 # Dragon Notes UNDER CONSTRUCTION
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# Randomness & Probability:Solved Problems

$$\bb{Pr1}\$$Find PDF of $$Y=e^{X}$$ for $$X\t{~}N(0,1)$$
$$S_Y =\{y:y>0\}$$. Let $$\ y=e^x$$; then, $$x=\ln{y}\Rightarrow g^{-1}(y)=\ln{y}$$
\ds p_Y(y)=p_X(\ln{y})\left|\frac{d\ln{y}}{dy}\right|= \left\{ \begin{array}{c} \begin{align} & p_X(\ln{y})\sfrac{1}{y},\ \ y > 0 \\ & 0, \hspace{86px} y\leq 0 \end{align} \end{array} \right. \ds =\bxred{\left\{ \begin{array}{c} \begin{align} & \sfrac{1}{\sqrt{2\pi}y}\t{exp}[-\sfrac{1}{2}(\ln{y})^2],\ \ y > 0 \\ & 0, \hspace{167px} y\leq 0 \end{align} \end{array} \right.}
$$\bb{Pr2}\$$Find PDF of $$Y=X^2$$ for $$X\t{~}N(0,1)\vplLup$$
The solutions to $$y=x^2$$ are $$x_1=-\sqrt{y}=g^{-1}_1(y),$$ and $$x_2=\sqrt{y}=g^{-1}_2(y)$$. We must add the PDFs (since both the $$x$$-intervals map into the same $$y$$-interval and the $$x$$-intervals are disjoint) to yield
$$p_Y(y)=p_X(g^{-1}_1(y))\left|\lfrac{dg^{-1}_1(y)}{dy}\right|+p_X(g^{-1}_2(y))\left|\lfrac{dg^{-1}_2(y)}{dy}\right|$$ = \left\{ \begin{array}{c} \begin{align} & \small{}\left[\frac{1}{\sqrt{2\pi}}e^{-y/2}\right]\frac{1}{2\sqrt{y}}+\left[\frac{1}{\sqrt{2\pi}}e^{-y/2}\right]\frac{1}{2\sqrt{y}},\ \ y \geq 0 \\ & \small{} 0, \hspace{265px} y < 0 \end{align} \end{array} \right.= \small{} \bxred{\left\{ \begin{array}{c} \begin{align} & \frac{1}{\sqrt{2\pi y}}e^{-y/2},\ \ y \geq 0 \\ & \small{} 0, \hspace{78px} y < 0 \end{align} \end{array} \right.}
$$\bb{Pr3}\$$Consider a Rayleigh rv that is input to a device that limits its output (such as a thermometer, limiting to a max value; all temps. above this maximum will read as the maximum). Determine $$p_Y(y)$$.
The effect of the device can be represented by the transformation$$\vplup$$
\vplup y=g(x)= \left\{ \begin{array}{c} \begin{align} & x,\hspace{40px} 0\leq x < x_{\t{max}} \\ & x_{\t{max}}, \hspace{11px} \ x\geq x_{\t{max}} \end{align} \end{array} \right.
$$p_Y(y)=0$$ for $$y<0$$ since $$X$$ can only take on non-negative values. For $$0\leq y < x_{\t{max}},\ g^{-1}(y)=y.$$ Lastly, for $$y\geq x_{\t{max}}$$, we have an infinite number of solutions $$x\in [x_{\t{max}},\infty)$$ (left fig.). In region two, we have
$$\vplup p_Y(y)=p_X(g^{-1}(y))\left|\lfrac{dg^{-1}(y)}{dy}\right|=p_X(y)$$
In region three, $$Y$$ cannot exceed $$x_{\t{max}}$$ and so $$y=x_{\t{max}}$$ is the only possible value. Its probability is equal to the probability that $$X\geq x_{\t{max}}$$:
$$\vplup \ds P[Y=x_{\t{max}}]=\int_{x_{\t{max}}}^{\infty}p_X(x)dx$$
since, as shown in bot-left fig, the $$x$$-interval $$[x_{\t{max}},\infty)$$ is mapped into the $$y$$-point $$y=x_{\t{max}}$$ is nonzero, we represent it in the PDF by using an impulse as
$$\ds \vplup p_Y(y)=\left[\int_{x_{\t{max}}}^{\infty}p_X(x)dx\right] \delta (y-x_{\t{max}}),\ \ y=x_{\t{max}}$$
Since it contains an impulse, $$p_Y(y)$$ is the PDF of a mixed rv. Finally, for $$x\geq 0$$, the Rayleigh PDF for $$\sigma^2=1$$ is
$$\ds \vplup p_X(x)=x\exp{-x^2/2},\ \ \t{so that the PDF of }Y\t{ becomes (shown in bot-right fig)}$$
\ds \vplup p_Y(y)= \left\{ \begin{array}{c} \begin{align} & 0, && y<0 \\ & y\exp{-y^2/2}, && 0\leq y < x_{\t{max}} \\ & \left[\int_{x_{\t{max}}}^{\infty}x\exp{-x^2/2}dx\right] \delta (y-x_{\t{max}}), && y=x_{\t{max}} \\ & 0, && y > x_{\t{max}} \end{align} \end{array} \right.
=\bxred{ \left\{ \begin{array}{c} \begin{align} & 0, && y<0 \\ & y\exp{-y^2/2}, && 0\leq y < x_{\t{max}} \\ & \exp{x_{\t{max}}^2}\delta (y-x_{\t{max}}), && y=x_{\t{max}} \hspace{89px}\\ & 0, && y>x_{\t{max}} \end{align} \end{array} \right.}  $$\bb{Pr4}\$$Let $$(X,Y)$$ have a standard bivariate Gaussian PDF. Given the transformation
$$\ds \mtx{W}{Z}=\mtxx{\sigma_W}{0}{0}{\sigma_Z}\mtx{X}{Y}+\mtx{\mu_W}{\mu_Z}\vphantom{\frac{A^{A^A}}{A}}$$,
determine the transformed PDF $$p_{W,Z}(w,z)$$.

