Exercises

Under $H_0$: $r \mid H_0 \sim \mathcal{N}(1, 1)$, so $$f(r{=}0.5 \mid H_0) = \frac{1}{\sqrt{2\pi}}\exp\!\Bigl(-\frac{(0.5 - 1)^2}{2}\Bigr) = \frac{1}{\sqrt{2\pi}}\,e^{-0.125}.$$

Under $H_1$: $r \mid H_1 \sim \mathcal{N}(-1, 1)$, so $$f(r{=}0.5 \mid H_1) = \frac{1}{\sqrt{2\pi}}\exp\!\Bigl(-\frac{(0.5 + 1)^2}{2}\Bigr) = \frac{1}{\sqrt{2\pi}}\,e^{-1.125}.$$

$f(r{=}0.5) = f(r \mid H_0)\,P(H_0) + f(r \mid H_1)\,P(H_1) = \frac{1}{\sqrt{2\pi}}\bigl(0.6\,e^{-0.125} + 0.4\,e^{-1.125}\bigr).KATEXPLACEHOLDER0ENDf(r{=}0.5) \approx \frac{1}{\sqrt{2\pi}}(0.5295 + 0.1299) = \frac{0.6594}{\sqrt{2\pi}}.$$

$P(H_0 \mid r{=}0.5) = \frac{0.6 \times 0.8825}{0.6594} = \frac{0.5295}{0.6594} \approx 0.8031.KATEXPLACEHOLDER0ENDP(H_1 \mid r{=}0.5) = \frac{0.4 \times 0.3247}{0.6594} = \frac{0.1299}{0.6594} \approx 0.1969.$$Sanity check:$0.8031 + 0.1969 = 1.0000$. The observation$r = 0.5$is closer to$+1$than to$-1$, so the posterior strongly favors$H_0$.$\square$

For an exponential random variable with rate $\lambda = 2$: $$f_X(x) = 2\,e^{-2x}, \quad x \ge 0,$$ $$F_X(x) = 1 - e^{-2x}, \quad x \ge 0.$$

The mean of an $\text{Exp}(\lambda)$ random variable is $$E[X] = \frac{1}{\lambda} = \frac{1}{2} = 0.5\;\text{ms}.$$

The variance is $$\operatorname{Var}(X) = \frac{1}{\lambda^2} = \frac{1}{4} = 0.25\;\text{ms}^2.$$

$P(X > 1) = 1 - F_X(1) = e^{-2 \cdot 1} = e^{-2} \approx 0.1353.$$Thus there is approximately a$13.5%$chance that the inter-arrival time exceeds$1$ms.$\square$

By LOTUS: $$E[Y] = E[X^2] = \int_{-\infty}^{\infty} x^2 \cdot \frac{1}{\sqrt{2\pi \cdot 4}}\, e^{-x^2/8}\,dx = \sigma^2 = 4.$$

This follows directly from the definition of variance for a zero-mean RV: $\operatorname{Var}(X) = E[X^2] - (E[X])^2 = E[X^2] = \sigma^2$.

The fourth central moment of a Gaussian $\mathcal{N}(0,\sigma^2)$ is $$E[X^4] = 3\sigma^4.$$

This can be derived via LOTUS and integration by parts, or by differentiating the moment generating function twice. With $\sigma^2 = 4$: $$E[Y^2] = E[X^4] = 3 \cdot 16 = 48.$$

$\operatorname{Var}(Y) = E[Y^2] - (E[Y])^2 = 48 - 16 = 32.$$Equivalently, since$Y = X^2$and$X \sim \mathcal{N}(0,4)$, we have$Y/4 \sim \chi^2_1$(chi-squared with one degree of freedom), whose variance is$2$. Therefore$\operatorname{Var}(Y) = 16 \cdot 2 = 32$.$\square$

$f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dy = \frac{1}{2\pi \cdot 1 \cdot 2 \cdot \sqrt{1-0.25}} \int_{-\infty}^{\infty} \exp\!\Bigl(-\frac{1}{2 \cdot 0.75} \Bigl[x^2 - \frac{2 \cdot 0.5\,x\,y}{2} + \frac{y^2}{4}\Bigr]\Bigr)\,dy.$$Simplifying the prefactor:$2\pi \cdot 2\sqrt{0.75} = 4\pi\sqrt{3}/2 = 2\pi\sqrt{3}$.

