Question4 Suppose that Y is a continuous random variable with the following cumu

Question

Question4 Suppose that Y is a continuous random variable with the following cumulative distribution function: 32" Let Y1, Y2.,Ys be independent and identically distributed random variables whose distribution is the same as Y. a) Develop a random-variate generator for the random variable Y. Generate a sample of size two from the distribution of Y by using the pseudo random numbers R1=0.30 and R2=0.50. b) Develop two distinct methods to generate random variates for the random variable Ymax max{Yi,Y2,...,Ys^. Discuss the relative efficiency of these two generation methods? How many inverse-transformations do your methods use to generate one sample for each random variable Ymax?

Explanation / Answer

FY(y)2   = y5/32   , 0y2

a) 0y2

y lies betweeen 0 and 2

a)random variable generator = using the excel function =RANDBETWEEN(smallest number,greatest number)

R1 =0.30    , F(R1) = 0.305/32 = 0.00243/32 = 7.59375E-05 ;E is Exponential

R2= 0.60        , F(R2) = 0.605/32 = 0.07776/32 = 0.00243

sample of size two = 7.59375E-05 , 0.00243

b) lets assume our data set between 0 and 2 as (with one decimal point)

0.1    1.1

0.2    1.2

0.3     1.3

0.4      1.4

0.5     1.5

0.6       1.6

0.7       1.7

0.8        1.8

0.9        1.9

1.0        2.0

method 1 = make a dataset of all the even number or odd number from above data set and take mean

method 2= take every 4th number (choose any random number) in the above dataset and find median

these two methods are equally efficient as the result will be a random number and there are no biases.

Inverse transformation method can be used to find the value of ymax from a dataset.

for ex 1) inverse of every data point is taken into account =1/0.1 , 1/0.2, 1/0/3.........

then make a data set and calculate y max

         2) inverse of square of every data point is taken into account =1/(0.1)2 ,1/(0.2)2, 1/(0.3)2.........

then make a data set and calculate y max

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