[Sol] First solve for $$x,y$$ as

$$\ds x=\frac{w-\mu_W}{\sigma_W},\quad y=\frac{z-\mu_Z}{\sigma_Z}$$
The inverse Jacobian matrix becomes
$$\ds \frac{\partial (x,y)}{\partial (w,z)}=\mtxx{1/\sigma_W}{0}{0}{1/\sigma_Z}$$
Applying $$\bb{15.2}$$ to $$\bb{6}$$, we obtain
$$\ds p_{W,Z}(w,z)=\frac{1}{2\pi\sqrt{1-\rho^2}}\cdot\t{exp}\lrbra{-\frac{1}{2(1-\rho^2)}\lrpar{\Frac{w-\mu_W}{\sigma_W}^2-2\rho\Frac{w-\mu_W}{\sigma_W}\Frac{z-\mu_Z}{\sigma_Z}+\Frac{z-\mu_Z}{\sigma_Z}^2}}\frac{1}{\sigma_W\sigma_Z}\Rightarrow$$
$$\ds \bxred{p_{W,Z}(w,z)=\frac{1}{2\pi\sqrt{(1-\rho^2)\sigma_W^2\sigma_Z^2}}\ \t{exp}\lrbra{-\frac{1}{2(1-\rho^2)}\lrpar{\Frac{w-\mu_W}{\sigma_W}^2-2\rho\Frac{w-\mu_W}{\sigma_W}\Frac{z-\mu_Z}{\sigma_Z}+\Frac{z-\mu_Z}{\sigma_Z}^2}}}$$ $$\bb{Pr5}\$$[Dart worthy?] Suppose you're challenged to a dart throwing contest by a champion player. The champion scores a bullseye 85% of the time. Suppose also that you aren't very skilled at dart-throwing, and that your darts are equally likely to land anywhere on the dartboard. Assume that bullseye radius is $$1/4\t{th}$$ the dartboard's - hence your odds of a bullseye are $$25\%$$.
$$\vplLup$$To make the challenge reasonable, the champion proposes the following. He gets one shot per game. For you, if your dart lands outside the region $$|x|\leq\Delta x/2$$, then you get to go again - until your dart lands within the region $$|x|\leq \Delta x/2$$, as shown in (a) below. You even get to pick the value of $$\Delta x$$ (note: only darts in the chosen region are counted). You may thus reason it best to exclude regions outside the bullseye - and choose as in (b).
$$\vplLup$$The champion bets you that, even with this aid, you will not outscore him. To solidify his confidence, he offers to return double any amount you bet - so if you stake $$5$$ and win, you get $$10$$.
$$\vplLup$$  Should you take this bet? If so, at which value of $$\Delta x$$ do odds turn to your favor? [Sol] To find the probability of your throw scoring a bullseye, we recall that your dart is equally likely to land anywhere on the dartboard. Hence, we
have
$$\ds P[\t{bullseye}|-\Delta x/2\leq X \leq \Delta x /2]=\frac{P[\t{bullseye,}-\Delta x/2\leq X \leq \Delta x/2]}{P[-\Delta x/2\leq X \leq \Delta x/2]}$$
$$\vplup$$Since $$\Delta x/2$$ is small, we can assume it to be $$\ll 1/4$$; therefore, the cross-hatched regions can be approximated by rectangles and so
\ds \begin{align} P[\t{bullseye}|-\Delta x/2\leq X \leq \Delta x/2] &= \frac{\vphantom{\int^{A^A}}P[\t{double cross-hatched region}]}{P[\t{double cross-hatched region}+P[\t{single cross-hatched region}]} \\ &= \frac{\Delta x(1/2)/\pi}{\Delta x(2)/\pi}\quad \t{(probability = rectangle area / dartboard area)} \\ &= \bxred{0.25 < 0.85} \end{align}
(The approximations are due to the use of rectangular approximations to the cross-hatched regions, which become exact as $$\Delta x\rightarrow 0$$)
$$\vplup$$Hence, the champion will have a higher probability of winning for any $$\Delta x$$, now matter how small it is chosen.
Even with rewards doubled (or tripled), you end up at a loss.
$$\bb{Pr6}\$$[rv Decorrelation] Decorrelate rv's $$X_1,X_2$$, whose PMF is given as
 $$x_2=-8\quad x_2=0\quad x_2=2\quad x_2=6$$ $$p_{X1}[x_1]$$ $$\mtXxxx{\ \ \ x_1=-8\vphantom{\frac{1}{4}}}{x_1=8\vphantom{\frac{1}{4}}}{x_1=2\vphantom{\frac{1}{4}}}{x_1=6\vphantom{\frac{1}{4}}}$$ $$\hspace{20px}\mtXxxx{\vphantom{\frac{1}{4}}0}{\frac{1}{4}}{\vphantom{\frac{1}{4}}0}{\vphantom{\frac{1}{4}}0}\hspace{60px}\mtXxxx{\frac{1}{4}}{\vphantom{\frac{1}{4}}0}{\vphantom{\frac{1}{4}}0}{\vphantom{\frac{1}{4}}0}\hspace{50px}\mtXxxx{\vphantom{\frac{1}{4}}0}{\vphantom{\frac{1}{4}}0}{\vphantom{\frac{1}{4}}0}{\frac{1}{4}}\hspace{50px}\mtXxxx{\vphantom{\frac{1}{4}}0}{\vphantom{\frac{1}{4}}0}{\frac{1}{4}}{\vphantom{\frac{1}{4}}0}$$ $$\quad \mtXxxx{\frac{1}{4}}{\frac{1}{4}}{\frac{1}{4}}{\frac{1}{4}}$$ $$p_{X_2}[x_2]$$ $$\hspace{22px}\frac{1}{4}\hspace{66px}\frac{1}{4}\hspace{58px}\frac{1}{4}\hspace{54px}\frac{1}{4}$$

