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HKN ECE 313 Exam 2 Review Session
Corey Snyder Ahnaf Masroor Kanad Sarkar
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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*
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Continuous-type Random variables
Cumulative Distribution Functions (CDFs) Must be non-decreasing πΉ π ββ =0, πΉ π +β =1 Must be right continuous πΉ π π = ββ π π π₯ π’ ππ’ π π=π =0 π π<πβ€π = πΉ π π β πΉ π π = π π π π π’ ππ’ πΈ π = π π₯ = ββ +β π’ π π₯ π’ ππ’
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Uniform Distribution ππππ(π,π): All values between π and π are equally likely pdf:π π’ = 1 πβπ πβ€π’β€π πππ π mean: π+π 2 variance: (πβπ ) 2 12
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Exponential Distribution
Exponential(π): limit of scaled geometric random variables pdf:π π‘ =π π βππ‘ π‘β₯0 mean: 1 π variance: 1 π 2 Memoryless Property π πβ₯π +π‘ πβ₯π } = π{πβ₯π‘} π ,π‘β₯0
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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 and time duration π‘, the number of counts N t ~ππππ (ππ‘) pdf: π π π‘ π = π βππ‘ ( ππ‘) π π! mean: ππ‘ variance: ππ‘ Furthermore, π π‘ β π π ~ ππππ π π‘βπ , π‘>π , Disjoint intervals, e.g. π =[0,2] and π‘=[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!)
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Linear scaling of pdfs πΌπ π=ππ+π: πΈ π =ππΈ π +π πππ π = π 2 πππ π
πππ π = π 2 πππ π π π¦ π£ = π π₯ ( π£βπ π ) 1 π
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Gaussian distribution
Gaussian (or Normal) Distribution ~ π(π, π 2 ) Standard Gaussian, π : π=0, π=1 πππ:π π’ = π π 2 π β (π’βπ ) 2 2 π 2 ππππ:π π£πππππππ: π 2 If we standardize our Gaussian, where: π = πβπ π Ξ¦ π = ββ π π π’ ππ’ π π =1βΞ¦ π = π β π π’ ππ’ Ξ¦ βπ =π(π)
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Ml parameter estimation
Suppose we have a random variable with a given distribution/pdf that depends on a parameter, π. By taking trials of the random variable, we can estimate π by finding the value that maximizes the likelihood of the observed event, π ML. There are a few ways we can find π 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!
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Functions of random variables
Suppose π = π(π), and we want to be able to describe the distribution of π Step 1: Identify the support of π. Sketch the pdf of π and π. Identify the support of π. Determine whether π is a Continuous or Discrete RV Take a deep breath! Youβve done some important work here. Step 2 (for CRV): Use the definition of the CDF to find the CDF of π: πΉ π π =π πβ€π =π{π π β€π} Step 2 (for DRV): Find the pmf of π directly using the definition of the pmf π π π£ =π π=π£ =π π π =π£ Step 3 (for CRV): Differentiate the CDF of π in order to find the pdf of π
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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, πΉ π π =1β π βπ =π’;then find πΉ π β1 (π)
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Failure rate functions
We can assess the probability of a failure in a system through a failure rate function, β(π‘). πΉ π‘ =1β π β 0 π‘ β π ππ Two popular failure rate functions: Consistent lifetime βBath tubβ
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Binary hypothesis testing
Similar to BHT with Discrete RVs Maximum Likelihood (ML) Rule Ξ k = π 1 (π) π 0 (π) Ξ π = >1 πππππππ π» 1 ππ π‘ππ’π <1 πππππππ π» 0 ππ π‘ππ’π Maximum a Posteriori (MAP) Rule Prior probabilities: π 0 =π π» 0 , π 1 =π (π» 1 ) π» 1 π‘ππ’π ππ π 1 π 1 π > π 0 π 0 π , same as Ξ k = π 1 (π) π 0 (π) >π π€βπππ π= π 0 π 1 Probabilities of False Alarm and Miss π ππππ π πππππ =π πππ¦ π» π» 0 ππ π‘ππ’π) π πππ π =π πππ¦ π» π» 1 ππ π‘ππ’π) π π = π 1 π πππ π + π π π ππππ π πππππ
