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HKN ECE 313 Exam 2 Review Session

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1 HKN ECE 313 Exam 2 Review Session
Corey Snyder Ahnaf Masroor Kanad Sarkar

2 Exam 2 topics Continuous-type Random Variables (mean and variance of CRVs) Uniform Distribution Exponential Distribution Poisson Process Linear Scaling of PDFs Gaussian Distribution ML Parameter Estimation for Continuous Random Variables Functions of a random variable Failure Rate Functions Binary Hypothesis Testing Joint CDFs, PMFs, and PDFs Independence of Random Variables Distributions of sums of random variables*

3 Continuous-type Random variables
Cumulative Distribution Functions (CDFs) Must be non-decreasing 𝐹 𝑋 βˆ’βˆž =0, 𝐹 𝑋 +∞ =1 Must be right continuous 𝐹 𝑋 𝑐 = βˆ’βˆž 𝑐 𝑓 π‘₯ 𝑒 𝑑𝑒 𝑃 𝑋=π‘Ž =0 𝑃 π‘Ž<𝑋≀𝑏 = 𝐹 𝑋 𝑏 βˆ’ 𝐹 𝑋 π‘Ž = π‘Ž 𝑏 𝑓 𝑋 𝑒 𝑑𝑒 𝐸 𝑋 = πœ‡ π‘₯ = βˆ’βˆž +∞ 𝑒 𝑓 π‘₯ 𝑒 𝑑𝑒

4 Uniform Distribution π‘ˆπ‘›π‘–π‘“(π‘Ž,𝑏): All values between π‘Ž and 𝑏 are equally likely pdf:𝑓 𝑒 = 1 π‘βˆ’π‘Ž π‘Žβ‰€π‘’β‰€π‘ 𝑒𝑙𝑠𝑒 mean: π‘Ž+𝑏 2 variance: (π‘βˆ’π‘Ž ) 2 12

5 Exponential Distribution
Exponential(πœ†): limit of scaled geometric random variables pdf:𝑓 𝑑 =πœ† 𝑒 βˆ’πœ†π‘‘ 𝑑β‰₯0 mean: 1 πœ† variance: 1 πœ† 2 Memoryless Property 𝑃 𝑇β‰₯𝑠+𝑑 𝑇β‰₯𝑠} = 𝑃{𝑇β‰₯𝑑} 𝑠,𝑑β‰₯0

6 Poisson process Poisson process is used to model the number of counts in a time interval. Similar to how the exponential distribution is a limit of the geometric distribution, the Poisson process is the limit of the Bernoulli process. Bonus: A Poisson process is a collection of i.i.d exponential random variables. If we have a rate πœ† and time duration 𝑑, the number of counts N t ~π‘ƒπ‘œπ‘–π‘ (πœ†π‘‘) pdf: 𝑓 𝑁 𝑑 π‘˜ = 𝑒 βˆ’πœ†π‘‘ ( πœ†π‘‘) π‘˜ π‘˜! mean: πœ†π‘‘ variance: πœ†π‘‘ Furthermore, 𝑁 𝑑 βˆ’ 𝑁 𝑠 ~ π‘ƒπ‘œπ‘–π‘  πœ† π‘‘βˆ’π‘  , 𝑑>𝑠 , Disjoint intervals, e.g. 𝑠=[0,2] and 𝑑=[2,3], are independent. This property is both important and remarkable. In fact, we shape our analysis of Poisson processes around this frequently. (More on this later!)

7 Linear scaling of pdfs 𝐼𝑓 π‘Œ=π‘Žπ‘‹+𝑏: 𝐸 π‘Œ =π‘ŽπΈ 𝑋 +𝑏 π‘‰π‘Žπ‘Ÿ π‘Œ = π‘Ž 2 π‘‰π‘Žπ‘Ÿ 𝑋
π‘‰π‘Žπ‘Ÿ π‘Œ = π‘Ž 2 π‘‰π‘Žπ‘Ÿ 𝑋 𝑓 𝑦 𝑣 = 𝑓 π‘₯ ( π‘£βˆ’π‘ π‘Ž ) 1 π‘Ž

