

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Most realistic signals are both random and periodic. For example, you can modulate a harmonic oscillator with a slow enough random signal that moves its frequency around a $\mu_{f}, \sigma_f$. This looks like: $$y= \sin \left( \frac{2 \pi \mathcal{N_s}(\mu_f, \sigma_f) n}{Fs} \right )$$ Where $\mathcal{N_s}(\cdot)$, denotes a normally distributed random ...

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When we say the stochastic process $X_{t_1}X_{t_2}\cdots X_{t_i}$ is a Gaussian Process with mean $\mu$ and variance $\sigma^2$ it only means $$X_{t_i}\sim\mathcal{N}(\mu,\sigma^2)$$ So still we know nothing about the relation of the $X_{t_i}$'s with each other. When we add that the process is also white, it means the $X_{t_i}$'s are uncorrelated. That is $$... 7 Calculate the autocorrelation of the process.$$\begin{align} R_{xx}[n] &=\mathbb{E}[(W[k] + c W[k-1])(W[k-n] + c W[k-1-n])] \\ &=\mathbb{E}[W[k]W[k-n]]+ \mathbb{E}[cW[k]W[k-1-n]]+\mathbb{E}[cW[k-1]W[k-n]]+\mathbb{E}[c^2W[k-1]W[k-1-n]] \\ &=\sigma^2\delta[n]+c\sigma^2\delta[n+1]+c\sigma^2\delta[n-1]+c^2\sigma^2\delta[n]\\ &=\sigma^2(1+c^2)\...

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No. Quoting Wikipedia's article Independence (probability theory): If $X$ and $Y$ are independent random variables, then the expectation operator $\operatorname{E}$ has the property $$\operatorname{E}[X Y] = \operatorname{E}[X]\operatorname{E}[Y].$$ Consider your $X(t_1)$ and $Y(t_2)$ as $X$ and $Y$ in this answer. If both $\operatorname{E}[X] \ne ... 6 From the wikipedia article: a stochastic process is said to be ergodic if its statistical properties can be deduced from a single, sufficiently long, random sample of the process. In other words: the time-ensemble statistical properties are the same as the realization-ensemble statistical properties. Maybe we need to take a step back and talk about what ... 6 In the usual sense of the term, first-order stationarity means that the first-order distribution of all the random variables is the same: each$X_t$has the same CDF, and so the same pdf (or pmf) too if the random variables are continuous (or discrete). If the random variables have a mean, then they all have the same mean. But, a sequence of independent ... 6 The definition of the autocorrelation function$R_x(\tau)$depends on the nature of your$x$. If$x$is a deterministic signal with finite energy then: $$R_x(\tau)=\int_{-\infty}^{+\infty}x(t)x^*(t-\tau)dt$$ If$x$is a deterministic signal with finite average power$^{(1)}$then: $$R_x(\tau)=\lim_{T\to+\infty}\frac{1}{T}\int_{-T/2}^{+T/2}x(t)x^*(t-\tau)dt$$... 6 A binary symmetric channel (BSC) can be characterized by its complemented probability$p$. Its well-known capacity is $$C = 1 - H(p) = 1 - (-p\log(p) - (1-p)\log(1-p))$$ where$H(p)$is binary entropy function: A$L-$concatenated BSC, which is also a BSC characterized by$p_L$, can be visualized as in the figure below The complemented probability$p_L$... 6 If you are talking about a given signal as "a deterministic realization of a phenomenon", it can be periodic, but not really random. However, some physical systems are prone to produce randomness and periodicity, like rotating machines, gears, cyclic engines, that produce signals similar to: Naturally rotating bodies (stars, planets) also produce random ... 5$\sin(t)$is no random process, because there's nothing random about it. You could add a random amplitude to get a random process: $$x(t)=A\sin(t)\tag{1}$$ This is a random process because$A$is a random variable. However,$x(t)$is not stationary, but it is cyclostationary, i.e., its statistical properties vary periodically. You can make the process$x(t)...

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Note that the condition $$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty\tag{1}$$ (i.e., that the signal $f(t)$ has finite energy) is very restrictive when we try to model signals, even though obviously any actually occurring signal must have finite energy. Modeling signals as random processes means that we ignore condition $(1)$. Models are always ...

