Exercises

$\mathbf{X} = \mathbf{M}[A, B]^\mathsf{T}$ where $[\mathbf{M}]_{k,:} = [t_k, 1]$. Since $[A,B]^\mathsf{T} \sim \mathcal{N}(\mathbf{0}, \text{diag}(1,4))$, $\mathbf{X}$ is jointly Gaussian. So $\{X(t)\}$ is a GP.

$\mu_X(t) = \mathbb{E}[At + B] = 0$. $C_X(s,t) = \mathbb{E}[(As+B)(At+B)] = st + 4$.

$C(\tau) = \frac{1}{2}e^{-|\tau|}(e^{j2\pi\tau} + e^{-j2\pi\tau})$. The Fourier transform of $e^{-|\tau|}e^{j2\pi f_0\tau}$ is $\frac{2}{1+(2\pi(f-f_0))^2} \cdot \frac{1}{2\pi}$... Actually, $\mathcal{F}\{e^{-|\tau|}\} = \frac{2}{1+(2\pi f)^2}$, so by the modulation property: $$P(f) = \frac{1}{1+(2\pi(f-1))^2} + \frac{1}{1+(2\pi(f+1))^2}$$

Both terms are nonnegative for all $f$, so $P(f) \geq 0$. Therefore $C(\tau)$ is a valid covariance function. $\checkmark$

For any $t_1, \ldots, t_N$, $[Z(t_1), \ldots, Z(t_N)]^\mathsf{T} = [X(t_1), \ldots, X(t_N)]^\mathsf{T} + [Y(t_1), \ldots, Y(t_N)]^\mathsf{T}$. The sum of independent Gaussian vectors is Gaussian. So $\{Z(t)\}$ is a GP.

$\mu_Z(t) = \mu_X(t) + \mu_Y(t)$. $C_Z(s,t) = C_X(s,t) + C_Y(s,t)$ (by independence, cross-covariance is zero).

$(X(t_1), X(t_2)) \sim \mathcal{N}\left(\mathbf{0}, \sigma^2\begin{bmatrix} 1 & \rho \\ \rho & 1\end{bmatrix}\right)$ where $\rho = e^{-\alpha|\tau|}$.

$\mathbb{P}(X(t_1) > 0, X(t_2) > 0) = \frac{1}{4} + \frac{1}{2\pi}\arcsin(\rho)$$This is the well-known result for the orthant probability of a bivariate Gaussian. As$\tau \to 0$,$\rho \to 1$and the probability$\to 1/2$. As$\tau \to \infty$,$\rho \to 0$and the probability$\to 1/4$ (independence).

$\mathcal{P}_N = N_0W = 2 \times 10^{-6} \times 10^6 = 2$ W.

$R_{N_B}(\tau) = N_0W\,\text{sinc}(2W\tau) = 2\,\text{sinc}(1) = 0$ W. (Since $2W\tau = 2 \times 10^6 \times 0.5 \times 10^{-6} = 1$ and $\text{sinc}(1) = 0$.)

$\dot{X}(t) = \lim_{h \to 0} \frac{X(t+h) - X(t)}{h}$ in mean square.

For any $t_1, \ldots, t_N$, the difference quotients $\frac{X(t_k+h)-X(t_k)}{h}$ are linear combinations of the Gaussian vector $[X(t_1+h), X(t_1), \ldots, X(t_N+h), X(t_N)]$, hence jointly Gaussian.

As $h \to 0$, each $\dot{X}(t_k)$ is the m.s. limit of Gaussian RVs. The characteristic function converges pointwise, and since the limit of Gaussian CFs is Gaussian, $[\dot{X}(t_1), \ldots, \dot{X}(t_N)]$ is jointly Gaussian. $\blacksquare$

$\check{h}(f) = \frac{\alpha}{\alpha + j2\pi f}$, so $|\check{h}(f)|^2 = \frac{\alpha^2}{\alpha^2 + (2\pi f)^2}$.

$P_Y(f) = \frac{\alpha^2 N_0/2}{\alpha^2 + (2\pi f)^2}$.

Inverse Fourier transform: $R_Y(\tau) = \frac{\alpha N_0}{4} e^{-\alpha|\tau|}$. This depends only on $|\tau|$, so $Y(t)$ is WSS. The variance is $R_Y(0) = \alphaN_0/4$.

$\Lambda_N = \sum_{k=1}^N s(t_k) N(t_k) \Delta t$. This is a linear combination of the jointly Gaussian vector $[N(t_1), \ldots, N(t_N)]$, hence Gaussian with zero mean.