8 Gaussian distribution
Gaussian (or Normal) Distribution ~ 𝑁(πœ‡, 𝜎 2 ) Standard Gaussian, 𝑋 : πœ‡=0, 𝜎=1 𝑝𝑑𝑓:𝑓 𝑒 = πœ‹ 𝜎 2 𝑒 βˆ’ (π‘’βˆ’πœ‡ ) 2 2 𝜎 2 π‘šπ‘’π‘Žπ‘›:πœ‡ π‘£π‘Žπ‘Ÿπ‘–π‘Žπ‘›π‘π‘’: 𝜎 2 If we standardize our Gaussian, where: 𝑋 = π‘‹βˆ’πœ‡ 𝜎 Ξ¦ 𝑐 = βˆ’βˆž 𝑐 𝑓 𝑒 𝑑𝑒 𝑄 𝑐 =1βˆ’Ξ¦ 𝑐 = 𝑐 ∞ 𝑓 𝑒 𝑑𝑒 Ξ¦ βˆ’π‘ =𝑄(𝑐)

9 Ml parameter estimation
Suppose we have a random variable with a given distribution/pdf that depends on a parameter, πœƒ. By taking trials of the random variable, we can estimate πœƒ by finding the value that maximizes the likelihood of the observed event, πœƒ ML. There are a few ways we can find πœƒ ML Take derivative of provided pdf and set it equal to zero (maximization) Observe the intervals where the likelihood increases and decreases, and find the maximum between these intervals Intuition!

10 Functions of random variables
Suppose π‘Œ = 𝑔(𝑋), and we want to be able to describe the distribution of π‘Œ Step 1: Identify the support of 𝑋. Sketch the pdf of 𝑋 and 𝑔. Identify the support of π‘Œ. Determine whether π‘Œ is a Continuous or Discrete RV Take a deep breath! You’ve done some important work here. Step 2 (for CRV): Use the definition of the CDF to find the CDF of π‘Œ: 𝐹 π‘Œ 𝑐 =𝑃 π‘Œβ‰€π‘ =𝑃{𝑔 𝑋 ≀𝑐} Step 2 (for DRV): Find the pmf of π‘Œ directly using the definition of the pmf 𝑝 π‘Œ 𝑣 =𝑃 π‘Œ=𝑣 =𝑃 𝑔 𝑋 =𝑣 Step 3 (for CRV): Differentiate the CDF of π‘Œ in order to find the pdf of π‘Œ

11 Generating a RV with a Specified distribution
We can generate any distribution by applying a function to a uniform distribution This function should be the inverse of the CDF of the desired distribution Ex: if we want an exponential distribution, 𝐹 𝑋 𝑐 =1βˆ’ 𝑒 βˆ’π‘ =𝑒;then find 𝐹 𝑋 βˆ’1 (𝑐)

12 Failure rate functions
We can assess the probability of a failure in a system through a failure rate function, β„Ž(𝑑). 𝐹 𝑑 =1βˆ’ 𝑒 βˆ’ 0 𝑑 β„Ž 𝑠 𝑑𝑠 Two popular failure rate functions: Consistent lifetime β€œBath tub”

13 Binary hypothesis testing
Similar to BHT with Discrete RVs Maximum Likelihood (ML) Rule Ξ› k = 𝑓 1 (π‘˜) 𝑓 0 (π‘˜) Ξ› π‘˜ = >1 π‘‘π‘’π‘π‘™π‘Žπ‘Ÿπ‘’ 𝐻 1 𝑖𝑠 π‘‘π‘Ÿπ‘’π‘’ <1 π‘‘π‘’π‘π‘™π‘Žπ‘Ÿπ‘’ 𝐻 0 𝑖𝑠 π‘‘π‘Ÿπ‘’π‘’ Maximum a Posteriori (MAP) Rule Prior probabilities: πœ‹ 0 =𝑃 𝐻 0 , πœ‹ 1 =𝑃 (𝐻 1 ) 𝐻 1 π‘‘π‘Ÿπ‘’π‘’ 𝑖𝑓 πœ‹ 1 𝑓 1 π‘˜ > πœ‹ 0 𝑓 0 π‘˜ , same as Ξ› k = 𝑓 1 (π‘˜) 𝑓 0 (π‘˜) >𝜏 π‘€β„Žπ‘’π‘Ÿπ‘’ 𝜏= πœ‹ 0 πœ‹ 1 Probabilities of False Alarm and Miss 𝑝 π‘“π‘Žπ‘™π‘ π‘’ π‘Žπ‘™π‘Žπ‘Ÿπ‘š =𝑃 π‘†π‘Žπ‘¦ 𝐻 𝐻 0 𝑖𝑠 π‘‘π‘Ÿπ‘’π‘’) 𝑝 π‘šπ‘–π‘ π‘  =𝑃 π‘†π‘Žπ‘¦ 𝐻 𝐻 1 𝑖𝑠 π‘‘π‘Ÿπ‘’π‘’) 𝑝 𝑒 = πœ‹ 1 𝑝 π‘šπ‘–π‘ π‘  + πœ‹ π‘œ 𝑝 π‘“π‘Žπ‘™π‘ π‘’ π‘Žπ‘™π‘Žπ‘Ÿπ‘š

