Contents
6.2 MRB 1994 Final
1. (15 pts.)
Three boxes that appear identical contain the following combinations of coins: Box X - 2 quaters; Box Y - 1 quaters, 2 dimes; Box Z - 1 quater, 1 dime. One of the boxes is selected at random, and a coin is selected at random from that box. The coin selected is a quater. What is the probability that the box selected contains at least one dime?
Solution
• We can define the events as
– A = Box selected at random contains at least one dime.
– Q = Coin drawn from box selected is a quater.
– X = Box X is selected.
– Y = Box Y is selected.
– Z = Box Z is selected.
• From the given information, we have $ P\left(X\right)=P\left(Y\right)=P\left(Z\right)=1/3 . P\left(Q|X\right)=1,P\left(Q|Y\right)=1/3,P\left(Q|Z\right)=1/2 $ .
• By using Bayes' theorem, $ P\left(A|Q\right) $ is
$ P\left(A|Q\right)=\frac{P\left(A\cap Q\right)}{P\left(Q\right)}=\frac{P\left(\left(Y\cup Z\right)\cap Q\right)}{P\left(Q\right)}=\frac{P\left(Y\cap Q\right)\cup P\left(Z\cap Q\right)}{P\left(Q\cap X\right)+P\left(Q\cap Y\right)+P\left(Q\cap Z\right)} $$ =\frac{P\left(Y\cap Q\right)+P\left(Z\cap Q\right)}{P\left(Q|X\right)P\left(X\right)+P\left(Q|Y\right)P\left(Y\right)+P\left(Q|Z\right)P\left(Z\right)} $$ =\frac{P\left(Q|Y\right)P\left(Y\right)+P\left(Q|Z\right)P\left(Z\right)}{P\left(Q|X\right)P\left(X\right)+P\left(Q|Y\right)P\left(Y\right)+P\left(Q|Z\right)P\left(Z\right)} $$ =\frac{P\left(Q|Y\right)+P\left(Q|Z\right)}{P\left(Q|X\right)+P\left(Q|Y\right)+P\left(Q|Z\right)}=\frac{\frac{1}{3}+\frac{1}{2}}{1+\frac{1}{3}+\frac{1}{2}}=\frac{\frac{5}{6}}{\frac{11}{6}}=\frac{5}{11}. $
2. (20 pts.)
Multiple Choice Problems: Select the single best answer to each of the following four multiple choice questions by circling the letter in fron of the answer. There is space to work out the problems on the next page if needed.
I.
Consider a random experiment with probability space $ \left(\mathcal{S},\mathcal{F},P\right) $ and let $ A\in\mathcal{F} $ and $ B\in\mathcal{F} $ . Let $ P\left(A\right)=1/3 , P\left(B\right)=1/3 $ , and $ P\left(A\cap B\right)=1/4 $ . Find $ P\left(A|\bar{B}\right) $ .
A. 1/12 B. 1/8 C. 1/6 D. 1/4 E. 1/3
Solution
$ P\left(A|\bar{B}\right)=\frac{P\left(A\cap\bar{B}\right)}{P\left(\bar{B}\right)}=\frac{P\left(A\cap\bar{B}\right)}{1-P\left(B\right)}=\frac{P\left(A\right)-P\left(A\cap B\right)}{1-P\left(B\right)}=\frac{\frac{1}{3}-\frac{1}{4}}{1-\frac{1}{3}}=\frac{\frac{1}{12}}{\frac{2}{3}}=\frac{1}{12}\cdot\frac{3}{2}=\frac{1}{8}. $
II.
Let $ \mathbf{X} $ and $ \mathbf{Y} $ be two jointly distributed random variables with joint pdf
$ f\left(x,y\right)=\left\{ \begin{array}{lll} e^{-\left(x+y\right)} \text{, for }x\geq0\text{ and }y\geq0,\\ 0 \text{, elsewhere.} \end{array}\right. $
Find $ P\left(\left\{ \mathbf{X}>\mathbf{Y}\right\} |\left\{ \mathbf{X}>1/2\right\} \right) $ .