[Sol] First determine the covariance matrix $$\bn{C}_X$$ and then $$\bn{A}$$ such that $$\bn{Y}=\bn{AX}$$ consists of uncorrelated variables. From the table, we have
\ds \begin{align} \vplLup E_{X_1}[X_1] &= E_{X_2}[X_2]=0 \\ E_{X_2}[X_2] &= E_{X_2}[X_2^2]=26 \\ E_{X_1,X_2}[X_1X_2] &= 6 \\ &\Rightarrow \\ \var{X_1} &= \var{X_2}=26 \\ \cov{X_1,X_2} &= 6 \\ &\Rightarrow \end{align}
$$\ds \boxed{\bn{C}_X=\mtxx{26}{6}{6}{26}}$$
$$\vplup$$Next, find the eigenvectors:
$$\ds \t{det}\left(\mtxx{26-\lambda}{6}{6}{26-\lambda}\right)=0 \Rightarrow (26-\lambda)(26-\lambda)-36=0 \Rightarrow$$
$$\vplup\boxed{\lambda_1=20,\ \lambda_2=32}$$

$$\vplup$$Solving for the corresponding eigenvectors and normalizing yields
$$\ds \vplup (\bn{C}_X-\lambda_1\bn{I})\bn{v}_1=\mtxx{6}{\hspace{10px}6}{6}{\hspace{10px}6}\mtx{v_1}{v_2}=\mtx{0}{0}\Rightarrow \boxed{\bn{v}_1=\mtx{1/\sqrt{2}}{-1/\sqrt{2}}}$$
$$\ds \vplup (\bn{C}_X-\lambda_2\bn{I})\bn{v}_2=\mtxx{-6}{6}{6}{-6}\mtx{v_1}{v_2}=\mtx{0}{0}\Rightarrow \boxed{\bn{v}_2=\mtx{1/\sqrt{2}}{1/\sqrt{2}}}$$