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Joint cdf, pmf, and pdf Discrete Random Variables
Continuous Random Variables Joint CDF πΉ π,π π’ π , π£ π =π{πβ€ π’ π , πβ€ π£ π } Joint PMF π π,π π’ π , π£ π =π π= π’ π , π= π£ π Marginal PMFs π π π’ = π π π,π (π’, π£ π ) π π π£ = π π π,π ( π’ π ,π£) Conditional PMFs π π π π£ π’ π =π π=π£ π= π’ π = π π,π π’ π ,π£ π π π’ π Joint CDF πΉ π,π π’ π , π£ π =π{πβ€ π’ π , πβ€ π£ π } Joint PDF πΉ π,π π’ π , π£ π = π π,π π’,π£ ππ£ππ’ Marginal PDFs π π π’ = ββ +β π π,π π’,π£ ππ£ π π (π£)= ββ +β π π,π π’,π£ ππ’ Conditional PDFs π π π π£ π’ π = π π,π π’ π ,π£ π π π’ π
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Independence of joint distributions
We can check independence in a joint distribution in a couple ways: π π,π π’,π£ = π π π’ π π π£ The support of π π,π π’,π£ is a product set Product set must have the swap property, which is satisfied if: π,π πππ π,π βπ π’πππππ‘ π π,π , πππ π‘βππ π,π πππ π,π πππ πππ πβπ π’πππππ‘( π π,π ) 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
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Distribution of sums of random variables
Suppose we want to find the distribution from the sum of two independent random variables where π = π + π The pdf or pmf of π is the convolution of the two pdfs/pmfs So...
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What Is Convolution? A beautiful, extraordinary linear operator that describes natural phenomena in a fundamental and concise manner.
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What Is Convolution? A beautiful, extraordinary linear operator that describes natural phenomena in a fundamental and concise manner. But for real, given π=π+π Discrete (pmfs): π π§ π’ = π π₯ π’ β π π¦ (π’) π π§ π’ = π=ββ β π π₯ π π π¦ (π’βπ) = π=ββ β π π₯ π’βπ π π¦ (π) Continuous (pdfs): π π§ π’ = π π₯ π’ β π π¦ (π’) π π§ π’ = ββ β π π₯ π π π¦ π’βπ ππ = ββ β π π₯ π’βπ π π¦ π ππ
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Fa15 Problem 2 Let π π‘ be a Poisson process with rate π>0
a. Obtain π{ π 3 =5} b. Obtain π{ π 7 β π 4 =5} and πΈ[ π 7 β π 4 ] c. Obtain π{ π 7 β π 4 =5| π 6 β π 4 =2} d. Obtain π{ π 6 β π 4 =2| π 7 β π 4 =5}
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FA14 Problem 3
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Fa15 problem 4 Let the joint pdf for the pair (π,π) be:
π π,π π₯,π¦ = ππ₯π¦, 0β€π₯β€1, 0β€π¦β€1, π₯+π¦β€1 0, ππ‘βπππ€ππ π a. Compute the marginal π π (π₯). You can leave it in terms of π. b. Obtain the value of the constant π for π π,π to be a valid joint pdf. c. Obtain π{π+π< 1 2 } d. Are π and π independent? Explain why or why not.
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Su16 Problem 3 Let π be Gaussian with mean β1 and variance 16
a. Express π{ π 3 β€β8} in terms of the Ξ¦ function b. Let π= 1 2 π Sketch the pdf of π, π π¦ π£ , clearly marking important points c. Express π{πβ₯ 1 2 } in terms of the Ξ¦ function
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SP15 Problem 4
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Fa15 problem 3 Let π be a continuous-type random variable taking values in [0,1]. Under hypothesis π» 0 the pdf of π is π 0 . Under hypothesis π» 1 the pdf of π is π 1 . Both pdfs are plotted below. The priors are known to be π 0 =0.6 and π 1 =0.4. a. Find the value of π b. Specify the maximum a posteriori (MAP) decision rule for testing π» 0 vs. π» 1 . c. Find the error probabilities π ππππ π πππππ , π πππ π πππ‘πππ‘πππ , and the average probability of error π π for the MAP rule.
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FA12 problem 6
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