$\text{Var}(\Lambda_N) = \sum_{k,\ell} s(t_k) s(t_\ell) \mathbb{E}[N(t_k)N(t_\ell)] \Delta t^2KATEXPLACEHOLDER0END= \frac{N_0}{2} \sum_k s(t_k)^2 \Delta t \xrightarrow{N\to\infty} \frac{N_0}{2}\int_0^T s(t)^2\,dt = \frac{E_s N_0}{2}$$

The m.s. limit $\Lambda$ is Gaussian (limit of Gaussian RVs) with mean $0$ and variance $E_sN_0/2$. $\blacksquare$

$P_Y(f) = |j2\pi f|^2 \cdot \frac{2\alpha}{\alpha^2 + (2\pi f)^2} = \frac{2\alpha(2\pi f)^2}{\alpha^2 + (2\pi f)^2}$.

As $|f| \to \infty$, $P_Y(f) \to 2\alpha$. So $\int_{-\infty}^{\infty} P_Y(f)\,df = \infty$. The output has infinite power — the differentiator amplifies high-frequency components without bound. The derivative of this process does not exist in mean square.

Expand $Y(t)$ in an orthonormal basis $\{\phi_k(t)\}$ including $\phi_1(t) = s(t)/\sqrt{E_s}$. The coefficients $Y_k = \int Y(t)\phi_k(t)\,dt$ are independent Gaussian under both hypotheses.

Under $H_1$: $Y_1 \sim \mathcal{N}(\sqrt{E_s}, N_0/2)$, $Y_k \sim \mathcal{N}(0, N_0/2)$ for $k \geq 2$. Under $H_0$: $Y_k \sim \mathcal{N}(0, N_0/2)$ for all $k$. The likelihood ratio is $L = \frac{f_1(Y_1, Y_2, \ldots)}{f_0(Y_1, Y_2, \ldots)} = \exp\left(\frac{2\sqrt{E_s} Y_1 - E_s}{N_0}\right)$, depending only on $Y_1 \propto \Lambda$.

By the Neyman-Fisher factorization theorem, $\Lambda$ is a sufficient statistic. $\blacksquare$

$P_x(f) = \mathcal{F}\{e^{-|\tau|}\} = \frac{2}{1 + (2\pi f)^2}$.

$P_Y(f) = P_x(f)$ for $|f| \leq 5$ and $0$ otherwise. At $f = 0$: $P_Y(0) = \frac{2}{1+0} = 2$ W/Hz.

$W(5) - W(3)$ is an increment over $[3,5]$, independent of $W(3)$ (which depends only on the process up to time 3). Therefore: $\mathbb{E}[W(3)(W(5)-W(3))] = \mathbb{E}[W(3)]\,\mathbb{E}[W(5)-W(3)] = 0 \cdot 0 = 0$.

$X(t) = W(t) - tW(1)$ is a linear combination of the Gaussian process $W$, hence Gaussian. $X(0) = 0$, $X(1) = W(1) - W(1) = 0$.

$C_X(s,t) = \mathbb{E}[X(s)X(t)] = \mathbb{E}[(W(s)-sW(1))(W(t)-tW(1))]KATEXPLACEHOLDER0END= \min(s,t) - s\min(1,t) - t\min(s,1) + st\min(1,1)KATEXPLACEHOLDER1END= \min(s,t) - st - st + st = \min(s,t) - st$$For$s \leq t$:$C_X(s,t) = s(1-t)$.

$\mathbb{E}[Y(t)] = e^{-t/2}\mathbb{E}[W(e^t)] = 0$.

$R_Y(s,t) = e^{-s/2}e^{-t/2}\mathbb{E}[W(e^s)W(e^t)] = e^{-(s+t)/2}\min(e^s, e^t)$$For$s \leq t$:$= e^{-(s+t)/2} \cdot e^s = e^{(s-t)/2} = e^{-|t-s|/2}$.

$R_Y(s,t) = e^{-|t-s|/2}$ depends only on $|t-s|$, so $Y(t)$ is WSS. The autocorrelation is $r_{xx}(\tau) = e^{-|\tau|/2}$, exponentially decaying.

$\mathbb{E}\left[\int_0^T W(t)\,dt\right] = \int_0^T \mathbb{E}[W(t)]\,dt = 0$.