A. $ 1/\sqrt{e} $ B. $ \left(2\sqrt{e}-1\right)/2\sqrt{e} $ C. $ \left(1-2\sqrt{e}\right)/2\sqrt{e} $ D. $ 2\sqrt{e}/\left(2\sqrt{e}-1\right) $ E. $ 2\sqrt{e}/\left(1-2\sqrt{e}\right) $
Solution
By using Bayes' theroem,
$ P\left(\left\{ \mathbf{X}>\mathbf{Y}\right\} |\left\{ \mathbf{X}>\frac{1}{2}\right\} \right)=\frac{P\left(\left\{ \mathbf{X}>\mathbf{Y}\right\} \cap\left\{ \mathbf{X}>\frac{1}{2}\right\} \right)}{P\left(\left\{ \mathbf{X}>\frac{1}{2}\right\} \right)}. $
$ P\left(\left\{ \mathbf{X}>\mathbf{Y}\right\} \cap\left\{ \mathbf{X}>\frac{1}{2}\right\} \right)=\int_{1/2}^{\infty}\int_{0}^{x}e^{-\left(x+y\right)}dydx=\int_{1/2}^{\infty}-e^{-\left(x+y\right)}\Bigl|_{0}^{x}dx $$ =\int_{1/2}^{\infty}\left[e^{-x}-e^{-2x}\right]dx=-e^{-x}+\frac{1}{2}e^{-2x}\Bigl|_{1/2}^{\infty}=e^{-\frac{1}{2}}-\frac{1}{2}e^{-1}. $
$ P\left(\left\{ \mathbf{X}>\frac{1}{2}\right\} \right)=\int_{1/2}^{\infty}e^{-\left(x+y\right)}dx=-e^{-\left(x+y\right)}\Bigl|_{1/2}^{\infty}=e^{-\left(\frac{1}{2}+y\right)}. $
$ P\left(\left\{ \mathbf{X}>\mathbf{Y}\right\} |\left\{ \mathbf{X}>\frac{1}{2}\right\} \right)=\frac{e^{-\frac{1}{2}}-\frac{1}{2}e^{-1}}{e^{-\frac{1}{2}}}=\frac{2\sqrt{e}-1}{2\sqrt{e}}. $
III.
Let $ \mathbf{X} $ be a random variable with mean 2 , variance 8 , and moment generating function $ \phi_{\mathbf{X}}\left(s\right)=E\left\{ e^{s\mathbf{X}}\right\} $ . Find the first three terms in the series expansion of $ \phi_{\mathbf{X}}\left(s\right) $ about zero. (Hint: The moment generating theorem).
A. $ 2s+2s^{2} $ B. $ 1+2s+6s^{2} $ C. $ 1+2s+2s^{2} $ D. $ 1+2s+4s^{2} $ E. $ 1+2s+12s^{2} $
Recall
According to the series expansion, $ e^{\lambda}=\sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!} $ .
Solution
$ \phi_{\mathbf{X}}\left(s\right)=E\left[e^{s\mathbf{X}}\right]=E\left[1+s\mathbf{X}+\frac{s^{2}\mathbf{X}^{2}}{2}+\cdots\right]=1+E\left[\mathbf{X}\right]s+\frac{E\left[\mathbf{X}^{2}\right]s^{2}}{2}+\cdots=1+2s+6s^{2}. $
Since $ E\left[\mathbf{X}^{2}\right]=Var\left[\mathbf{X}\right]+\left(E\left[\mathbf{X}\right]\right)^{2}=8+4=12 $ .
IV.
Find the characteristic function $ \Phi\left(\omega\right) $ of an exponentially distributed random variable with mean $ \mu $ .
A. $ \exp\left\{ \mu\left(e^{i\omega}-1\right)\right\} $ B. $ \exp\left\{ \mu e^{i\omega}\right\} -1 $ C. $ \left(1-i\omega\mu\right)^{-1} $ D. $ \left(1+i\omega\mu\right)^{-2} $ E. $ e^{i\omega\mu}e^{-\frac{1}{2}\omega^{2}\mu^{2}} $
Solution
$ \Phi\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\int_{0}^{\infty}\frac{1}{\mu}e^{-\frac{x}{\mu}}e^{i\omega x}dx=\frac{1}{\mu}\int_{0}^{\infty}e^{-x\left(1/\mu-i\omega\right)}=\frac{1}{\mu}\cdot\frac{e^{-x\left(1/\mu-i\omega\right)}}{-\left(1/\mu-i\omega\right)}\biggl|_{0}^{\infty} $$ =\frac{1}{\mu}\cdot\frac{1}{\left(1/\mu-i\omega\right)}=\frac{1}{1-i\omega\mu}=\left(1-i\omega\mu\right)^{-1}. $
$ \because\frac{1}{\mu}-i\omega>0 $ because $ i\omega $ is a imaginary term.
