RW6

Worksheet 06 - Expected Value Directions : Please upload a PDF to Gradescope that includes both your written responses and corresponding R code inputs/outputs (if requested) for each problem. Special Directions Be sure to demonstrate the correct use of improper integral notation (and mathematical notation in general) in your work. Probem 1. The Pareto random variable with parameters α > 0 and β > 0 has the probability density function f X ( x ) = βα β x β +1 , where α < x < ∞ . Problem 1 Part a) Verify that f X ( x ) is a density function. Hint : Recall that the area under the density function over its entire domain must be one. If you are rusty on improper integral notation, see an example in part (c) first. NOTE : Upload an image of your work for part (a) by replacing "upload_image.jpg" with your appropriately titled .jpg file in the R chunk below. Problem 1 Part b) Determine P ( X > 2 α ) . Hint : Recall that P ( X > 2 α ) would be the area under the density function over the portion of its domain that is greater than 2 α . If you are successful, you will see that P ( X > 2 α ) is simply a power of 2. NOTE : Upload an image of your work for part (b) by replacing "upload_image.jpg" with your appropriately titled .jpg file in the R chunk below. 1

Problem 1 Part c) Determine the variance of X , given E [ X ] = αβ β − 1 . What restriction do we require on β for convergence in the computation of the variance? Hint : The calculation of E [ X ] , the mean of X , is given below. You do not need to repeat it in your work. Notice that we require β > 1 for convergence. Hint : There is a different restriction on β for the variance. Additionally, you will need to determine E [ X 2 ] before you can determine Var ( X ) . You will not be required to simplify your final answer for Var ( X ) , but if you do then you will see that Var ( X ) = α 2 β ( β − 1) 2 ( β − 2) Calculation of E [ X ] : E [ X ] = [[[[[[[ ∞ α x · βα β x β +1 dx = βα β lim t →∞ [[[[[[[ t α x − β dx = βα β lim t →∞ x 1 − β 1 − β t α = α β β 1 − β lim t →∞ t 1 − β − α 1 − β so we need 1 − β < 0 ⇒ β > 1 for convergence. = α β β 1 − β 0 − α 1 − β = − α β β 1 − β · α 1 − β = − αβ 1 − β = αβ β − 1 2

Problem 1 Part e) Using R and the probability transform in part (d), simulate 1000 Pareto random variables with α = 1 and β = 4 . Find the sample mean and variance of this simulation. Next, compute the population mean and variance, based on earlier work in part (c), where α = 1 and β = 4 . Finally, comment on the comparison of the population mean and variance with the sample mean and variance. Hint : Recall, when describing how close simulated values are to the true population value, be sure to comment using specific values and consider how many standard deviations the simulated mean is from the population mean. set.seed( 2022 ) # for repeatability # NOTE : simulate 1000 uniformly distributed random variables on [0, 1] u = runif( 1000 , min = 0 , max = 1 ) # NOTE : use the probability transform to find the quantiles NOTE : these # quantiles are the simulated Pareto random variables pareto_sim = ( 1 /( 1 - u))ˆ( 1 / 6 ) # NOTE : compute the mean and variance of the simulated values NOTE : we ask R to # print the output of these lines by wrapping them in parenthesis (mean(pareto_sim)) ## [1] 1.200218 (var(pareto_sim)) ## [1] 0.06808687 Space for response: After plugging in alpha = 1 and beta = 4 into the formula we derived in part d. From the sample mean we get a value of 1.2 which is close to 1.33. We can conclude from this that the variance 0f 0.068 is the avg distance from our obtained mean of 1.2. 4