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− | ==7.8 QE 2004 January== | + | ==7.8 [[ECE_PhD_Qualifying_Exams|QE]] 2004 January== |
'''1. (30 pts.)''' | '''1. (30 pts.)''' | ||
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'''(a)''' | '''(a)''' | ||
− | Assume that <math>\mathcal{S}</math> is the sample space of a random experiment and that <math>\mathcal{F}_{1}</math> and <math>\mathcal{F}_{2}</math> are <math>\sigma</math> -fields (valid event spaces) on <math>\mathcal{S}</math> . Prove that <math>\mathcal{F}_{1}\cap\mathcal{F}_{2}</math> is also a <math>\sigma</math> -field on <math>S</math> . | + | Assume that <math class="inline">\mathcal{S}</math> is the sample space of a random experiment and that <math class="inline">\mathcal{F}_{1}</math> and <math class="inline">\mathcal{F}_{2}</math> are <math class="inline">\sigma</math> -fields (valid event spaces) on <math class="inline">\mathcal{S}</math> . Prove that <math class="inline">\mathcal{F}_{1}\cap\mathcal{F}_{2}</math> is also a <math class="inline">\sigma</math> -field on <math class="inline">S</math> . |
'''(b)''' | '''(b)''' | ||
− | Consider a sample space <math>\mathcal{S}</math> and corresponding event space <math>\mathcal{F}</math> . Suppose that <math>P_{1}</math> and <math>P_{2}</math> are both balid probability measures defined on <math>\mathcal{F}</math> . Prove that <math>P</math> defined by <math>P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right),\qquad\forall A\in\mathcal{F}</math> is also a valid probability measure on <math>\mathcal{F}</math> if <math>\alpha_{1},\;\alpha_{2}\geq0</math> and <math>\alpha_{1}+\alpha_{2}=1</math> . | + | Consider a sample space <math class="inline">\mathcal{S}</math> and corresponding event space <math class="inline">\mathcal{F}</math> . Suppose that <math class="inline">P_{1}</math> and <math class="inline">P_{2}</math> are both balid probability measures defined on <math class="inline">\mathcal{F}</math> . Prove that <math class="inline">P</math> defined by <math class="inline">P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right),\qquad\forall A\in\mathcal{F}</math> is also a valid probability measure on <math class="inline">\mathcal{F}</math> if <math class="inline">\alpha_{1},\;\alpha_{2}\geq0</math> and <math class="inline">\alpha_{1}+\alpha_{2}=1</math> . |
'''Answer''' | '''Answer''' | ||
− | • Because <math>P_{1}</math> and <math>P_{2}</math> are valid probability measures, we know that they satisfy the axioms of probability: | + | • Because <math class="inline">P_{1}</math> and <math class="inline">P_{2}</math> are valid probability measures, we know that they satisfy the axioms of probability: |
− | 1. <math>P_{1}\left(A\right)\geq0</math> and <math>P_{2}\left(A\right)\geq0</math> , <math>\forall A\in\mathcal{F}\left(\mathcal{S}\right)</math> . | + | 1. <math class="inline">P_{1}\left(A\right)\geq0</math> and <math class="inline">P_{2}\left(A\right)\geq0</math> , <math class="inline">\forall A\in\mathcal{F}\left(\mathcal{S}\right)</math> . |
− | 2. <math>P_{1}\left(\mathcal{S}\right)=1</math> and <math>P_{2}\left(\mathcal{S}\right)=1</math> . | + | 2. <math class="inline">P_{1}\left(\mathcal{S}\right)=1</math> and <math class="inline">P_{2}\left(\mathcal{S}\right)=1</math> . |
− | 3. If <math>A_{1}</math> and <math>A_{2}\in\mathcal{F}\left(\mathcal{S}\right)</math> are disjoint events, then <math>P_{1}\left(A_{1}\cup A_{2}\right)=P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right)</math> and <math>P_{2}\left(A_{1}\cup A_{2}\right)=P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right)</math> . | + | 3. If <math class="inline">A_{1}</math> and <math class="inline">A_{2}\in\mathcal{F}\left(\mathcal{S}\right)</math> are disjoint events, then <math class="inline">P_{1}\left(A_{1}\cup A_{2}\right)=P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right)</math> and <math class="inline">P_{2}\left(A_{1}\cup A_{2}\right)=P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right)</math> . |