$\text{Var}\left(\int_0^T W(t)\,dt\right) = \int_0^T\int_0^T \min(s,t)\,ds\,dtKATEXPLACEHOLDER0END= 2\int_0^T\int_0^t s\,ds\,dt = 2\int_0^T \frac{t^2}{2}\,dt = \int_0^T t^2\,dt = \frac{T^3}{3}$$

$\text{CPE} = \Phi(T_s) - \Phi(0) = \sigma(W(T_s) - W(0)) = \sigma W(T_s) \sim \mathcal{N}(0, \sigma^2 T_s)$.

$\mathbb{P}(|\text{CPE}| > \pi/4) = 2Q\left(\frac{\pi/4}{\sigma\sqrt{T_s}}\right)$$For QPSK to work reliably, this must be small, requiring$\sigma\sqrt{T_s} \ll \pi/4$, i.e.,$\sigma^2 T_s \ll (\pi/4)^2 \approx 0.62$.

Try $\phi_k(t) = \sqrt{2}\sin(k\pi t)$. These satisfy the boundary conditions $\phi_k(0) = \phi_k(1) = 0$.

$\int_0^1 (\min(s,t) - st)\sqrt{2}\sin(k\pi s)\,ds$$Split at$s = t$:$\int_0^t s(1-t)\sqrt{2}\sin(k\pi s),ds + \int_t^1 t(1-s)\sqrt{2}\sin(k\pi s),ds$. After integration by parts (twice each), this equals$\frac{1}{k^2\pi^2}\sqrt{2}\sin(k\pi t) = \frac{1}{k^2\pi^2}\phi_k(t)$.

$\lambda_k = \frac{1}{k^2\pi^2}$, $k = 1, 2, 3, \ldots$

The KL expansion is: $X(t) = \sum_{k=1}^{\infty} Z_k \frac{\sqrt{2}\sin(k\pi t)}{k\pi}$ where $Z_k \sim \mathcal{N}(0,1)$ are i.i.d.

$\tilde{\mathbf{N}} = \mathbf{F}\mathbf{N}$ where $\mathbf{F}$ is the $N \times N$ unitary DFT matrix ($[\mathbf{F}]_{k,n} = e^{-j2\pi kn/N}/\sqrt{N}$).

$\mathbb{E}[\tilde{\mathbf{N}}\tilde{\mathbf{N}}^H] = \mathbf{F}\,\mathbb{E}[\mathbf{N}\mathbf{N}^H]\,\mathbf{F}^H = \mathbf{F}(\sigma^2\mathbf{I})\mathbf{F}^H = \sigma^2\mathbf{I}$.

A linear transformation of a Gaussian vector is Gaussian. The covariance $\sigma^2\mathbf{I}$ means the components are uncorrelated, and for Gaussian RVs, uncorrelated implies independent. So $\tilde{N}[k] \sim \mathcal{CN}(0, \sigma^2)$ i.i.d. $\blacksquare$

$[\mathbf{K}]_{ij} = \exp(-(t_i-t_j)^2/4.5)$. $\mathbf{K} = \begin{bmatrix} 1 & 0.8007 & 0.1353 \\ 0.8007 & 1 & 0.3679 \\ 0.1353 & 0.3679 & 1\end{bmatrix}$

$[\mathbf{k}_*]_i = \exp(-(2-t_i)^2/4.5)$, giving $\mathbf{k}_* = [0.4111, 0.8007, 0.8007]^\mathsf{T}$.

$\mu_*(2) = \mathbf{k}_*^\mathsf{T}\mathbf{K}^{-1}\mathbf{y} \approx -0.06$

$\sigma_*^2(2) = 1 - \mathbf{k}_*^\mathsf{T}\mathbf{K}^{-1}\mathbf{k}_* \approx 0.14$

(Numerical values from matrix inversion.)

Write $W(t) = W(s) + (W(t) - W(s))$, where $W(t)-W(s) \sim \mathcal{N}(0, t-s)$ is independent of $\mathcal{F}_s$.

$\mathbb{E}[W(t)^2 \mid \mathcal{F}_s] = \mathbb{E}[(W(s) + (W(t)-W(s)))^2 \mid \mathcal{F}_s]KATEXPLACEHOLDER0END= W(s)^2 + 2W(s)\mathbb{E}[W(t)-W(s)] + \mathbb{E}[(W(t)-W(s))^2]KATEXPLACEHOLDER1END= W(s)^2 + 0 + (t-s)$$

$\mathbb{E}[W(t)^2 - t \mid \mathcal{F}_s] = W(s)^2 + (t-s) - t = W(s)^2 - s$. So $M(t) = W(t)^2 - t$ satisfies $\mathbb{E}[M(t) \mid \mathcal{F}_s] = M(s)$. $\blacksquare$