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I want to present a question which seems to predict a theorem I know. Hence I guess I'm missing something and would be happy to understand what I'm missing.

Here's is the system: enter image description here
The signals $ {X}_{1}, {X}_{2} $ are Independent White Gaussian Noise Process.
The filters $ {H}_{1}(f), {H}_{2}(f) $ are identical, namely, $ {H}_{1}(f) = {H}_{2} $.
The phase, $ \theta \sim U[0, 2\pi] $ and independent of any other parameter or signal.

The questions are

  1. Are $ {Y}_{1}(t), \ {Y}_{2}(t) $ Gaussian Processes?
  2. Is $ r(t) $ Gaussian Process?

The answers, according to a book are $ {Y}_{1}, \ {Y}_{2} $ aren't Gaussian Processes and $ r(t) $ is a Gaussian Process.
Yet $ {Y}_{1}, \ {Y}_{2} $ are obviously independent (Edit: They are dependent, see answers and comments below) and according to Cramer's Theorem:
http://en.wikipedia.org/wiki/Normal_distribution#Infinite_divisibility_and_Cram.C3.A9r.27s_theorem
If the sum of two independent R.V. is Gaussian then each of them must be Gaussian.
Yet it's pretty clear to me that Y1 / Y2 aren't Gaussian.
I wrote the following MATLAB Simulation:

numRealizations = 1e4;
numSamples      = 1e3;

h1Coef = ones(10, 1) / 10;
h2Coef = h1Coef;

harmonicFreq    = 10; %<! [Hz]
timeVector      = [0:(numSamples - 1)];
randomPhaseVec  = 2 * pi * rand(numRealizations, 1);

% White Noise is Ergodic
% Each Row is a realization (Over the Time Vector)
x1SignalMatrix  = randn(numRealizations, numSamples);
x2SignalMatrix  = randn(numRealizations, numSamples);

% Filtering along the columns
z1SignalMatrix = filter(h1Coef, 1, x1SignalMatrix, [], 2);
z2SignalMatrix = filter(h2Coef, 1, x2SignalMatrix, [], 2);

y1SignalMatrix = z1SignalMatrix .* cos(bsxfun(@plus, (2 * pi * harmonicFreq * timeVector), randomPhaseVec));
y2SignalMatrix = z2SignalMatrix .* sin(bsxfun(@plus, (2 * pi * harmonicFreq * timeVector), randomPhaseVec));

rSignalMatrix = y1SignalMatrix + y2SignalMatrix;

timeIndex           = randi([1, numSamples], [1, 1]);
realizationIndex    = randi([1, numRealizations], [1, 1]);

figure;
hist(x1SignalMatrix(:, timeIndex), 500);
figure;
normplot(x1SignalMatrix(:, timeIndex));


figure;
hist(z1SignalMatrix(:, timeIndex), 500);
figure;
normplot(z1SignalMatrix(:, timeIndex));


figure;
hist(y1SignalMatrix(:, timeIndex), 500);
figure;
normplot(y1SignalMatrix(:, timeIndex));

figure;
hist(y1SignalMatrix(realizationIndex, :), 500);
figure;
normplot(y1SignalMatrix(realizationIndex, :));


figure;
hist(rSignalMatrix(:, timeIndex), 500);
figure;
normplot(rSignalMatrix(:, timeIndex));

figure;
hist(rSignalMatrix(realizationIndex, :), 500);
figure;
normplot(rSignalMatrix(realizationIndex, :));

This is the distribution of Y1 over the Ensemble: The Distribution of Y1 over the Ensemble

Which is clearly not Gaussian as I expected.

This is the distribution of Y2 over the time (One Realization): The Distribution of Y2 over Time - 1 Realization

Now, the results for $ r(t) $:

The Distribution of the Ensemble: enter image description here

The Distribution of 1 Realization: enter image description here

Well, for Y1, I can see why the ensemble won't Gaussian as for each 't' the sample is multiplied by a different factor hence each of them, though is Gaussian, is coming from a different distribution.
I even wrote a proof why it doesn't comply with Gaussian Distribution properties (The fourth Moment) - http://imgur.com/l017RrR. Yet I can't get how $ r(t) $ is Gaussian.
What am I missing?