Inside the exponent, group terms involving $y$: $$\frac{y^2}{4} - \frac{0.5\,x\,y}{1} = \frac{1}{4}\Bigl(y^2 - 2xy\Bigr) = \frac{1}{4}\Bigl[(y - x)^2 - x^2\Bigr].$$

The full exponent becomes $$-\frac{1}{1.5}\Bigl[x^2 + \frac{(y-x)^2}{4} - \frac{x^2}{4}\Bigr] = -\frac{1}{1.5}\cdot\frac{3x^2}{4} - \frac{(y - x)^2}{6}.$$

That is, $-\frac{x^2}{2} - \frac{(y - x)^2}{6}$.

$f_X(x) = \frac{e^{-x^2/2}}{2\pi\sqrt{3}} \int_{-\infty}^{\infty} e^{-(y-x)^2/6}\,dy.KATEXPLACEHOLDER0ENDf_X(x) = \frac{e^{-x^2/2}}{2\pi\sqrt{3}} \cdot \sqrt{6\pi} = \frac{e^{-x^2/2}}{2\pi\sqrt{3}} \cdot \sqrt{6\pi} = \frac{e^{-x^2/2}}{\sqrt{2\pi}}.$$Therefore$f_X(x) = \frac{1}{\sqrt{2\pi}},e^{-x^2/2}$, confirming that$X \sim \mathcal{N}(0, 1)$, consistent with the marginal parameters.$\square$

With $\lambda t = 5 \cdot 1 = 5$: $$P(N(1) = 3) = \frac{5^3}{3!}\,e^{-5} = \frac{125}{6}\,e^{-5} \approx 20.833 \times 0.006738 \approx 0.1404.$$

With $\lambda t = 5 \cdot 0.5 = 2.5$: $$P(N(0.5) = 0) = \frac{2.5^0}{0!}\,e^{-2.5} = e^{-2.5} \approx 0.0821.$$

For a Poisson process, $E[N(t)] = \operatorname{Var}(N(t)) = \lambda t$. With $t = 10$: $$E[N(10)] = 5 \times 10 = 50\;\text{packets},$$ $$\operatorname{Var}(N(10)) = 50\;\text{packets}^2.$$

The standard deviation is $\sqrt{50} \approx 7.07$ packets. $\square$

Since $C_2$ only activates when $C_1$ declares $D_1$: $$P(D_2 \mid H_1) = P(D_2 \mid D_1, H_1)\,P(D_1 \mid H_1) = 0.98 \times 0.95 = 0.931.$$

$$P(D_2 \mid H_0) = P(D_2 \mid D_1, H_0)\,P(D_1 \mid H_0) = 0.05 \times 0.10 = 0.005.$$

By the total probability theorem: $$P(D_2) = P(D_2 \mid H_1)\,P(H_1) + P(D_2 \mid H_0)\,P(H_0) = 0.931 \times 0.01 + 0.005 \times 0.99 = 0.00931 + 0.00495 = 0.01426.$$

$P(H_1 \mid D_2) = \frac{P(D_2 \mid H_1)\,P(H_1)}{P(D_2)} = \frac{0.00931}{0.01426} \approx 0.6529.$$The two-stage system raises the posterior from the prior of$1%$to approximately$65%$. The cascaded detection dramatically reduces the false alarm rate from$10%$to$0.5%$, which is essential in radar and cognitive radio applications.$\square$

Since $X$ and $Y$ are independent $\mathcal{N}(0, \sigma^2)$: $$f_{X,Y}(x,y) = f_X(x)\,f_Y(y) = \frac{1}{2\pi\sigma^2} \exp\!\Bigl(-\frac{x^2 + y^2}{2\sigma^2}\Bigr).$$

Let $x = \rho\cos\theta$ and $y = \rho\sin\theta$, with Jacobian $|J| = \rho$. The joint PDF in polar form is $$f_{R,\Theta}(\rho,\theta) = \frac{\rho}{2\pi\sigma^2} \exp\!\Bigl(-\frac{\rho^2}{2\sigma^2}\Bigr), \quad \rho \ge 0,\; 0 \le \theta < 2\pi.$$

$f_R(r) = \int_0^{2\pi} f_{R,\Theta}(r,\theta)\,d\theta = \frac{r}{2\pi\sigma^2}\,e^{-r^2/(2\sigma^2)} \cdot 2\pi = \frac{r}{\sigma^2}\,e^{-r^2/(2\sigma^2)}, \quad r \ge 0.$$This is the **Rayleigh PDF** with parameter$\sigma$. Its mean is$E[R] = \sigma\sqrt{\pi/2}$and its second moment is$E[R^2] = 2\sigma^2$. The Rayleigh distribution is fundamental in wireless communications for modeling the envelope of multipath fading.$\square$