$$\vplup$$The modal matrix becomes
$$\ds \bn{V}=[\bn{v_1}\ \ \bn{v_2}]=\mtxx{1/\sqrt{2}}{1/\sqrt{2}}{-1/\sqrt{2}}{1/\sqrt{2}},\t{ and therefore}$$
$$\ds \bn{V}^T=\boxed{\bn{A}=\mtxx{1/\sqrt{2}}{-1/\sqrt{2}}{1/\sqrt{2}}{1/\sqrt{2}}}$$
$$\vplup$$Hence, the transformed r-vector $$\bn{Y}=\bn{AX}$$ is explicitly
$$\ds \vplLup^{\vplup}\vplup \bxred{Y_1=\frac{1}{\sqrt{2}}X_1-\frac{1}{\sqrt{2}}X_2} \quad$$ $$\ds \vplup \bxred{Y_2=\frac{1}{\sqrt{2}}X_1+\frac{1}{\sqrt{2}}X_2}$$
$$\vplup$$$$Y_1$$ and $$Y_2$$ are hence uncorrelated rv's with
$$\ds \vplup E_{\bn{Y}}[\bn{Y}] = E_{\bn{Y}}[\bn{AX}]=\bn{A}E_{\bn{X}}[\bn{X}]=\bn{0}$$
$$\bn{C}_Y = \bn{A}\bn{C}_X\bn{A}^T = \bn{V}^T\bn{C}_X\bn{V} = \bn{\Lambda} = \mtxx{20}{0}{0}{32}$$
$$\vplLup$$Note that the solution works by rotating every datapoint by a fixed angle - effectively rotating the regression line, and changing its slope (the
correlation coefficient). The $$\bn{A}$$ found, and in general decorelation, is the rotation matrix:
$$\ds \bn{A}=\mtxx{\cos{\theta}}{-\sin{\theta}}{\sin{\theta}}{\cos{\theta}}$$
$$\vplup$$ - where $$\theta=\pi/4$$ in this example. As shown in the figure below, the values of $$\bn{X}$$ become the values of $$\bn{Y}$$ via a $$\deg{45}$$ rotation. $$\bb{Pr7}\$$[Bivariate Gaussian realization] Obtain a bivariate Gaussian realization $$(W,Z)$$ by transforming the standard bivariate Gaussian $$(X,Y)$$

[Sol] Given $$X,Y\sim N(0,1)$$, the distribution $$N\lrpar{[\mu_X\ \mu_Y],[\bn{C}]}$$ can be obtained by applying the affine transformation

$$\ds \mtx{W}{Z}=\bn{G}\mtx{X}{Y}+\mtx{a}{b}\quad \bb{1^*}$$
$$\vplup$$ The mean vector and covariance matrix of $$[W\ Z]^T$$ transform according to

$$\ds E\lrbra{\mtx{W}{Z}}=\bn{G}E\lrbra{\mtx{X}{Y}}+\mtx{a}{b}$$
$$\ds \vplup \bn{C}_{W,Z}=\bn{G}\bn{C}_{X,Y}\bn{G}^T\quad \bb{2^*}$$
Then, $$(W,Z)$$ with a given mean $$[\mu_W\ \mu_Z]^T$$ and covariance matrix $$\bn{C}_{W,Z}$$ can be obtained by applying $$\bb{1^*}$$ with a suitable $$\bn{G}$$ and $$[a\ b]^T$$ so that

$$\ds E\lrbra{\mtx{W}{Z}}=\mtx{\mu_W}{\mu_Z},\quad \bn{C}_{W,Z}=\mtxx{\sigma_W^2}{\rho\sigma_W\sigma_Z}{\rho\sigma_Z\sigma_W}{\sigma_W^2}\quad \bb{3^*}$$
Since $$(X,Y)$$ are zero-mean, $$E[[X\ Y]^T]=0$$ - and so we choose $$a=\mu_W$$ and $$b=\mu_Z$$. Also, since $$X$$ & $$Y$$ are independent, hence uncorrelated, and
with unity variances, we have
$$\ds \bn{C}_{X,Y}=\mtxx{1}{\hspace{10px} 0}{0}{\hspace{10px}1}=\bn{I}$$
To obtain $$\bn{C}_{W,Z}$$ using $$\bb{2^*}$$, construct $$\bn{G}$$ as follows; let $$\bn{G}$$ be a lower triangular matrix
$$\ds \bn{G}=\mtxx{a}{\hspace{10px}0}{b}{\hspace{10px}c}$$
$$\Rightarrow$$
$$\bn{G}\bn{G}^T=\mtxx{a}{\hspace{10px}0}{b}{\hspace{10px}c}=\hspace{5px}$$$$\mtxx{a^2}{ab}{ab}{b^2+c^2}$$$$\mtxx{a}{\hspace{10px}b}{0}{\hspace{10px}c}$$
Next, equate the elements of $$\bn{C}_{W,Z}$$ in $$\bn{3^*}$$ to those of $$\bn{G}\bn{G}^T$$ above:
$$\ds a=\sigma_W,\quad b=\rho\sigma_Z,\quad c=\sigma_Z\sqrt{1-\rho^2}$$
Hence, we have
$$\ds \bn{G}=\mtxx{\sigma_W}{0}{\rho\sigma_Z}{\sigma_Z\sqrt{1-\rho^2}}$$
- and the complete transformation is

