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6

First of all there is some serious "cheating" in the MFCC reconstruction experiment you linked to: not only the MFCCs are used, but also the voiced/unvoiced bit and the pitch. MFCC are not speaker independent. In fact, they are used for speaker identification/verification tasks! The speaker "idiosyncracies" are both in their prosody (preserved by this ...

5

Inferring whether two signals are independent is very hard to do (given finite observations) without any prior knowledge/assumptions. Two random variable $X$ and $Y$ are independent if the value of $X$ doesn't give any information about the value of $Y$ (i.e. doesn't affect our prior probability distribution for $Y$). This is equivalent to any nonlinear ...

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[0] &= v[0] \\ x[1] &= a x[0] + v[1] \\ x[2] &= a x[1] + v[2] = a^2 x[0] + a v[1] + v[2] = a^2 v[0] + a v[1] + v[2] \\ x[2] &= \left( a^2 v[0] + a v[1] \right) + v[2] \\ ... &= ... \\ x[k] &= \left( a^k v[0] + a^{... 3 No, NO and \mathbf{NO}. Knowing that E[X]=E[Y]=0 and E[XY]=0 (or more generally that E[XY]=E[X]E[Y] or equivalently, E[XY]-E[X]E[Y]=0) does not help in the least in proving or deducing that X and Y are independent random variables. The conditions stated in the previous sentence are necessary for X and Y to be deemed independent random ... 3 As mentioned above, both signals 3 and 5 appear to be quite correlated and have a similar period. We can think of two signals being correlated if we can shift one of the sources left or right and increase or decrease its amplitude so that it fits on top of the other source. Note we are not changing the frequency of the source, we are just performing a phase ... 2 The components are instantaneously uncorrelated if E[y_i(t_1) y_j(t_2)] = 0 $$for all i \not= j and for all t_1,t_2 and when i = j :$$ E[y_i(t_1) y_i(t_2)] = 0 $$for all t_1 \not= t_2. The components are spatially uncorrelated if$$ E[y_i(t) y_j(t)] = 0 $$for all i \not= j (assuming a change in index corresponds to a spatial change). Note ... 2 One must be careful when asking questions about the relationships between the elements of a complex random vector. The short answer to your question is that you cannot say much for either cases simply by considering the covariance (or correlation) matrix. Actually, the covariance (correlation) matrix is not enough to capture all the relationships that ... 2 For the first case, as you wrote, it means the elements are not correlated. Since this is a Gaussian Random Vector it means the elements are independent. It means that at most only one element of  \boldsymbol{\mu}  is not zero. Since if there were more than 1, the matrix  \boldsymbol{R}  wasn't diagonal. Update Let's define  \hat{\boldsymbol{x}} = \... 2 In the additive model y=s+n, when the signal is deterministic, it adds coherently over the "realizations". Hence, its variance V(\sum s_n) = V(N s) = N^2 V( s). And when the noise w is independent identically distributed (IID), then V(\sum w_n) = NV( n) . This is a classical result on the Variance of Uncorrelated Variables. You can find this ... 2 How did the professor state this? Did they say that the noise is iid? That means that the samples are independent, identically distributed. That means they are independent by definition (they are assumed to be so). White noise, while a little more vague, also makes the assumption of independence. Again, this means that the noise is assumed to be ... 1 Let's look at things in the following perspective; Statistical independence is an assumption imposed on events A,B or random variables X,Y, whose (the assumption's) mathematical consequence is expressed as;$$ P( A \cap B) = P(A) P(B)$$and$$ E\{X Y\} = E\{X\}E\{Y\} $$So these are the consequences of independence, but not the cause of it. (This is ... 1 Essentially E\{XY\}=0 (assuming E\{X\}=E\{Y\}=0) means that X and Y are uncorrelated, but uncorrelated is a weaker property than probabilistic independence. A collection of independent random variables will always be uncorrelated, but being uncorrelated does not mean two random variables are independent. 1 The sequence x[k] only depends on the current input v[k] and on past inputs v[k-l], l>0. Consequently, x[k-1] only depends on past inputs v[k-l], l>0. And since E\{v[k]v[k-l]\}=0 for l>0, you also have E\{v[k]x[k-1]\}=0. 1 Hint 1: you can rewrite products of the exponential argument:$$ f_{k,l}=\frac{1}{N}\left(e^{\frac{j2\pi(k-1)}{N}}\right)^{l-1} and recognize a Vandermonde matrix. Hint 2: consider the dimension of the subspace spanned, and the linear independence.

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