3. (15 pts.)
Let $ \left\{ t_{k}\right\} $ be a set of Poisson points with parameter \lambda on the positive real line such that if $ \mathbf{N}\left(t_{1},t_{2}\right) $ is defined as the number of points in the interval $ \left(t_{1},t_{2}\right] $ , then
$ P\left(\left\{ \mathbf{N}\left(t_{1},t_{2}\right)=k\right\} \right)=\frac{\left[\lambda\left(t_{2}-t_{1}\right)\right]^{k}e^{-\lambda\left(t_{2}-t_{1}\right)}}{k!},\quad k=0,1,2,\cdots,\quad t_{2}>t_{1}\geq0. $
Let $ \mathbf{X}\left(t\right)=\mathbf{N}\left(0,t\right) $ be the Poisson counting process associated with these points for $ t>0\;\left(n.b.,\;\mathbf{X}\left(0\right)=0\right) $
(a) Find the mean of $ \mathbf{X}\left(t\right) $ .
$ \Phi_{\mathbf{X}\left(t\right)}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}\left(t\right)}\right]=\sum_{k=0}^{\infty}e^{i\omega k}\frac{\left(\lambda t\right)^{k}e^{-\lambda t}}{k!}=e^{-\lambda t}\sum_{k=0}^{\infty}\frac{\left[e^{i\omega}\lambda t\right]^{k}}{k!}=e^{\lambda t\left[e^{i\omega}-1\right]}. $
$ \phi_{\mathbf{X}}\left(s\right)=e^{\lambda t\left[e^{s}-1\right]}. $
$ E\left[\mathbf{X}\left(t\right)\right]=\frac{d\phi_{\mathbf{X}}\left(s\right)}{ds}\biggl|_{s=0}=e^{\lambda t\left[e^{s}-1\right]}\lambda te^{s}\biggl|_{s=0}=\lambda t. $
Alternative solution: $ \mathbf{X}\left(t\right) $ is Poisson process with parameter $ \lambda t \Rightarrow E\left[\mathbf{X}\left(t\right)\right]=Var\left[\mathbf{X}\left(t\right)\right]=\lambda t $ .
(b) Find the variance of $ \mathbf{X}\left(t\right) $ .
$ E\left[\mathbf{X}\left(t\right)^{2}\right]=\frac{d^{2}\phi_{\mathbf{X}\left(t\right)}\left(s\right)}{ds^{2}}\left|_{s=0}\right.=\frac{d}{ds}\left[e^{\lambda t\left[e^{s}-1\right]}\lambda te^{s}\right]\left|_{s=0}\right.=\lambda te^{s}e^{\lambda t\left[e^{s}-1\right]}\lambda te^{s}+\lambda te^{s}e^{\lambda t\left[e^{s}-1\right]}\left|_{s=0}\right.=\left(\lambda t\right)^{2}+\lambda t. $
$ Var\left[\mathbf{X}\left(t\right)\right]=E\left[\mathbf{X}\left(t\right)^{2}\right]-\left(E\left[\mathbf{X}\left(t\right)\right]\right)^{2}=\left(\lambda t\right)^{2}+\lambda t-\left(\lambda t\right)^{2}=\lambda t. $
Alternative solution: $ \mathbf{X}\left(t\right) $ is Poisson process with parameter $ \lambda t \Rightarrow E\left[\mathbf{X}\left(t\right)\right]=Var\left[\mathbf{X}\left(t\right)\right]=\lambda t . $
(c) Derive an expression for the autocorrelation function of $ \mathbf{X}\left(t\right) $ .
Assume $ t_{2}>t_{1} $ .
$ R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{X}\left(t_{2}\right)\right]=E\left[\mathbf{X}\left(t_{1}\right)\left[\mathbf{X}\left(t_{2}\right)-\mathbf{X}\left(t_{1}\right)+\mathbf{X}\left(t_{1}\right)\right]\right] $$ =E\left[\mathbf{X}\left(t_{1}\right)\left[\mathbf{X}\left(t_{2}\right)-\mathbf{X}\left(t_{1}\right)\right]\right]+E\left[\mathbf{X}^{2}\left(t_{1}\right)\right] $$ =E\left[\mathbf{X}\left(t_{1}\right)\right]E\left[\mathbf{X}\left(t_{2}\right)-\mathbf{X}\left(t_{1}\right)\right]+E\left[\mathbf{X}^{2}\left(t_{1}\right)\right] $$ =\left(\lambda t_{1}\right)\lambda\left(t_{2}-t_{1}\right)+\lambda^{2}t_{1}^{2}+\lambda t_{1}=\lambda^{2}t_{1}t_{2}-\lambda^{2}t_{1}^{2}+\lambda^{2}t_{1}^{2}+\lambda t_{1}=\lambda^{2}t_{1}t_{2}+\lambda t_{1}. $
Similarly, for $ t_{2}<t_{1} $ , $ R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\lambda^{2}t_{1}t_{2}+\lambda t_{2} $ .