Thank You.

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    $\begingroup$ Readers should be aware of a previous version of this question on dsp.SE. $\endgroup$ – Dilip Sarwate Sep 22 '13 at 16:45
  • $\begingroup$ @DilipSarwate, They are connected, yet, as we discussed there, The ensemble of the system aren't linear. Hence I don't see why you made the connection. Do you have any solution for this problem? As your first guess of a Linear System isn't right. $\endgroup$ – Royi Sep 22 '13 at 17:04
  • $\begingroup$ It is interesting that you claim that the question you posed here is different from the one that you posted on dsp.SE and yet you have posted the same answer here as well as on dsp.SE and have accepted both as the best answers to the allegedly different questions. $\endgroup$ – Dilip Sarwate Sep 29 '13 at 15:13
  • $\begingroup$ It is the same question. Only different views of the problem. On the Signal Processing forum I tried going by proving it is an LTI or Linear to the least to show the distribution. Since I saw we couldn't solve it by this approach I went the Random Process approach. At the end I found the solution using the "Characteristic Function". Which is the right solution. I would have marked your answer, yet I don't agree with the assumption "Let Z=XcosΘ+YsinΘ. Since X and Y are independent of Θ, their conditional joint density given that Θ=θ is the same as their unconditional density.". $\endgroup$ – Royi Sep 29 '13 at 22:01
  • $\begingroup$ @DilipSarwate, If you could clarify your meaning by "Let Z=XcosΘ+YsinΘ. Since X and Y are independent of Θ, their conditional joint density given that Θ=θ is the same as their unconditional density." and prove it I'd mark your answer as the answer. $\endgroup$ – Royi Sep 29 '13 at 22:02
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I believe that, stripped to its essentials, this question is asking the following.

Let $X, Y, \Theta$ denote independent random variables and suppose that $X, Y \sim N(0,1)$. Then,

  • Is $X\cos \Theta + Y\sin \Theta$ a normal random variable?

  • Are $X\cos \Theta$ and $Y\sin\Theta$ normal random variables? Are they independent random variables? Are they uncorrelated random variables?

  • Do any of the answers to the above questions change if $\Theta$ has a specific distribution, e.g. $\Theta \sim U[0,2\pi)$?


Let $Z = X\cos\Theta+Y\sin\Theta$. Since $X$ and $Y$ are independent of $\Theta$, their conditional joint density given that $\Theta = \theta$ is the same as their unconditional joint density. Thus, $$f_{X,Y\mid \Theta}(x,y\mid \Theta=\theta) = \frac{1}{2\pi}\exp\left(-\frac{x^2+y^2}{2}\right), ~ -\infty < x, y < \infty.\tag{1}$$



Edit added in response to OP's question:

Proof of claim: If $(X,Y)$ is given to be independent of $\Theta$, we have that $$f_{X,Y,\Theta}(x,y,\theta) = f_{X,Y}(x,y)f_\Theta(\theta), \qquad \text{definition of independence}$$ Consequently, $$\begin{align} f_{X,Y\mid \Theta}(x,y\mid \Theta=\theta) &= \frac{f_{X,Y,\Theta}(x,y,\theta)}{f_\Theta(\theta)} &\text{definition of conditional density}\\ &= \frac{f_{X,Y}(x,y)f_\Theta(\theta)}{f_\Theta(\theta)} &\text{substitute from above}\\ &= f_{X,Y}(x,y)\end{align}$$ Thus, $f_{X,Y\mid \Theta}(x,y\mid \Theta=\theta)$, the conditional joint density of $(X,Y)$ given $\Theta = \theta$ is the same as $f_{X,Y}(x,y)$, the unconditional joint density of random variables $X$ and $Y$ that have never heard of $\Theta$ and don't know that they are independent of $\Theta$. Note that it is not necessary to assume that $X$ and $Y$ are independent of each other: just that they are independent of $\Theta$. But, in this case, $X$ and $Y$ are assumed to be independent standard normal random variables with joint density $f_{X,Y}(x,y)$ given by the right side of $(1)$, and thus $f_{X,Y\mid \Theta = \theta}(x,y\mid\Theta = \theta)$ equals the right side of $(1)$.

