# Tag Info

9

Avoid complicated calculations, let $X$ and $Y$ be i.i.d. standard normal random variables, your random variable $Z$ has the same distribution of $V$ $$V \triangleq \left(\frac{X}{\sqrt{X^2+Y^2}},\frac{Y}{\sqrt{X^2+Y^2}} \right)$$ (easy to see $\lVert V\rVert=1$ and the angle of $V$ is equivalent to the angle of a circularly symmetric Normal hence uniform). ...

6

Since the real and imaginary parts are very much dependent on one another (if you have the value of one, you know the value of the other exactly), it seems like you could apply the marginal pdf of the real part $r$, given a value of the imaginary part $i$: $$f_{ri}(r, i) = f_{r | i}(r\ |\ i) f_i(i)$$ You noted the pdf of the real and imaginary parts ...

5

No, it will be Rician Distribution. Rayleigh distribution is a special case of Rician Distribution when the normal random variables involved are of zero mean and equal variances. I have assumed you meant :$$y=x_r + j \cdot x_i + n_r + j \cdot n_i$$ where $x_r, x_i$ are deterministic variables. Basically, you now have $y_r = (n_r + x_r),\ \sim \mathcal N(... 4 Based on the existing answers, which opened my eyes for what's going on here, I would like to present yet another very simple expression for the solution, which is just slightly different from the one in AlexTP's answer (and which turned out to be equivalent to the one given in Jason R's answer, as shown below in the EDIT-part). [EDIT: now that AlexTP has ... 3 No, NO and$\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 In general, the characteristic function of a random variable is related to the fourier transform of the distribution as follows: $$\varphi_Z(-\omega) = \mathscr F \big\{ f_Z(z) \big\}$$ Why? Because, fourier transform of a PDF$f_Z(z)$is: $$\mathscr F \big\{ f_Z(z) \big\} = \int^{\infty}_{-\infty} f_Z(z)\cdot e^{-j\omega z} \ \mathrm dz$$ And ... 2 With reference to$N_o$this usually is the symbol for the power spectral density of thermal noise, where$N_o = kT$, where k is Boltzmann's Constant and T is the temperature in Kelvin. The thermal noise signal is a complex, white Gaussian distributed noise, with half of the power in the real component and half the power in the imaginary component (but power ... 2 @Dan Boschen gives a precise answer in the comment. I would like to add to that a simple derivation using the formal definition as $$S_{xx}(f) = E \vert x(t) \vert^2$$ Where$E$stands for expectation and the process$x(t)$is the one you describe in OP. Since it is complex, we can decompose it as $$x(t) = r(t) + j.i(t)$$ Where$r(t),i(t)$are the real ... 2 The matrices$\hat{\mathbf{M}}$and$\tilde{\mathbf{M}}$are constructed in such a way that the relation$\mathbf{M}\mathbf{x}=\mathbf{y}$implies$\hat{\mathbf{M}}\hat{\mathbf{x}}=\hat{\mathbf{y}}$and$\tilde{\mathbf{M}}\tilde{\mathbf{x}}=\tilde{\mathbf{y}}$. Consequently, for constructing the matrix$\tilde{\mathbf{M}}$, each element$m_{kl}$of$\...

1

The envelope |y| follows a Rice/Rician distribution if the variances are the same. a Beckmann distribution if the variances are different. To my knowledge, there are not simplified closed-form expressions for the PDF and the moments. However, in On the Distribution Function of the Generalized Beckmann Random Variable and Its Applications in Communications ...

1

You have $$E[s_ks_k^*]=E[|s_k|^2]=1$$ because the complex random variables $s_k$ have zero mean and unit variance. That means that all elements of the main diagonal of $E[\mathbf{s}\mathbf{s}^H]$ equal $1$. Furthermore, with $s_k=x_k+jy_k$we have \begin{align}E[s_ks^*_l]&=E[x_kx_l+y_ky_l+j(x_ly_k-x_ky_l)]\\&=E[x_kx_l]+E[y_ky_l]+jE[x_ly_k]-jE[... 1 Assume Y = g(X) be the function of RV X, then by using the followingE\{ g(X) \} = \int g(x) f_X(x) dx $$variance of Y can be computed without the computation of pdf f_Y(y) as:$$ \begin{align} \text{Var(Y)} &= E\{ (Y-\mu_Y)^2 \} = E\{ Y^2 \} - (\mu_Y)^2 \\ & = E\{ g^2(X) \} - E\{ g(X) \}^2 \\ & = \int g^2(x) f_x(x) dx - \left(\...

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 ...

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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.

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