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Let \(X(t)=Y \cos (\theta t)+Z \sin (\theta t)\) where \(Y\) and \(Z\) are independent \(N(0,1)\) random variables, and let \(\tilde{X}(t)=R \cos (\theta t+\Psi)\) where \(R\) and \(\Psi\) are independent. Find distributions for \(R\) and \(\Psi\) such that the processes \(X\) and \(\tilde{X}\) have the same fdds.

Short Answer

Expert verified
R ~ Rayleigh, 唯 ~ Uniform[-蟺, 蟺].

Step by step solution

01

Expression of Amplitude and Phase of X(t)

The expression \(X(t) = Y \cos(\theta t) + Z \sin(\theta t)\) resembles a sinusoidal function with an amplitude and phase. We can express it in the form \(R \cos(\theta t + \Psi)\), where \(R\) is the amplitude and \(\Psi\) the phase shif. To express it like this, we use the identity:\[ R \cos(\theta t + \Psi) = R ( \cos(\Psi)\cos(\theta t) - \sin(\Psi)\sin(\theta t) ). \] By comparing terms, we have that \(Y = R \cos(\Psi)\) and \(Z = -R \sin(\Psi)\).
02

Derive expression for R

The amplitude \(R\) can be found by comparing the coefficients: \[ R = \sqrt{Y^2 + Z^2} \]This expression results from considering the sum of squares of the equations \(Y = R\cos(\Psi)\) and \(Z = -R\sin(\Psi)\).
03

Distribution of R

Given that \(Y\) and \(Z\) are independent \(N(0,1)\) random variables, \(Y^2 + Z^2\) follows a chi-squared distribution with 2 degrees of freedom (also known as an exponential distribution). Thus, the distribution for \(R\) is \(\sqrt{Y^2 + Z^2}\), which follows a Rayleigh distribution.
04

Derive expression for 饾毧

From the expressions \(Y = R \cos(\Psi)\) and \(Z = -R \sin(\Psi)\),\[ \tan(\Psi) = -\frac{Z}{Y} \]\(\Psi\) is given by \(\textrm{atan2}(-Z, Y)\) where \(\textrm{atan2}\) is the inverse tangent function that takes into consideration the signs of both arguments to determine the correct quadrant.
05

Distribution of 饾毧

For independent standard normal random variables \(Y\) and \(Z\), the distribution of \(\Psi = \textrm{atan2}(-Z, Y)\) is uniform on \([-\pi, \pi]\) due to rotational symmetry. Thus, \(\Psi\) is uniformly distributed.
06

Conclusion: \(\tilde{X}(t) = R \cos(\theta t + \Psi)\) matches \(X(t)\)

With \(R\) having a Rayleigh distribution and \(\Psi\) uniformly distributed on \([-\pi, \pi]\), process \(\tilde{X}(t)\)has the same finite-dimensional distributions as \(X(t)\), thus making them equivalent in distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables
Random variables are an essential concept in probability and statistics that represent values resulting from random phenomena. To understand this, imagine a process like rolling a die. The result of each roll is uncertain, and each potential outcome (like 1, 2, 3, etc.) can be seen as a random variable.

In the context of the given exercise, random variables are utilized to model components of a sinusoidal function. Here, two standard normal random variables, denoted as \(Y\) and \(Z\), are used. Both \(Y\) and \(Z\) are independent, meaning the outcome of \(Y\) has no influence on \(Z\) and vice versa. They each follow a normal distribution with a mean of 0 and a variance of 1, denoted as \(N(0,1)\). This type of distribution is also known as a Gaussian distribution, which is symmetric around the mean.

Understanding these random variables helps in analyzing the behavior of more complex stochastic processes, such as the one in the exercise. Their independence and distribution play a crucial role in deriving the distributions of derived quantities like \(R\) and \(\Psi\). This is an important insight since knowing how random variables operate gives us the tools to predict and characterize the behavior of stochastic processes.
Rayleigh Distribution
The Rayleigh distribution is a continuous probability distribution commonly used to model the magnitude of a vector in two-dimensional space, especially when components are normally distributed. Imagine walking in a direction determined randomly and your steps are represented by independent normally distributed variables. The distance from the start, modeled as the vector length, will follow a Rayleigh distribution.

In the exercise, we derived the amplitude \(R\) from \(Y\) and \(Z\), two independent standard normal random variables. By computing \(R = \sqrt{Y^2 + Z^2}\), we found it follows a Rayleigh distribution. This follows from the properties of chi-squared distributions, where the sum of squares of independent standard normal variables gives us a chi-squared distribution. When we take the square root of this sum (as in \(R\)), it becomes Rayleigh-distributed.

Numerous real-life phenomena use the Rayleigh distribution as a model, such as signal strength in wireless communications or wave heights. Understanding this distribution helps in comprehending how amplitude changes in fluctuating environments, illustrating the importance of the derivation in the exercise.
Uniform Distribution
The uniform distribution is a simple yet fundamental concept in statistics where all outcomes are equally likely. Consider spinning a perfectly balanced roulette wheel; each number on the wheel has an equal chance of being the result. This is an example of a uniform distribution.

In our exercise, the phase angle \(\Psi\) was found to follow a uniform distribution over the interval \([-\pi, \pi]\). This outcome arises due to the rotational symmetry present in the trigonometric components \(Y = R \cos(\Psi)\) and \(Z = -R \sin(\Psi)\). The use of the function \(\textrm{atan2}(-Z, Y)\) ensures \(\Psi\) is calculated accurately, considering the signs of \(Y\) and \(Z\) to identify the correct quadrant.

Recognizing that all angles are equally probable provides uniformity in the directionality of the process, capturing rotational symmetry. This understanding is pivotal in ensuring the processes \(X(t)\) and \(\tilde{X}(t)\) do have equivalent finite-dimensional distributions, as required by the task.

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