For future reference, note that since $$\begin{align} f_{X,Y\mid \Theta = \theta}(x,y\mid \Theta = \theta) &= \frac{1}{2\pi}\exp\left(-\frac{x^2+y^2}{2}\right),\\ &= \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right) \times \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{y^2}{2}\right),\\ &= f_X(x)f_Y(y)\\ &=f_{X\mid \Theta=\theta}(x\mid \Theta = \theta) f_{Y\mid \Theta=\theta}(y\mid \Theta =\theta) \end{align}$$ where the last equality follows from the observations that, by independence, $$\begin{align} f_{X\mid \Theta=\theta}(x\mid \Theta = \theta) &= \frac{f_{X,\Theta}(x,\theta)}{f_\Theta(\theta)} = \frac{f_X(x)f_\Theta(\theta)}{f_\Theta(\theta)}=f_X(x)\\ f_{Y\mid \Theta=\theta}(y\mid \Theta = \theta) &= \frac{f_{Y,\Theta}(y,\theta)}{f_\Theta(\theta)} = \frac{f_Y(y)f_\Theta(\theta)}{f_\Theta(\theta)}=f_Y(y) \end{align}$$

Thu, $X$ and $Y$ are conditionally independent of each other given $\theta$, and of course they are unconditionally independent as well.

End of edit added in response to OP's question:



Thus, conditioned on $\Theta = \theta$, $Z = X\cos\theta + Y\sin\theta$ is a weighted sum of two independent standard normal random variables. So, the conditional distribution of $Z$ is a normal distribution, and in fact, a standard normal distribution. This follows readily from $$\begin{align} E[Z\mid \Theta=\theta] &= E[X\cos\theta+Y\sin\theta] = 0,\\ \operatorname{var}(Z\mid\Theta=\theta) &= \operatorname{var}(X)\cos^2\theta + \operatorname{var}(Y)\sin^2\theta = 1\cdot\cos^2\theta + 1\cdot\sin^2\theta = 1. \end{align}$$

Since the conditional density of $Z$ given $\Theta = \theta$ is the standard normal density regardless of the value of $\Theta$, it follows that

$\quad$ the unconditional density of $Z = X\cos\Theta + Y\sin\Theta$ is the standard normal density.

Note that the distribution of $\Theta$ is immaterial. Indeed, $\Theta$ could be a degenerate random variable that takes on a fixed value $\theta$ with probability $1$, and the unconditional density of $Z$ would still be the standard normal density. In particular, $X\cos\Theta+Y\sin\Theta$ is a standard normal random variable when $\Theta \sim U[0,2\pi)$.