$ \therefore R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\lambda^{2}t_{1}t_{2}+\lambda\min\left(t_{1},t_{2}\right). $
$ \because $ Recall: $ Var\left[\mathbf{X}\left(t_{1}\right)\right]=E\left[\mathbf{X}^{2}\left(t_{1}\right)\right]-\left(E\left[\mathbf{X}\left(t_{1}\right)\right]\right)^{2}\Longrightarrow E\left[\mathbf{X}^{2}\left(t_{1}\right)\right]=Var\left[\mathbf{X}\left(t_{1}\right)\right]+\left(E\left[\mathbf{X}\left(t_{1}\right)\right]\right)^{2} $ .
(d) Is $ \mathbf{X}\left(t\right) $ wide-sense stationary? Explain your answer.
No, $ \mathbf{X}\left(t\right) $ is not WSS, because $ E\left[\mathbf{X}\left(t\right)\right]=\lambda t $ is not constant.
4. (15 pts.)
Let $ \mathbf{X} $ be a continuous random variable with pdf $ f_{\mathbf{X}}\left(x\right) $ , mean $ \mu $ , and variance $ \sigma^{2} $ . Prove the Chebyshev Inequality:$ P\left(\left\{ \mathbf{X}-\mu\right\} \geq\epsilon\right)\leq\frac{\sigma^{2}}{\epsilon^{2}} $, where $ \epsilon $ is any positive constant.
Solution
You can find the proof of Chebyshev inequality [CS1ChebyshevInequality].
5. (15 pts.)
Let $ \mathbf{X}\left(t\right) $ be a zero-mean wide-sense stationary Gaussian white noise process with autocorrelation function $ R_{\mathbf{XX}}\left(\tau\right)=S_{0}\delta\left(\tau\right) $ . Suppose that $ \mathbf{X}\left(t\right) $ is the input to a linear time invariant system with impulse response $ h\left(t\right)=e^{-\alpha t}\cdot1_{\left[0,\infty\right)}\left(t\right) $, where $ \alpha $ is a positive constant. Let $ \mathbf{Y}\left(t\right) $ be the output of the system and assume that the input has been applied to the system for all time.
(a) What is the mean of $ \mathbf{Y}\left(t\right) $ ?
$ E\left[\mathbf{Y}\left(t\right)\right]=E\left[\int_{-\infty}^{\infty}h\left(\tau\right)\mathbf{X}\left(t-\tau\right)d\tau\right]=\int_{-\infty}^{\infty}h\left(\tau\right)E\left[\mathbf{X}\left(t-\tau\right)\right]d\tau=\int_{-\infty}^{\infty}h\left(\tau\right)\cdot0d\tau=0. $
(b) What is the power spectral density of $ \mathbf{Y}\left(t\right) $ ?
$ S_{\mathbf{XX}}\left(\omega\right)=\int_{-\infty}^{\infty}S_{0}\delta\left(\tau\right)e^{-i\omega\tau}d\tau=S_{0}. $
$ H\left(\omega\right)=\int_{-\infty}^{\infty}h\left(t\right)e^{-i\omega t}dt=\int_{0}^{\infty}e^{-\alpha t}e^{-i\omega t}dt=\int_{0}^{\infty}e^{-\left(\alpha+i\omega\right)t}dt=\frac{e^{-\left(\alpha+i\omega\right)t}}{-\left(\alpha+i\omega\right)}\biggl|_{0}^{\infty}=\frac{1}{\alpha+i\omega}. $
$ S_{\mathbf{YY}}\left(\omega\right)=S_{\mathbf{XX}}\left(\omega\right)\left|H\left(\omega\right)\right|^{2}=S_{\mathbf{XX}}\left(\omega\right)H\left(\omega\right)H^{*}\left(\omega\right)=S_{0}\cdot\frac{1}{\alpha+i\omega}\cdot\frac{1}{\alpha-i\omega}=\frac{S_{0}}{\alpha^{2}+\omega^{2}}. $
(c) What is the autocorrelation function of $ \mathbf{Y}\left(t\right) $ ?