− | 4. If <math>A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right)</math> is countable collection of disjoint events, then <math>P_{1}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)</math> and <math>P_{2}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right)</math> . | + | 4. If <math class="inline">A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right)</math> is countable collection of disjoint events, then <math class="inline">P_{1}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)</math> and <math class="inline">P_{2}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right)</math> . |
• Now, we check each condition to become a valid probability measure: | • Now, we check each condition to become a valid probability measure: | ||
− | 1. <math>P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)\geq0 , \forall A\in\mathcal{F}\left(\mathcal{S}\right)</math> . | + | 1. <math class="inline">P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)\geq0 , \forall A\in\mathcal{F}\left(\mathcal{S}\right)</math> . |
− | – <math>\because\alpha_{1}\geq0,\;\alpha_{2}\geq0,\; P_{1}\left(A\right)\geq0,\text{ and }P_{2}\left(A\right)\geq0</math> . | + | – <math class="inline">\because\alpha_{1}\geq0,\;\alpha_{2}\geq0,\; P_{1}\left(A\right)\geq0,\text{ and }P_{2}\left(A\right)\geq0</math> . |
− | 2. <math>P\left(S\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)=\alpha_{1}+\alpha_{2}=1</math> . | + | 2. <math class="inline">P\left(S\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)=\alpha_{1}+\alpha_{2}=1</math> . |
− | 3. If <math>A_{1}</math> and <math>A_{2}\in\mathcal{F}\left(\mathcal{S}\right)</math> are disjoint events, then <math>P\left(A_{1}\cup A_{2}\right)=\alpha_{1}P_{1}\left(A_{1}\cup A_{2}\right)+\alpha_{2}P_{2}\left(A_{1}\cup A_{2}\right)=\alpha_{1}\left\{ P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right)\right\} +\alpha_{2}\left\{ P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right)\right\}</math><math> =\alpha_{1}P_{1}\left(A_{1}\right)+\alpha_{2}P_{2}\left(A_{1}\right)+\alpha_{1}P_{1}\left(A_{2}\right)+\alpha_{2}P_{2}\left(A_{2}\right)=P\left(A_{1}\right)+P\left(A_{2}\right).</math> | + | 3. If <math class="inline">A_{1}</math> and <math class="inline">A_{2}\in\mathcal{F}\left(\mathcal{S}\right)</math> are disjoint events, then <math class="inline">P\left(A_{1}\cup A_{2}\right)=\alpha_{1}P_{1}\left(A_{1}\cup A_{2}\right)+\alpha_{2}P_{2}\left(A_{1}\cup A_{2}\right)=\alpha_{1}\left\{ P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right)\right\} +\alpha_{2}\left\{ P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right)\right\}</math><math class="inline"> =\alpha_{1}P_{1}\left(A_{1}\right)+\alpha_{2}P_{2}\left(A_{1}\right)+\alpha_{1}P_{1}\left(A_{2}\right)+\alpha_{2}P_{2}\left(A_{2}\right)=P\left(A_{1}\right)+P\left(A_{2}\right).</math> |
− | 4. If <math>A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right)</math> is countable collection of disjoint events, then <math>P\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}P_{1}\left(\cup_{i=0}^{\infty}A_{i}\right)+\alpha_{2}P_{2}\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)+\alpha_{2}\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right)</math><math>=\sum_{i=1}^{\infty}\left\{ \alpha_{1}P_{1}\left(A_{i}\right)+\alpha_{2}P_{2}\left(A_{i}\right)\right\} =\sum_{i=1}^{\infty}P\left(A_{i}\right).</math> | + | 4. If <math class="inline">A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right)</math> is countable collection of disjoint events, then <math class="inline">P\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}P_{1}\left(\cup_{i=0}^{\infty}A_{i}\right)+\alpha_{2}P_{2}\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)+\alpha_{2}\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right)</math><math class="inline">=\sum_{i=1}^{\infty}\left\{ \alpha_{1}P_{1}\left(A_{i}\right)+\alpha_{2}P_{2}\left(A_{i}\right)\right\} =\sum_{i=1}^{\infty}P\left(A_{i}\right).</math> |