$$\ds \bxred{\mtx{W}{Z}=\mtxx{\sigma_W}{0}{\rho\sigma_Z}{\sigma_Z\sqrt{1-\rho^2}}\mtx{X}{Y}+\mtx{\mu_W}{\mu_Z}}$$

$$\bb{Pr8}\$$[Second-order joint moments of multivariate Gaussian PDF]
Derive the second-order moments of $$\bn{X}\sim N(\bn{0},\bn{C})$$. The characteristic function is
$\phi_\bn{X}(\bn{\omega})=\exp{-\frac{1}{2}\bn{\omega}^T\bn{C\omega}}$
[Sol]
$$\ds\vplup \t{Let }Q(\bn{\omega})=\bn{\omega}^T\bn{C\omega}=\sum_{m=1}^{N}\sum_{n=1}^{N}\omega_m\omega_n\bb{C}_{mn}$$
- where $$\bb{C}_{mn}=c_mn$$ for simplification. Note that $$Q$$ has a quadratic form. Hence, with $$l_i=l_j=1$$ and the other $$l\t{'s}=0$$, we have

\ds\vplup \begin{align} E_{X_1,X_2,...,X_N}[X_1^{l_1}X_2^{l_2}...X_N^{l_N}]&=\frac{1}{j^{l_1+l_2+...+l_N}}\frac{\partial^{l_1+l_2+...+l_N}} {\partial \omega_1^{l_1}\partial \omega_2^{l_2}...\partial \omega_N^{l_N}}\phi_{X_1,X_2,...,X_N}(\omega_1,\omega_2,...,\omega_N) \vert {}_{\omega_1=\omega_2=...=\omega_N=0}\Rightarrow \\ E_{X_iX_j}[X_iX_j]&=\frac{1}{j^2}\frac{\partial^2}{\partial\omega_i\partial\omega_j}\exp{-Q(\bn{\omega})}{\vert}_{\bn{\omega}=0}\\ &\Rightarrow \\ \frac{\partial\t{exp}[-Q(\bn{\omega})/2]}{\partial\omega_i} &= -\frac{1}{2}\frac{\partial Q(\bn{\omega})}{\partial \omega_i}\exp{-Q(\bn{\omega})}\\ \frac{\partial^2\t{exp}[-Q(\bn{\omega})/2]}{\partial\omega_i\partial\omega_j}&=\frac{1}{4}\frac{\partial Q(\bn{\omega})}{\partial \omega_i} \frac{\partial Q(\bn{\omega})}{\partial\omega_j}\exp{-Q(\bn{\omega})}-\frac{1}{2}\frac{\partial^2 Q(\bn{\omega})}{\partial\omega_i\partial\omega_j}\exp{-Q(\bn{\omega})}\ \bb{1^*}\t{. But,}\\ \frac{\partial Q(\bn{\omega})}{\partial\omega_i}{\vert}_{\bn{\omega}=0} &= \sum_{m=1}^{N}\sum_{n=1}^{N}\frac{\partial \omega_m\omega_n}{\partial\omega_i}\bb{C}_{mn}{\vert}_{\bn{\omega}=0}\\ &= \sum_{m=1}^{N}\sum_{n=1}^{N} \lrbra{\omega_m\frac{\partial\omega_m}{\partial\omega_i}\bb{C}_{mn}+\omega_n\frac{\partial\omega_m}{\partial\omega_n}\bb{C}_{mn}}_{\bn{\omega}=0}\ \bb{2^*}\t{, and also}\\ \frac{\partial^2Q(\bn{\omega})}{\partial\omega_i\partial\omega_j}{\vert}_{\bn{\omega}=0}&=\sum_{m=1}^{N}\sum_{n=1}^{N}\frac{\partial^2\omega_m\omega_n}{\partial\omega_i\omega_j}\bb{C}_{mn}{\vert}_{\bn{\omega}=0}\t{. But}\\ \frac{\partial^2\omega_m\omega_n}{\partial\omega_i} &= \omega_m\frac{\partial\omega_n}{\partial\omega_n}+\omega_n\frac{\partial\omega_m}{\partial\omega_i}\\ &= \omega_m\delta_{ni}+\omega_n\delta_{mi}\t{,}\\ \t{where }\delta_{ij}\t{ is the Kronecker}&\t{ delta, defined to be }1\t{ if }i=j\t{ and }0\t{ otherwise. Hence,}\\ \frac{\partial^2\omega_m\omega_n}{\partial\omega_i\partial\omega_j} &= \delta_{mj}\delta_{ni}+\delta_{nj}\delta_{mi} \end{align}
and $$\delta_{mj}\delta_{ni}=1$$ if $$(m,n)=(j,i)$$ and $$=0$$ otherwise, and $$\delta_{nj}\delta_{mi}=1$$ if $$(m,n)=(i,j)$$ and $$=0$$ otherwise. Thus,
\ds \vplup \begin{align} \frac{\partial^2 Q(\bn{\omega})}{\partial\omega_i\omega_j}\vert{}_{\bn{\omega}=0} &= c_{ji}+c_{ij}\\ &= 2c_{ij}\ \ (\bn{C}^T=\bn{C})\end{align}
Then, we have the expected result from $$\bb{1^*}$$ and $$\bb{2^*}$$ that
\ds \vplup \begin{align} E_{X_i X_j} &=\frac{1}{j^2}\lrbra{-\frac{1}{2}\frac{\partial^2 Q(\bn{\omega})}{\partial\omega_i\omega_j}\exp{-Q(\bn{\omega})/2} }\vert {}_{\bn{\omega}=\bn{0}}\\ &= \frac{1}{j^2}\lrpar{-\frac{1}{2}}(2c_{ij})=c_{ij}=\bb{C}_{ij}\end{align}
Lastly, we extend the characteristic function approach to determining the PDF for a sum of IID rv's. Letting $$Y=\sum_{i=1}^{N}X_i$$, the char. function of $$Y$$ is
defined by
$$\ds \vplup \phi_Y(\omega)=E_Y[\t{exp}(j\omega Y)]$$
With $$g(X_1,X_2,...,X_N)=\t{exp}[j\omega\sum_{i=1}^{N}X_i]$$, evaluating real and imaginary integrals separately, we have
\ds \vplup \begin{align} E_{X1,X_2,...X_N}[g(X_1,X_2,...,X_N)] &= \pnint\pnint\cdots\pnint g(x_1,x_2,...,x_N)p_{X_1,X_2,...,X_N}(x_1,x_2,...,x_N)dx_1dx_2...dx_N \Rightarrow\\ \phi_Y(\omega) &= E_{X_1,X_2,...,X_N}\lrbra{\exp{j\omega\sum_{i=1}^{N}X_i}}\\ &= E_{X_1,X_2,...,X_N}\lrbra{\prod_{i=1}^{N}\t{exp}(j\omega X_i) }.\t{ Now, since } X_i\t{'s are IID, we have that}\\ \phi_Y(\omega) &= \prod_{i=1}^{N}E_{X_i}[\t{exp}(j\omega X_i)]\t{ (independence)}\\ &= \prod_{i=1}^{N}\phi_{X_i}(\omega) \\ &= [\phi_X(\omega)]^N\t{ (identically distributed)} \end{align}
where $$\vplup\phi_X(\omega)$$ is the common char. function of the rv's. To finally obtain the PDF of the sum rv, we apply the inverse Fourier transform:
$\bxred{p_Y(y)=\frac{1}{2\pi}\pnint [\phi_X(\omega)]^N\t{exp}(j\omega y)d\omega}$
$$\bb{Pr9}\$$[Series/parallel system failure rates]