14 Joint cdf, pmf, and pdf Discrete Random Variables
Continuous Random Variables Joint CDF 𝐹 𝑋,π‘Œ 𝑒 π‘œ , 𝑣 π‘œ =𝑃{𝑋≀ 𝑒 π‘œ , π‘Œβ‰€ 𝑣 π‘œ } Joint PMF 𝑝 𝑋,π‘Œ 𝑒 π‘œ , 𝑣 π‘œ =𝑃 𝑋= 𝑒 π‘œ , π‘Œ= 𝑣 π‘œ Marginal PMFs 𝑝 𝑋 𝑒 = 𝑗 𝑝 𝑋,π‘Œ (𝑒, 𝑣 𝑗 ) 𝑝 π‘Œ 𝑣 = 𝑖 𝑝 𝑋,π‘Œ ( 𝑒 𝑖 ,𝑣) Conditional PMFs 𝑝 π‘Œ 𝑋 𝑣 𝑒 π‘œ =𝑃 π‘Œ=𝑣 𝑋= 𝑒 π‘œ = 𝑝 𝑋,π‘Œ 𝑒 π‘œ ,𝑣 𝑝 𝑋 𝑒 π‘œ Joint CDF 𝐹 𝑋,π‘Œ 𝑒 π‘œ , 𝑣 π‘œ =𝑃{𝑋≀ 𝑒 π‘œ , π‘Œβ‰€ 𝑣 π‘œ } Joint PDF 𝐹 𝑋,π‘Œ 𝑒 π‘œ , 𝑣 π‘œ = 𝑓 𝑋,π‘Œ 𝑒,𝑣 𝑑𝑣𝑑𝑒 Marginal PDFs 𝑓 𝑋 𝑒 = βˆ’βˆž +∞ 𝑓 𝑋,π‘Œ 𝑒,𝑣 𝑑𝑣 𝑓 π‘Œ (𝑣)= βˆ’βˆž +∞ 𝑓 𝑋,π‘Œ 𝑒,𝑣 𝑑𝑒 Conditional PDFs 𝑓 π‘Œ 𝑋 𝑣 𝑒 π‘œ = 𝑓 𝑋,π‘Œ 𝑒 π‘œ ,𝑣 𝑓 𝑋 𝑒 π‘œ

15 Independence of joint distributions
We can check independence in a joint distribution in a couple ways: 𝑓 𝑋,π‘Œ 𝑒,𝑣 = 𝑓 𝑋 𝑒 𝑓 π‘Œ 𝑣 The support of 𝑓 𝑋,π‘Œ 𝑒,𝑣 is a product set Product set must have the swap property, which is satisfied if: π‘Ž,𝑏 π‘Žπ‘›π‘‘ 𝑐,𝑑 βˆˆπ‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘ 𝑓 𝑋,π‘Œ , π‘Žπ‘›π‘‘ π‘‘β„Žπ‘’π‘› π‘Ž,𝑑 π‘Žπ‘›π‘‘ 𝑐,𝑏 π‘Žπ‘Ÿπ‘’ π‘Žπ‘™π‘ π‘œβˆˆπ‘ π‘’π‘π‘π‘œπ‘Ÿπ‘‘( 𝑓 𝑋,π‘Œ ) Checking for a product set is only sufficient to prove dependence. Saying that the support of the joint pdf is a product set is not sufficient to check independence

16 Distribution of sums of random variables
Suppose we want to find the distribution from the sum of two independent random variables where 𝑍 = 𝑋 + π‘Œ The pdf or pmf of 𝑍 is the convolution of the two pdfs/pmfs So...