'''2. (10 pts.)''' | '''2. (10 pts.)''' | ||
− | Identical twins come from the same egg and and hence are of the same sex. Fraternal twins have a probability <math>1/2</math> of being of the same sex. Among twins, the probability of a fraternal set is p and of an identical set is <math>q=1-p</math> . Given that a set of twins selected at random are of the same sex, what is the probability they are fraternal? (Simplify your answer as much as possible.) Sketch a plot of the conditional probability that the twins are fraternal given that they are of the same sex as a function of <math>q</math> (the probability that a set of twins are identical.) | + | Identical twins come from the same egg and and hence are of the same sex. Fraternal twins have a probability <math class="inline">1/2</math> of being of the same sex. Among twins, the probability of a fraternal set is p and of an identical set is <math class="inline">q=1-p</math> . Given that a set of twins selected at random are of the same sex, what is the probability they are fraternal? (Simplify your answer as much as possible.) Sketch a plot of the conditional probability that the twins are fraternal given that they are of the same sex as a function of <math class="inline">q</math> (the probability that a set of twins are identical.) |
'''Note''' | '''Note''' | ||
− | This problem is identical to the problem | + | This problem is identical to the problem in the [[ECE 600 Finals MRB 2004 Final|MBR 2004 Spring Final]]. |
'''3. (30 pts.)''' | '''3. (30 pts.)''' | ||
− | Let <math>\mathbf{X}\left(t\right)</math> be a real continuous-time Gaussian random process. Show that its probabilistic behavior is completely characterized by its mean <math>\mu_{\mathbf{X}}\left(t\right)=E\left[\mathbf{X}\left(t\right)\right]</math> and its autocorrelation function <math>R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{X}\left(t_{2}\right)\right].</math> | + | Let <math class="inline">\mathbf{X}\left(t\right)</math> be a real continuous-time Gaussian random process. Show that its probabilistic behavior is completely characterized by its mean <math class="inline">\mu_{\mathbf{X}}\left(t\right)=E\left[\mathbf{X}\left(t\right)\right]</math> and its autocorrelation function <math class="inline">R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{X}\left(t_{2}\right)\right].</math> |
'''4. (30 pts.)''' | '''4. (30 pts.)''' | ||
− | Assume that <math>\mathbf{X}\left(t\right)</math> is a zero-mean, continuous-time, Gaussian white noise process with autocorrelation function <math>R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\delta\left(t_{1}-t_{2}\right)</math>. Let <math>\mathbf{Y}\left(t\right)</math> be a new random process defined by <math>\mathbf{Y}\left(t\right)=\frac{1}{T}\int_{t-T}^{t}\mathbf{X}\left(s\right)ds</math>, where <math>T>0</math> . | + | Assume that <math class="inline">\mathbf{X}\left(t\right)</math> is a zero-mean, continuous-time, Gaussian white noise process with autocorrelation function <math class="inline">R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\delta\left(t_{1}-t_{2}\right)</math>. Let <math class="inline">\mathbf{Y}\left(t\right)</math> be a new random process defined by <math class="inline">\mathbf{Y}\left(t\right)=\frac{1}{T}\int_{t-T}^{t}\mathbf{X}\left(s\right)ds</math>, where <math class="inline">T>0</math> . |
'''(a)''' | '''(a)''' | ||
− | What is the mean of <math>\mathbf{Y}\left(t\right)</math> ? | + | What is the mean of <math class="inline">\mathbf{Y}\left(t\right)</math> ? |
− | <math>E\left[\mathbf{Y}\left(t\right)\right]=E\left[\frac{1}{T}\int_{t-T}^{t}\mathbf{X}\left(s\right)ds\right]=\frac{1}{T}\int_{t-T}^{t}E\left[\mathbf{X}\left(s\right)\right]ds=\frac{1}{T}\int_{t-T}^{t}0ds=0.</math> | + | <math class="inline">E\left[\mathbf{Y}\left(t\right)\right]=E\left[\frac{1}{T}\int_{t-T}^{t}\mathbf{X}\left(s\right)ds\right]=\frac{1}{T}\int_{t-T}^{t}E\left[\mathbf{X}\left(s\right)\right]ds=\frac{1}{T}\int_{t-T}^{t}0ds=0.</math> |