$ S_{\mathbf{YY}}\left(\omega\right)=\frac{S_{0}}{\alpha^{2}+\omega^{2}}=\left(\frac{S_{0}}{2\alpha}\right)\frac{2\alpha}{\alpha^{2}+\omega^{2}}\leftrightarrow\left(\frac{S_{0}}{2\alpha}\right)e^{-\alpha\left|\tau\right|}=R_{\mathbf{YY}}\left(\tau\right). $
$ \because e^{-\alpha\left|\tau\right|}\leftrightarrow\frac{2\alpha}{\alpha^{2}+\omega^{2}}\text{ (on the table given)}. $
If there is no table, then
$ R_{\mathbf{YY}}\left(\tau\right)=\frac{1}{2\pi}\int_{-\infty}^{\infty}S_{\mathbf{YY}}\left(\omega\right)e^{i\omega\tau}d\omega=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{S_{0}}{\alpha^{2}+\omega^{2}}\cdot e^{i\omega\tau}d\omega. $
(d) Write an expression for the second-order density $ f_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}\left(y_{1},y_{2}\right) $ of $ \mathbf{Y}\left(t\right) $ .
$ \mathbf{Y}\left(t\right) $ is a WSS Gaussian random process with $ E\left[\mathbf{Y}\left(t\right)\right]=0 , \sigma_{\mathbf{Y}\left(t\right)}^{2}=R_{\mathbf{YY}}\left(0\right)=\frac{S_{0}}{2\alpha} $ .
$ r_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}=r\left(t_{1}-t_{2}\right)=\frac{C_{\mathbf{YY}}\left(t_{1}-t_{2}\right)}{\sqrt{\sigma_{\mathbf{Y}\left(t_{1}\right)}^{2}\sigma_{\mathbf{Y}\left(t_{2}\right)}^{2}}}=\frac{R_{\mathbf{YY}}\left(t_{1}-t_{2}\right)}{R_{\mathbf{YY}}\left(0\right)}=e^{-\alpha\left|t_{1}-t_{2}\right|}. $
$ f_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}\left(y_{1},y_{2}\right)=\frac{1}{2\pi\sigma_{\mathbf{Y}\left(t_{1}\right)}\sigma_{\mathbf{Y}\left(t_{2}\right)}\sqrt{1-r^{2}}}\exp\left\{ \frac{-1}{2\left(1-r^{2}\right)}\left[\frac{y_{1}^{2}}{\sigma_{\mathbf{Y}\left(t_{1}\right)}^{2}}-\frac{2ry_{1}y_{2}}{\sigma_{\mathbf{Y}\left(t_{1}\right)}\sigma_{\mathbf{Y}\left(t_{2}\right)}}+\frac{y_{2}^{2}}{\sigma_{\mathbf{Y}\left(t_{2}\right)}^{2}}\right]\right\} $$ =\frac{1}{2\pi\frac{S_{0}}{2\alpha}\sqrt{1-e^{-2\alpha\left|t_{1}-t_{2}\right|}}}\exp\left\{ \frac{-1}{2\left(1-e^{-2\alpha\left|t_{1}-t_{2}\right|}\right)}\left[\frac{y_{1}^{2}}{S_{0}/2\alpha}-\frac{2y_{1}y_{2}e^{-\alpha\left|t_{1}-t_{2}\right|}}{S_{0}/2\alpha}+\frac{y_{2}^{2}}{S_{0}/2\alpha}\right]\right\} $$ =\frac{\alpha}{\pi S_{0}\sqrt{1-e^{-2\alpha\left|t_{1}-t_{2}\right|}}}\exp\left\{ \frac{-\alpha}{S_{0}\left(1-e^{-2\alpha\left|t_{1}-t_{2}\right|}\right)}\left[y_{1}^{2}-2y_{1}y_{2}e^{-\alpha\left|t_{1}-t_{2}\right|}+y_{2}^{2}\right]\right\} $ .
6. (20 pts.)
(a)
Let A , B , and C be three events defined on a random experiment. If $ P\left(A\cap B\cap C\right)=P\left(A\right)P\left(B\right)P\left(C\right) $ , then A , B , and C are statistically independent.
Recall
Two events A and B are independent iff $ P\left(A\cap B\right)=P\left(A\right)P\left(B\right) $ .