$$\s{\vplup^{\vplup}}$$Let RVs $$\s{}X$$ & $$\s{}Y$$ equal the times of failure of two systems $$\s{}S_1$$ & $$\s{}S_2$$, respectively, and $$\s{}Z$$ the time of failure of a joint system $$\s{}S$$ comprised of $$\s{}S_1$$ & $$\s{}S_2$$.
$$\hspace{1px}$$Let $$\s{}A=$$ series-connected $$\s{}S_1$$ & $$\s{}S_2$$, $$\s{}B=$$ parallel-connected $$\s{}S_1$$ & $$\s{}S_2$$ – both as shown below. Assume $$\s{}X$$ & $$\s{}Y$$ are independent.
$$\vplup$$ Determine the PDF and CDF of $$\s{}Z$$ for $$\s{}S$$ in $$\s{}A$$ & $$\s{}B$$. [Sol]
The distribution $$\s{}F_X(t)$$ is the probability that $$\s{}S_1$$ will fail prior to time $$\s{}t$$ (asm. start at $$\s{}t=0$$); same with $$\s{}F_Y(t)$$ for $$\s{}S_2$$. The join distribution $$\s{}F_{X,Y}(t_1,t_2)$$ equals the probability that $$\s{}S_1$$ fails prior to $$\s{}t_1$$ and $$\s{}S_2$$ fails prior to $$\s{}t_2$$.
Series:
Two systems are said to be connected in series if the combined system $$\s{}S$$ fails when at least one of them fails. It then follows that $$\s{}Z$$ is the smaller of the two numbers $$\s{}X$$ & $$\s{}Y$$; hence,
$$\s{}\ds Z^{\t{SER}}=\min{X}{Y}$$
Parallel:
$$\s{}S$$ fails only when both $$\s{}S_1$$ & $$\s{}S_2$$ fail; thus,
$$\s{}Z^{\t{PAR}}=\max{X}{Y}$$
PDF & CDF (series)
For a given $$\s{}z$$, the region $$\s{}D_z$$ in the $$\s{}xy$$-plane is such that $$\s{}\min{x}{y} \leq z$$ – or, $$\s{}x≤z$$ or $$\s{}y≤z$$. Hence, to find $$\s{}F_Z(z)$$, it suffices to determine the mass in $$\s{}D_z;$$ $$\s{}F_Z(z)$$ is thus (see fig)
$$\s{}\ds F_Z(z) = F_X(z)+F_Y(z)-F_{X,Y}(z,z);$$
$$\s{}[\{X,Y\} = \t{indep.}] \rightarrow \bxred{F_Z^{\t{SER}}(z)=F_X(z)+F_Y(z)-F_X(z)F_Y(z)} \hspace{173px}$$
$$\s{}\ds \vplup \Rightarrow f_Z(z) = f_X(z) + f_Y(z) - f_X(z)F_Y(z)-F_X(z)f_Y(z)\Rightarrow$$
$$\s{}\ds \bxred{f_Z^{\t{SER}}(z)=f_X(z)[1-F_Y(z)]+f_Y(z)[1-F_X(z)]}$$
PDF & CDF (parallel)
$$\s{}D_z$$ in the $$\s{}xy$$-plane is such that $$\s{}\max{x}{y}\leq z$$ – or, $$\s{}x\leq z$$ and $$\s{}y\leq z$$. The mass of this region equals $$\s{}F_{X,Y}(z,z)$$ (see fig); hence,
$$\s{}\ds F_Z(z) = F_{X,Y}(z,z)\rightarrow$$
$$\s{}\bxred{F_Z^{\t{PAR}}(z)=F_X(z)F_Y(z)}$$
$$\s{}\ds \Rightarrow f_Z(z)=\pfrac{F_{X,Y}(z,z)}{x}+\pfrac{F_{X,Y}(z,z)}{y}=\intlim{-\infty}{z}f_{X,Y}(z,y)dy + \intlim{-\infty}{z}f_{X,Y}(z,y)dx\Rightarrow$$
$$\s{}\bxred{f_Z^{\t{PAR}}(z)=f_X(z)F_Y(z)+f_Y(z)F_X(z)}$$ $$\bb{Pr10}\$$[White Noise Power Spectral Density]
Derive the power spectral density of the white noise random process.

[Sol] Since $$X[n]$$ is white noise, it has a zero mean and ACS $$r_X[k]=\sigma^2\delta[k]$$. Then,
\ds\begin{align}P_X(f) &= \ilim{M}\frac{1}{2M+1}E\lrbra{\sum_{n=-M}^{M}X[n]\exp{j2\pi fn}\sum_{m=-M}^{M}X[m]\exp{-j2\pi fm}} \\ &= \ilim{M}\frac{1}{2M+1}\sum_{n=-M}^{M}\sum_{m=-M}^{M}\underbrace{E[X[n]X[m]]}_{r_X[m-n]}\t{exp}[-j2\pi f(m-n)]\quad \bb{1^*} \\ &= \ilim{M}\frac{1}{2M+1}\sum_{n=-M}^{M}\sum_{m=-M}^{M}\sigma^2 \delta[m-n]\t{exp}[-j2\pi f(m-n)] \\ &= \ilim{M}\frac{1}{2M+1}\sum_{n=-M}^{M}\sigma^2 \\ &= \ilim{M}\sigma^2 =\sigma^2 \end{align}
Hence, the PSD for white noise is
$$\ds \bxred{P_X(f)=\sigma^2 \quad -1/2\leq f \leq 1/2}$$
- illustrating that white noise contains equal contributions of average power at all frequencies.