For $X_i \sim \text{Exp}(\lambda)$: $$M_{X_i}(t) = E[e^{tX_i}] = \int_0^{\infty} e^{tx}\,\lambda e^{-\lambda x}\,dx = \lambda \int_0^{\infty} e^{-(\lambda - t)x}\,dx = \frac{\lambda}{\lambda - t}, \quad t < \lambda.$$

Since the $X_i$ are independent, the MGF of the sum factorizes: $$M_{S_n}(t) = \prod_{i=1}^n M_{X_i}(t) = \left(\frac{\lambda}{\lambda - t}\right)^n, \quad t < \lambda.$$

This is the MGF of the Erlang-$n$ distribution (equivalently, $\text{Gamma}(n, \lambda)$), with PDF $$f_{S_n}(s) = \frac{\lambda^n\,s^{n-1}}{(n-1)!}\,e^{-\lambda s}, \quad s \ge 0.$$

The Erlang-$n$ distribution has mean $n/\lambda$ and variance $n/\lambda^2$: $$E[S_3] = \frac{3}{2} = 1.5\;\text{time units},$$ $$\operatorname{Var}(S_3) = \frac{3}{4} = 0.75\;\text{time units}^2.$$

In a Poisson process context, $S_3$ is the arrival time of the third event — this is the connection between inter-arrival times and the counting process. $\square$

The characteristic equation is $$\det(\mathbf{C}_{\mathbf{X}} - \lambda \mathbf{I}) = (5-\lambda)^2 - 9 = \lambda^2 - 10\lambda + 16 = (\lambda - 8)(\lambda - 2) = 0.$$

Eigenvalues: $\lambda_1 = 8$, $\lambda_2 = 2$.

For $\lambda_1 = 8$: $(5-8)v_1 + 3v_2 = 0 \Rightarrow v_1 = v_2$, giving $\mathbf{q}_1 = \frac{1}{\sqrt{2}}[1, 1]^T$.

For $\lambda_2 = 2$: $(5-2)v_1 + 3v_2 = 0 \Rightarrow v_1 = -v_2$, giving $\mathbf{q}_2 = \frac{1}{\sqrt{2}}[1, -1]^T$.

With $\mathbf{Q} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ and $\mathbf{Y} = \mathbf{Q}^T \mathbf{X}$: $$\mathbf{C}_{\mathbf{Y}} = E[\mathbf{Y}\mathbf{Y}^T] = \mathbf{Q}^T \mathbf{C}_{\mathbf{X}} \mathbf{Q} = \mathbf{Q}^T \mathbf{Q} \mathbf{\Lambda} \mathbf{Q}^T \mathbf{Q} = \mathbf{\Lambda} = \begin{bmatrix} 8 & 0 \\ 0 & 2 \end{bmatrix}.$$

Since $\mathbf{C}_{\mathbf{Y}}$ is diagonal, $\operatorname{Cov}(Y_1, Y_2) = 0$: the components are uncorrelated.

In a $2 \times 2$ MIMO system, the received signals may be correlated due to insufficient antenna spacing or scattering geometry. The KLT rotates the coordinate system to align with the principal axes of the covariance ellipsoid.

After the transform, $Y_1$ (with variance $8$) carries the dominant signal energy along the sum direction $(\mathbf{q}_1)$, while $Y_2$ (variance $2$) captures the weaker difference component. This is equivalent to the whitening step in MIMO receivers, and when $\mathbf{X}$ is jointly Gaussian, uncorrelatedness implies independence — enabling independent per-stream detection. $\square$

For each $X_i \sim \text{Uniform}(0,1)$: $$E[X_i] = \frac{1}{2}, \quad \operatorname{Var}(X_i) = \frac{1}{12}.$$

By linearity of expectation and independence: $$E[S_n] = \frac{n}{2}, \quad \operatorname{Var}(S_n) = \frac{n}{12}.$$

For $n = 12$: $E[S_{12}] = 6$ and $\operatorname{Var}(S_{12}) = 1$, so $\sigma_{S_{12}} = 1$.

By the CLT: $$P(S_{12} > 7) \approx P\!\left(\frac{S_{12} - 6}{1} > \frac{7 - 6}{1}\right) = P(Z > 1) = 1 - \Phi(1) \approx 0.1587.$$

The exact distribution of $S_{12}$ is the Irwin-Hall distribution with $n = 12$, which has PDF given by the inclusion-exclusion formula: $$f_{S_{12}}(s) = \frac{1}{11!}\sum_{k=0}^{\lfloor s \rfloor} (-1)^k \binom{12}{k}(s - k)^{11}.$$

The exact value of $P(S_{12} > 7)$ is approximately $0.1573$.