Solution
False. Must also know $ P(A\cap B)=P\left(A\right)P\left(B\right) $ , $ P\left(B\cap C\right)=P\left(B\right)P\left(C\right) $ , and $ P\left(C\cap A\right)=P\left(C\right)P\left(A\right) $ .
(b)
If the autocorrelation function $ R_{\mathbf{X}}\left(t_{1},t_{2}\right) $ of random process $ \mathbf{X}\left(t\right) $ can be written as a function of the time difference $ t_{2}-t_{1} $ , then $ \mathbf{X}\left(t\right) $ is wide-sense stationary.
Solution
False. $ E\left[\mathbf{X}\left(t\right)\right] $ must also be constant.
(c)
All stationary random processes are wide-sense stationary.
Solution
True.
(d)
The autocorrelation function $ R_{\mathbf{XX}}\left(\tau\right) $ of a real wide-sense stationary random process $ \mathbf{X}\left(t\right) $ is nonnegative for all $ \tau $ .
Solution
False. $ R_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ is non-negative definite. However, it does not mean that $ R_{\mathbf{XX}}\left(t_{1},t_{2}\right) $ is nonnegative.
(e)
Let $ \mathbf{X}\left(t\right) $ and $ \mathbf{Y}\left(t\right) $ be two zero-mean statistically independent, jointly wide-sense stationary random processes. Then the cross-correlation function $ R_{\mathbf{XY}}\left(\tau\right)=0 $ for all $ \tau $ .
Solution
True.
$ R_{\mathbf{XY}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{Y}^{*}\left(t_{2}\right)\right]=E\left[\mathbf{X}\left(t_{1}\right)\right]E\left[\mathbf{Y}^{*}\left(t_{2}\right)\right]=0\cdot0=0. $
(f)
The cross-correlation function $ R_{\mathbf{XY}}\left(\tau\right) $ of two real, jointly wide-sense stationary random process $ \mathbf{X}\left(t\right) $ and $ \mathbf{Y}\left(t\right) $ has its peak value at $ \tau=0 $ .
Solution
False. Consider $ \mathbf{Y}\left(t\right)=\mathbf{X}\left(t-\delta\right) $ where $ \delta\neq0 $ .
(g)
The power spectral density of a real, wide-sense stationary random process $ \mathbf{X}\left(t\right) $ is a non-negative even function of $ \omega $ .
Solution
True.
(h)
If $ \mathbf{X} $ and $ \mathbf{Y} $ are two statistically independent random variables, then $ f_{\mathbf{X}}\left(x|y\right)=f_{\mathbf{X}}\left(x\right) $ .
Solution.
True. $ P\left(\mathbf{\left\{ X=x\right\} }|\left\{ \mathbf{Y}=y\right\} \right)=\frac{P\left(\left\{ \mathbf{X}=x\right\} \cap\left\{ \mathbf{Y}=y\right\} \right)}{P\left(\left\{ \mathbf{Y}=y\right\} \right)}=\frac{P\left(\left\{ \mathbf{X}=x\right\} \right)\cdot P\left(\left\{ \mathbf{Y}=y\right\} \right)}{P\left(\left\{ \mathbf{Y}=y\right\} \right)}=P\left(\left\{ \mathbf{X}=x\right\} \right). $
$ f_{\mathbf{X}}\left(x|y\right)=\frac{f_{\mathbf{XY}}\left(x,y\right)}{f_{\mathbf{Y}}\left(y\right)}=\frac{f_{\mathbf{X}}\left(x\right)\cdot f_{\mathbf{Y}}\left(y\right)}{f_{\mathbf{Y}}\left(y\right)}=f_{\mathbf{X}}\left(x\right). $
(i)
If $ \mathbf{X} $ and $ \mathbf{Y} $ are two random variables, and $ f_{X}\left(x|y\right)=f_{X}\left(x\right) $ , then $ \mathbf{X} $ and $ \mathbf{Y} $ are statistically independent.
Solution
True.
(j)
If $ \left\{ \mathbf{X}_{n}\right\} $ is a sequence of random variables that converges to a random variable $ \mathbf{X} $ as $ n\rightarrow\infty $ , then $ \left\{ \mathbf{X}_{n}\right\} $ converges to $ \mathbf{X} $ in the means-square sense.
Solution
False. The explanation at first is about converge in almost everywhere. $ \left(a.e.\right)\nRightarrow\left(m.s.\right) $ and $ \left(m.s.\right)\nRightarrow\left(a.e.\right) $ .