[Sol 2] A more straightforward approach utilizes the knowledge of ACS. From $$\bb{1^*}$$ we see
$$\ds P_X(f)=\ilim{M}\frac{1}{2M+1}\sum_{n=-M}^{M}\sum_{m=-M}^{M}r_X[m-n]\t{exp}[-j2\pi f(m-n)]\quad \bb{2^*}$$
This can be simplified using
$$\ds \sum_{n=-M}^{M}\sum_{m=-M}^{M}g[m-n]=\sum_{k=-2M}^{2M}(2M+1-|k|)g[k]$$
which results from considering $$g[m-n]$$ as an element of the $$(2M+1)\times(2M+1)$$ matrix $$\bn{G}$$ with elements $$\bb{G}_{mn}=g[m-n]$$ for $$m=-M,...,M$$ and $$n=-M,...,M$$ and then summing all the elements. Using the relationship in $$\bb{2^*}$$ produces
\ds \begin{align}P_X(f) &= \ilim{M}\frac{1}{2M+1}\sum_{k=-2M}^{2M}(2M+1-|k|)r_X[k]\exp{-j2\pi fk} \\ &= \ilim{M}\sum_{k=-2M}^{2M}\lrpar{1-\frac{|k|}{2M+1}}r_X[k]\exp{-j2\pi fk} \end{align}
Assuming that $$\sum_{k=-\infty}^{\infty}|r_X[k]|<\infty$$, the limit can be shown to produce the final result
$$\ds P_X(f)=\sum_{k=-\infty}^{\infty}r_X[k]\exp{-j2\pi fk}$$
which states that the PSD is the discrete-time Fourier transform of the ACS. Since $$r_X[k]=\sigma^2 \delta [k]$$,
$$\ds P_X(f) = \sum_{k=-\infty}^{\infty}\sigma^2 \delta [k]\exp{-j2\pi fk} \Rightarrow$$
$$\ds \bxred{P_X(f)=\sigma^2 \quad -1/2\leq f \leq 1/2}$$
- in agreement with previous solution. $$P_X(f)$$ is shown in the figure below; as can be seen, the total average power in $$X[n]$$, which is $$r_X=\sigma^2$$, is given by the area under the PSD curve. $$\bb{Pr11}\$$[Linear prediction of MA random process, one-step]
Find the optimal predictor $$\hat{X}[n_0+1]$$ and the minimum mean-square error $$\ts{mse}{min}=\sigma_U^2$$, given the zero-mean WSS rp
$$\ds X[n]=U[n]-bU[n-1],\ \ \abs{b}<1$$

[Sol] The solution shall be executed as folllows:

: Write the $$z$$-transform of the ACS as
$$\ds\mathcal{P}_X(z)=\frac{\sigma_U^2}{\mathcal{A}(z)\mathcal{A}(z^{-1})},$$
$$\ds\t{where }\mathcal{P}_X(z)=\iisum{k}r_X[k]z^{-k},\quad \mathcal{A}(z)=1-\iisum{k}a[k]z^{-k}$$
asm
$$\mathcal{A}(z)$$ has all its zeros inside the $$z$$-plane unit circle (sequence is stable and causal)

: For the impulse response, the solution of
$$\ds r_X[l+1]=\zisum{k}h[k]r_X[l-k],\quad l=0,1,...$$
$$\ds\t{is }h_{\t{opt}}[k]=a[k+1],\quad k=0,1,...$$

: The optimal predictor is then
$$\ds\hat{X}[n_0+1]=\zisum{k}a[k+1]X[n_0-k],\quad \t{with }\ts{mse}{min}=\sigma_U^2$$

First determine the PSD. Since the system function is $$\mathcal{H}(z)=1-bz^{-1}$$, the frequency response follows as $$H(f)=1-b\ \t{exp}(-j2\pi f)$$. Then,
$$P_X(f)=H(f)\cdot H^*(f)\sigma_U^2=(1-b\t{exp}(-j2\pi f))(1-b\t{exp}(j2\pi f))\sigma_U^2$$
Replacing $$\t{exp}(j2\pi f)$$ by $$z$$, we have
$$\ds\mathcal{P}_X(z)=(1-bz^{-1})(1-bz)\sigma_U^2\Rightarrow$$
$$\ds\boxed{\mathcal{A}(z)=\frac{1}{1-bz^{-1}}}$$
To convert to $$1-\sum_{k=1}^{\infty}a[k]z^{-k}$$, take the inverse $$z$$-transform, assuing a stable and causal sequence:
$$\ds\mathcal{Z}^{-1}\{\mathcal{A}(z)\}=\left\{\nmtx{b^k,\ \ k\geq 0}{0,\ \ k<0}\right.$$
and so $$a[k]=-b^k$$ for $$k\geq1$$. The optimal predictor is thus
\ds\begin{align}\hat{X}[n_0+1] &= \zisum{k}a[k+1]X[n_0+k] \\ &= \zisum{k}(-b^{k+1})X[n_0-k] \Rightarrow \end{align}
$$\ds\bxred{\hat{X}[n_0+1]= -bX[n_0]-b^2X[n_0-1]-b^3X[n_0-2]-...,\ \ \ \ts{mse}{min}=\sigma_U^2}$$

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