The CLT approximation of $0.1587$ has a relative error of about $0.9\%$, which is remarkably accurate for $n = 12$. This is partly because the uniform distribution is symmetric and bounded, so higher-order cumulants vanish or are small, accelerating CLT convergence. In telecommunications, the sum of $12$ uniform RVs is a classic method for generating approximate Gaussian samples. $\square$

The Fourier transform of $R_X(\tau) = e^{-|\tau|}$ is $$S_X(f) = \int_{-\infty}^{\infty} e^{-|\tau|}\,e^{-j2\pi f \tau}\,d\tau = \frac{2\alpha}{\alpha^2 + (2\pi f)^2}.$$

With $\alpha = 1$: $$S_X(f) = \frac{2}{1 + 4\pi^2 f^2}.$$

The Fourier transform of $h(t) = 2e^{-2t}u(t)$ is $$H(f) = \int_0^{\infty} 2\,e^{-2t}\,e^{-j2\pi f t}\,dt = \frac{2}{2 + j2\pi f} = \frac{\beta}{\beta + j2\pi f}.$$

The squared magnitude: $$|H(f)|^2 = \frac{4}{4 + 4\pi^2 f^2}.$$

The output PSD: $$S_Y(f) = |H(f)|^2\,S_X(f) = \frac{4}{4 + 4\pi^2 f^2} \cdot \frac{2}{1 + 4\pi^2 f^2} = \frac{8}{(4 + 4\pi^2 f^2)(1 + 4\pi^2 f^2)}.$$

Let $u = 4\pi^2 f^2$. We use partial fractions: $$\frac{8}{(4+u)(1+u)} = \frac{8}{3}\left(\frac{1}{1+u} - \frac{1}{4+u}\right).$$

The output power is $$E[Y^2(t)] = \int_{-\infty}^{\infty} S_Y(f)\,df = \frac{8}{3}\left[\int_{-\infty}^{\infty}\frac{df}{1+4\pi^2 f^2} - \int_{-\infty}^{\infty}\frac{df}{4+4\pi^2 f^2}\right].$$

Using $\int_{-\infty}^{\infty}\frac{df}{a^2+4\pi^2 f^2} = \frac{1}{2a}$:

• First integral ($a=1$): $1/2$.
• Second integral ($a=2$): $1/4$.

$$E[Y^2(t)] = \frac{8}{3}\left(\frac{1}{2} - \frac{1}{4}\right) = \frac{8}{3} \cdot \frac{1}{4} = \frac{2}{3} \approx 0.667.$$

The filter reduces the input power from $R_X(0) = 1$ to $2/3$, acting as a low-pass filter that attenuates high-frequency noise components. $\square$

All entries are non-negative. Row sums:

• Row $G$: $0.7 + 0.2 + 0.1 = 1.0$ $\checkmark$
• Row $F$: $0.3 + 0.5 + 0.2 = 1.0$ $\checkmark$
• Row $B$: $0.1 + 0.3 + 0.6 = 1.0$ $\checkmark$

Therefore $\mathbf{P}$ is a valid (right) stochastic matrix.

The equation $\boldsymbol{\pi}\mathbf{P} = \boldsymbol{\pi}$ gives: $$\pi_G = 0.7\pi_G + 0.3\pi_F + 0.1\pi_B, \tag{1}$$ $$\pi_F = 0.2\pi_G + 0.5\pi_F + 0.3\pi_B, \tag{2}$$ $$\pi_B = 0.1\pi_G + 0.2\pi_F + 0.6\pi_B. \tag{3}$$

From (1): $0.3\pi_G = 0.3\pi_F + 0.1\pi_B$, i.e., $3\pi_G = 3\pi_F + \pi_B$. $\quad(1')$

From (3): $0.4\pi_B = 0.1\pi_G + 0.2\pi_F$, i.e., $4\pi_B = \pi_G + 2\pi_F$. $\quad(3')$

From $(3')$: $\pi_G = 4\pi_B - 2\pi_F$.

Substitute into $(1')$: $3(4\pi_B - 2\pi_F) = 3\pi_F + \pi_B$, $12\pi_B - 6\pi_F = 3\pi_F + \pi_B$, $11\pi_B = 9\pi_F$, $\pi_F = \frac{11}{9}\pi_B$.

Then $\pi_G = 4\pi_B - \frac{22}{9}\pi_B = \frac{14}{9}\pi_B$.

Normalization: $\frac{14}{9}\pi_B + \frac{11}{9}\pi_B + \pi_B = 1$, so $\frac{34}{9}\pi_B = 1$, giving $\pi_B = \frac{9}{34}$.