'''(b)''' | '''(b)''' | ||
− | What is the autocorrelation function of <math>\mathbf{Y}\left(t\right)</math> ? | + | What is the autocorrelation function of <math class="inline">\mathbf{Y}\left(t\right)</math> ? |
− | <math>R_{\mathbf{YY}}\left(t_{1},t_{2}\right)=E\left[\mathbf{Y}\left(t_{1}\right)\mathbf{Y}^{*}\left(t_{2}\right)\right]=E\left[\right]</math> | + | <math class="inline">R_{\mathbf{YY}}\left(t_{1},t_{2}\right)=E\left[\mathbf{Y}\left(t_{1}\right)\mathbf{Y}^{*}\left(t_{2}\right)\right]=E\left[\right]</math> |
'''(c)''' | '''(c)''' | ||
− | Write an expression for the second-order pdf <math>f_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}\left(y_{1},y_{2}\right)</math> of <math>\mathbf{Y}\left(t\right)</math> . | + | Write an expression for the second-order pdf <math class="inline">f_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}\left(y_{1},y_{2}\right)</math> of <math class="inline">\mathbf{Y}\left(t\right)</math> . |
(d) | (d) | ||
− | Under what conditions on <math>t_{1}</math> and <math>t_{2}</math> will <math>\mathbf{Y}\left(t_{1}\right)</math> and <math>\mathbf{Y}\left(t_{2}\right)</math> be statistically independent? | + | Under what conditions on <math class="inline">t_{1}</math> and <math class="inline">t_{2}</math> will <math class="inline">\mathbf{Y}\left(t_{1}\right)</math> and <math class="inline">\mathbf{Y}\left(t_{2}\right)</math> be statistically independent? |
---- | ---- | ||
[[ECE600|Back to ECE600]] | [[ECE600|Back to ECE600]] | ||
− | [[ECE 600 QE|Back to ECE 600 QE]] | + | [[ECE 600 QE|Back to my ECE 600 QE page]] |
+ | |||
+ | [[ECE_PhD_Qualifying_Exams|Back to the general ECE PHD QE page]] (for problem discussion) |
Latest revision as of 07:31, 27 June 2012
7.8 QE 2004 January
1. (30 pts.)
This question consists of two separate short questions relating to the structure of probability space:
(a)
Assume that $ \mathcal{S} $ is the sample space of a random experiment and that $ \mathcal{F}_{1} $ and $ \mathcal{F}_{2} $ are $ \sigma $ -fields (valid event spaces) on $ \mathcal{S} $ . Prove that $ \mathcal{F}_{1}\cap\mathcal{F}_{2} $ is also a $ \sigma $ -field on $ S $ .
(b)
Consider a sample space $ \mathcal{S} $ and corresponding event space $ \mathcal{F} $ . Suppose that $ P_{1} $ and $ P_{2} $ are both balid probability measures defined on $ \mathcal{F} $ . Prove that $ P $ defined by $ P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right),\qquad\forall A\in\mathcal{F} $ is also a valid probability measure on $ \mathcal{F} $ if $ \alpha_{1},\;\alpha_{2}\geq0 $ and $ \alpha_{1}+\alpha_{2}=1 $ .
Answer
• Because $ P_{1} $ and $ P_{2} $ are valid probability measures, we know that they satisfy the axioms of probability:
1. $ P_{1}\left(A\right)\geq0 $ and $ P_{2}\left(A\right)\geq0 $ , $ \forall A\in\mathcal{F}\left(\mathcal{S}\right) $ .
2. $ P_{1}\left(\mathcal{S}\right)=1 $ and $ P_{2}\left(\mathcal{S}\right)=1 $ .
3. If $ A_{1} $ and $ A_{2}\in\mathcal{F}\left(\mathcal{S}\right) $ are disjoint events, then $ P_{1}\left(A_{1}\cup A_{2}\right)=P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right) $ and $ P_{2}\left(A_{1}\cup A_{2}\right)=P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right) $ .
4. If $ A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right) $ is countable collection of disjoint events, then $ P_{1}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right) $ and $ P_{2}\left(\cup_{i=1}^{\infty}A_{i}\right)=\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right) $ .
• Now, we check each condition to become a valid probability measure:
1. $ P\left(A\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)\geq0 , \forall A\in\mathcal{F}\left(\mathcal{S}\right) $ .
– $ \because\alpha_{1}\geq0,\;\alpha_{2}\geq0,\; P_{1}\left(A\right)\geq0,\text{ and }P_{2}\left(A\right)\geq0 $ .
2. $ P\left(S\right)=\alpha_{1}P_{1}\left(A\right)+\alpha_{2}P_{2}\left(A\right)=\alpha_{1}+\alpha_{2}=1 $ .