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The answer to the question (a counterexample) Properties of random processes will in general be time-dependent. They are not only when talking about stationary processes. Another related concept (not relevant here but that you might find interesting) is ergodicity. Most of us find it difficult to understand the difference between both concepts, but then you ...

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Suppose we have a discrete-time sequence $x[t]$ which is stationary, zero mean, white noise with variance $\sigma^2$. Then the autocorrelation of $x$ is: $$\begin{array} RR_{xx}[\tau] &=& E\left[ x[t] x[t+\tau] \right]\\ &=& \left \{ \begin{array} EE \left[ x[t]^2 \right], {\rm if\ }\tau=0 \\ 0, {\rm otherwise} \end{array} \right. \\ &=&... 4 Since the question has been raised as to whether the hint that I had given to the OP in a comment on the original question was appropriate for a newcomer to signal processing, here goes. Stripped of extraneous baggage and notation, the question is whether it is possible to determine the value of E[X^2Y^2] straightforwardly where X and Y are zero-... 4 If you sample a random process for a specific t, you will get one realization of a random variable. For another t, you get another realization of that random variable. This random variable has its statistics which is almost impossible to learn in real world because not all sample paths are observable. See the brown rectangle in the figure below. That ... 4 In easy words: A process is stationary if its stochastic properties are independent of the time you look at it. Think of it like this: A stochastic process is just a Random Variable (RV) that, instead of giving you e.g. a real value gives you a function every time you look at it. We call that realizations. If you now take a lot of these realizations, and ... 4 Mean of a signal can be practically visualized as the dc average value present in the signal (for a complete sinusoidal period), for e.g Variance of a signal is the difference between the normalized squared sum of instantaneous values with the mean value. In other words it provides you with the deviation of the signal from its mean value. It gives you the ... 4 Adding to Zeeshan's useful answer, here are some additional comments toward some their practical use (not limited to this but may helps add intuition into their use depending on your background): Variance: The average "AC" power quantity of a signal is directly proportional to the variance (simply the average of the squares; this relation to "power" is ... 4 Think it this way; assuming x[-1] = 0, then recusively compute the output x[k] for k \ge 0 such as$$\begin{align} x &= v \\ x &= a x + v \\ x &= a x + v = a^2 x + a v + v = a^2 v + a v + v \\ x &= \left( a^2 v + a v \right) + v \\ ... &= ... \\ x[k] &= \left( a^k v + a^{...

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It is usually more difficult to understand the non-ergodic case (that is why people look for examples of such processes more often). As an example of ergodic process, let the process $X(t)$ represent repeated coin flips. At each time $t$, we have a random variable $X$ that can choose between $0$ or $1$. If it is a fair coin, then the ensemble mean is $\frac{... 3 The limit$\lim_{\tau\to\infty} R_x(\tau)$, if it exists, equals$E^2[X(t)]$and so$E[X(t)]=0$in this case. More generally, the mean of a WSS process is nonzero only if the power spectral density has an impulse at the origin. This can be applied to periodic autocorrelation functions such as$\cos(\omega_0t)$pointed out in @MattL's comment. If the ... 3 For this problem you can't use the formula involving$|H(f)|^2$because it only applies to linear time-invariant (LTI) systems, and a squarer is obviously a non-linear system. The only way to solve this problem that I can think of is to use the formula $$E\{x^2y^2\}=E\{x^2\}E\{y^2\}+2E^2\{xy\}\tag{1}$$ which is valid for jointly Gaussian and zero mean ... 3 Joint behavior cannot always be deduced from individual behavior. For example, if$X$and$Y$are (nondegenerate) random variables with finite means, then$P\{X<E[X]\}$and$P\{Y<E[Y]\}$both are nonzero. This is because there must be probability mass to the left of the mean (and to the right of the mean too) except in the degenerate case when$X\$ has ...

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Stationarity is a statitical concept defined through stochastic processes. However the following is a practical account of why most typical images are nonstationary : Throughout the image window, consider a small rectangular test block, say 16x16, and observe the frequency spectrum estimation of that block. Now move this test block accross the image and re-...

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