Let $\hat{X} = X\cos\Theta$ and $\hat{Y} = Y\sin\Theta$. These are uncorrelated random variables since independence of $X,Y,\Theta$ gives us that $$\begin{align} E[\hat{X}] &= E[X\cos\Theta] = E[X]E[\cos\Theta]=0\cdot E[\cos\Theta] = 0,\\ E[\hat{Y}] &= E[Y\sin\Theta] = E[Y]E[\sin\Theta]=0\cdot E[\sin\Theta] = 0,\\ E[\hat{X}\hat{Y}] &= E[XY\cos \Theta\sin \Theta] = E[X]E[Y]E[\cos \Theta \sin \Theta] = 0\cdot 0\cdot E[\cos \Theta \sin \Theta]. \end{align}$$ Note that, conditioned on $\Theta = \theta$, $\hat{X}$ and $\hat{Y}$are conditionally independent zero-mean normal random variables with variances $\cos^2\theta$ and $\sin^2\theta$. They are thus also conditionally uncorrelated random variables. The conditional joint density is $$f_{\hat{X},\hat{Y}\mid \Theta}(\hat{x},\hat{y}\mid \Theta=\theta) = \frac{1}{2\pi\cos\theta\sin\theta}\exp\left(-\frac{\hat{x}^2}{2\cos^2\theta} +\frac{\hat{y}^2}{2\sin^2\theta}\right), ~ -\infty < x, y < \infty.$$ While $\hat{X}$ and $\hat{Y}$ are conditionally independent given $\Theta = \theta$, their joint density is very much dependent on $\theta$, and it is not immediately obvious that their unconditional density $$f_{\hat{X},\hat{Y}}(\hat{x},\hat{y}) = \int_{-\infty}^\infty f_{\hat{X},\hat{Y}\mid \Theta}(\hat{x},\hat{y}\mid \Theta=\theta)f_\Theta(\theta)\,\mathrm d\theta$$ factors into the product of the marginal densities $$\begin{align} f_{\hat{X}}(\hat{x}) &= \int_{-\infty}^\infty f_{\hat{X}\mid \Theta}(\hat{x}\mid \Theta=\theta)f_\Theta(\theta)\,\mathrm d\theta = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\cos\theta}\exp\left(-\frac{\hat{x}^2}{2\cos^2\theta} \right)f_\Theta(\theta)\,\mathrm d\theta\\ f_{\hat{Y}}(\hat{y}) &= \int_{-\infty}^\infty f_{\hat{Y}\mid \Theta}(\hat{y}\mid \Theta=\theta)f_\Theta(\theta)\,\mathrm d\theta = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sin\theta}\exp\left(-\frac{\hat{y}^2}{2\sin^2\theta} \right)f_\Theta(\theta)\,\mathrm d\theta \end{align}$$ as would be needed for $\hat{X}$ and $\hat{Y}$ to be unconditionally independent random variables. Nor is it immediately obvious that the marginal distributions are normal distributions. However, as noted above, $\hat{X}$ and $\hat{Y}$ are nonetheless unconditionally uncorrelated random variable.

Some special cases when $\hat{X}$ and $\hat{Y}$ are indeed unconditionally independent normal random variables are as follows.

  • $\Theta$ equals a constant $\theta$ with probability $1$. In this case, the unconditional density is the same as the conditional density and thus $\hat{X}$ and $\hat{Y}$ are independent zero-mean normal random variables with variances $\cos^2\theta$ and $\sin^2\theta$ respectively.

  • $\Theta$ is a discrete random variable taking on values $\pi/4, 3\pi/4, 5\pi/4, 7\pi/4$ with equal probability. In this case, for all $4$ values of $\Theta$, the conditional joint distribution is that of independent $N(0,\frac{1}{2})$ random variables and so the unconditional distribution also enjoys the same properties.

Note however that the latter choice does not extend to $\Theta$ being a uniform distribution over $N\geq 5$ phases, and so it is doubtful that the $U[0,2\pi)$) for $\Theta$ will yield either normal distributions for $\hat{X}$ and $\hat{Y}$ or that the joint distribution will be the product of the marginals as required for independence. The same comments apply to other distributions that one might choose for $\Theta$. But, $\hat{X}$ and $\hat{Y}$ will nonetheless be conditionally as well as unconditionally uncorrelated random variables regardless of the distribution of $\Theta$.

Thus, the OP's claim that "$Y1$ and $Y2$ are obviously independent" (emphasis added) should be taken with a considerably large grain of salt. If they are indeed independent, there is nothing obvious about it (to my poor brain, at least; YMMV).