3. If $ A_{1} $ and $ A_{2}\in\mathcal{F}\left(\mathcal{S}\right) $ are disjoint events, then $ P\left(A_{1}\cup A_{2}\right)=\alpha_{1}P_{1}\left(A_{1}\cup A_{2}\right)+\alpha_{2}P_{2}\left(A_{1}\cup A_{2}\right)=\alpha_{1}\left\{ P_{1}\left(A_{1}\right)+P_{1}\left(A_{2}\right)\right\} +\alpha_{2}\left\{ P_{2}\left(A_{1}\right)+P_{2}\left(A_{2}\right)\right\} $$ =\alpha_{1}P_{1}\left(A_{1}\right)+\alpha_{2}P_{2}\left(A_{1}\right)+\alpha_{1}P_{1}\left(A_{2}\right)+\alpha_{2}P_{2}\left(A_{2}\right)=P\left(A_{1}\right)+P\left(A_{2}\right). $
4. If $ A_{1},A_{2},\cdots,A_{n},\cdots\in\mathcal{F}\left(\mathcal{S}\right) $ is countable collection of disjoint events, then $ P\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}P_{1}\left(\cup_{i=0}^{\infty}A_{i}\right)+\alpha_{2}P_{2}\left(\cup_{i=0}^{\infty}A_{i}\right)=\alpha_{1}\sum_{i=1}^{\infty}P_{1}\left(A_{i}\right)+\alpha_{2}\sum_{i=1}^{\infty}P_{2}\left(A_{i}\right) $$ =\sum_{i=1}^{\infty}\left\{ \alpha_{1}P_{1}\left(A_{i}\right)+\alpha_{2}P_{2}\left(A_{i}\right)\right\} =\sum_{i=1}^{\infty}P\left(A_{i}\right). $
2. (10 pts.)
Identical twins come from the same egg and and hence are of the same sex. Fraternal twins have a probability $ 1/2 $ of being of the same sex. Among twins, the probability of a fraternal set is p and of an identical set is $ q=1-p $ . Given that a set of twins selected at random are of the same sex, what is the probability they are fraternal? (Simplify your answer as much as possible.) Sketch a plot of the conditional probability that the twins are fraternal given that they are of the same sex as a function of $ q $ (the probability that a set of twins are identical.)
Note
This problem is identical to the problem in the MBR 2004 Spring Final.
3. (30 pts.)
Let $ \mathbf{X}\left(t\right) $ be a real continuous-time Gaussian random process. Show that its probabilistic behavior is completely characterized by its mean $ \mu_{\mathbf{X}}\left(t\right)=E\left[\mathbf{X}\left(t\right)\right] $ and its autocorrelation function $ R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=E\left[\mathbf{X}\left(t_{1}\right)\mathbf{X}\left(t_{2}\right)\right]. $
4. (30 pts.)
Assume that $ \mathbf{X}\left(t\right) $ is a zero-mean, continuous-time, Gaussian white noise process with autocorrelation function $ R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\delta\left(t_{1}-t_{2}\right) $. Let $ \mathbf{Y}\left(t\right) $ be a new random process defined by $ \mathbf{Y}\left(t\right)=\frac{1}{T}\int_{t-T}^{t}\mathbf{X}\left(s\right)ds $, where $ T>0 $ .
(a)
What is the mean of $ \mathbf{Y}\left(t\right) $ ?
$ E\left[\mathbf{Y}\left(t\right)\right]=E\left[\frac{1}{T}\int_{t-T}^{t}\mathbf{X}\left(s\right)ds\right]=\frac{1}{T}\int_{t-T}^{t}E\left[\mathbf{X}\left(s\right)\right]ds=\frac{1}{T}\int_{t-T}^{t}0ds=0. $
(b)
What is the autocorrelation function of $ \mathbf{Y}\left(t\right) $ ?
$ R_{\mathbf{YY}}\left(t_{1},t_{2}\right)=E\left[\mathbf{Y}\left(t_{1}\right)\mathbf{Y}^{*}\left(t_{2}\right)\right]=E\left[\right] $
(c)
Write an expression for the second-order pdf $ 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) $ .
(d)
Under what conditions on $ t_{1} $ and $ t_{2} $ will $ \mathbf{Y}\left(t_{1}\right) $ and $ \mathbf{Y}\left(t_{2}\right) $ be statistically independent?
Back to the general ECE PHD QE page (for problem discussion)