17 What Is Convolution? A beautiful, extraordinary linear operator that describes natural phenomena in a fundamental and concise manner.

18 What Is Convolution? A beautiful, extraordinary linear operator that describes natural phenomena in a fundamental and concise manner. But for real, given 𝑍=𝑋+π‘Œ Discrete (pmfs): 𝑝 𝑧 𝑒 = 𝑝 π‘₯ 𝑒 βˆ— 𝑝 𝑦 (𝑒) 𝑝 𝑧 𝑒 = π‘˜=βˆ’βˆž ∞ 𝑝 π‘₯ π‘˜ 𝑝 𝑦 (π‘’βˆ’π‘˜) = π‘˜=βˆ’βˆž ∞ 𝑝 π‘₯ π‘’βˆ’π‘˜ 𝑝 𝑦 (π‘˜) Continuous (pdfs): 𝑓 𝑧 𝑒 = 𝑓 π‘₯ 𝑒 βˆ— 𝑓 𝑦 (𝑒) 𝑓 𝑧 𝑒 = βˆ’βˆž ∞ 𝑓 π‘₯ 𝜏 𝑓 𝑦 π‘’βˆ’πœ π‘‘πœ = βˆ’βˆž ∞ 𝑓 π‘₯ π‘’βˆ’πœ 𝑓 𝑦 𝜏 π‘‘πœ

19 Fa15 Problem 2 Let 𝑁 𝑑 be a Poisson process with rate πœ†>0
a. Obtain 𝑃{ 𝑁 3 =5} b. Obtain 𝑃{ 𝑁 7 βˆ’ 𝑁 4 =5} and 𝐸[ 𝑁 7 βˆ’ 𝑁 4 ] c. Obtain 𝑃{ 𝑁 7 βˆ’ 𝑁 4 =5| 𝑁 6 βˆ’ 𝑁 4 =2} d. Obtain 𝑃{ 𝑁 6 βˆ’ 𝑁 4 =2| 𝑁 7 βˆ’ 𝑁 4 =5}

20 FA14 Problem 3

21 Fa15 problem 4 Let the joint pdf for the pair (𝑋,π‘Œ) be:
𝑓 𝑋,π‘Œ π‘₯,𝑦 = 𝑐π‘₯𝑦, 0≀π‘₯≀1, 0≀𝑦≀1, π‘₯+𝑦≀1 0, π‘œπ‘‘β„Žπ‘’π‘Ÿπ‘€π‘–π‘ π‘’ a. Compute the marginal 𝑓 𝑋 (π‘₯). You can leave it in terms of 𝑐. b. Obtain the value of the constant 𝑐 for 𝑓 𝑋,π‘Œ to be a valid joint pdf. c. Obtain 𝑃{𝑋+π‘Œ< 1 2 } d. Are 𝑋 and π‘Œ independent? Explain why or why not.

22 Su16 Problem 3 Let 𝑋 be Gaussian with mean βˆ’1 and variance 16
a. Express 𝑃{ 𝑋 3 β‰€βˆ’8} in terms of the Ξ¦ function b. Let π‘Œ= 1 2 𝑋 Sketch the pdf of π‘Œ, 𝑓 𝑦 𝑣 , clearly marking important points c. Express 𝑃{π‘Œβ‰₯ 1 2 } in terms of the Ξ¦ function

23 SP15 Problem 4

24 Fa15 problem 3 Let 𝑋 be a continuous-type random variable taking values in [0,1]. Under hypothesis 𝐻 0 the pdf of 𝑋 is 𝑓 0 . Under hypothesis 𝐻 1 the pdf of 𝑋 is 𝑓 1 . Both pdfs are plotted below. The priors are known to be πœ‹ 0 =0.6 and πœ‹ 1 =0.4. a. Find the value of 𝑐 b. Specify the maximum a posteriori (MAP) decision rule for testing 𝐻 0 vs. 𝐻 1 . c. Find the error probabilities 𝑝 π‘“π‘Žπ‘™π‘ π‘’ π‘Žπ‘™π‘Žπ‘Ÿπ‘š , 𝑝 π‘šπ‘–π‘ π‘  π‘‘π‘’π‘‘π‘’π‘π‘‘π‘–π‘œπ‘› , and the average probability of error 𝑝 𝑒 for the MAP rule.

25 FA12 problem 6


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