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  • $\begingroup$ Since I agree I didn't find a contradiction to a proved theorem I agree, given a proof r(t) is Gaussian that the assumption Y1, Y2 are independent is wrong. I can't see why. This is the reason I'm posting it here. I didn't see how you claim that since the conditional density of Z is Gaussian it unconditional density is Gaussian as well. Thank you for your effort and answer. $\endgroup$ – Royi Sep 23 '13 at 8:22
  • $\begingroup$ Now I can see why $ {Y}_{1} $ and $ {Y}_{2} $ are dependent. Looking at it at the other way around, assume there's a uniformly distributed variable. It's Cosine and Sine functions are clearly dependent. Hence the multiplication of those function by a random number is dependent. $\endgroup$ – Royi Sep 29 '13 at 13:58
  • $\begingroup$ I could '-1' your answer since the argument it is based is unproven - "Let Z=XcosΘ+YsinΘ. Since X and Y are independent of Θ, their conditional joint density given that Θ=θ is the same as their unconditional density". There are examples that tough two random variables are independent a function of both isn't defined by only one of them. $\endgroup$ – Royi Sep 29 '13 at 21:55
  • $\begingroup$ @Drazick I could '-1' your answer since the argument it is based is unproven - "Let Z=XcosΘ+YsinΘ. Since X and Y are independent of Θ, their conditional joint density given that Θ=θ is the same as their unconditional density You should feel absolutely free to do so. Your action will reveal more about your understanding of the issues involved than you might want others to know. $\endgroup$ – Dilip Sarwate Sep 30 '13 at 1:43
  • $\begingroup$ You miss the point. I didn't do so because I give you the credit and think I might something. You on the other way, thinks you know what I meant when I wrote the the question and the answer and act like that. That not the way communities should work. I answered exactly what I asked. The upper and lower branches of this system aren't Linear and not Time Invariant Systems. Another point, as opposed to what you said, given Gaussian Input, if the Output is Gaussian doesn't mean it is a Linear System. $\endgroup$ – Royi Sep 30 '13 at 5:49
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Solution:

  1. The processes $ {Y}_{1} $ and $ {Y}_{2} $ are dependent.
    $ {Y}_{1} $ and $ {Y}_{2} $ are dependent intuitively by looking at the problem from a different angle.
    Given a uniformly distributed random variable $ \theta $ the random variables $ sin(\theta) $ and $ cos(\theta) $ are clearly dependent. Multiplying them by a Gaussian Random Variable doesn't decorrelate them.
  2. The processes $ {Y}_{1} $ and $ {Y}_{2} $ aren't Gaussian (And not Ergodic).
    Clearly over time (The Phase is constant) the process $ {Y}_{i} $ is Gaussian. Yet the ensemble isn't Gaussian since each of its realization is retrieved from Gaussian Distribution with different parameters.
  3. The process $ r(t) $ is indeed Gaussian.
    This could be proved by using the Characteristic Function of the random process.
    $ X \sim N(0, 1) $, $ Y \sim N(0, 1) $, $ \theta \sim U(0, 2\pi) $ all are independent.
    Let $ Z = X sin(\theta) + Y cos(\theta) $. Looking at its Characteristic Function and applying the Smoothing Theorem yields: \begin{align} { \varphi }_{Z}(t) & = & \mathbb{E}[{e}^{itZ}] \\ & = & \mathbb{E} [ \mathbb{E}[{e}^{itZ} | \theta] ] \\ & = & \mathbb{E} [ \mathbb{E}[{e}^{it(X cos(\theta) + Y sin(\theta))} | \theta] ] \end{align}
    Looking at the last equation per realization of $ \theta $:
    $$ \mathbb{E}[{e}^{it(X cos(\theta) + Y sin(\theta))} | \Theta = \theta] = \mathbb{E}[{e}^{itX sin(\theta)}] \mathbb{E}[{e}^{itY cos(\theta)}] $$ Each of the item is the Characteristic Function of a scaled Normally Distributed Random Process: $$ \mathbb{E}[{e}^{itX sin(\theta)}] \mathbb{E}[{e}^{itY cos(\theta)}] = {e}^{-\frac{{t}^{2}}{2} ({sin}^{2}(\theta) + {cos}^{2}(\theta))} = {e}^{-\frac{{t}^{2}}{2}} $$ Hence we get: $$ { \varphi }_{Z}(t) = \mathbb{E} [ \mathbb{E}[{e}^{itZ} | \Theta = \theta] ] = \mathbb{E} [{e}^{-\frac{{t}^{2}}{2}}] = {e}^{-\frac{{t}^{2}}{2}} $$ Namely, it has the Characteristic Function of a Normalized Gaussian Variable -> $ Z \sim N(0, 1) $.
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