R
rcaus
Member
Dear Hubbers
I am clear with the answer . However, in the exam I would have instead tried to find the f(x,y) = f(x) * (fy/x=x). Basically the joint distribution of the two normal distribution. And then use the random number as subject of formula and basically then get confused.
Pls help me to clarify the logic behind the simple answer which the examiner gave. Also what does the relevance of the uniform distribution
As a final point > see Sep 07/3 - very simple question where it gave u a CDF and three random numbers . The question ask you to simulate the observations of X. Can you help me to understand why we cannot use the same approach above.
April 06
Qu 6
One variable of interest, T, in the description of a physical process can be modelled as
T = XY where X and Y are random variables such that X ~ N(200, 100) and Y depends
on X in such a way that Y|X = x ~ N(x, 1).
Simulate two observations of T, using the following pairs of random numbers
(observations of a uniform (0, 1) random variable), explaining your method and
calculations clearly:
Random numbers
0.5714 , 0.8238
0.3192 , 0.6844
Ans6
Solving P(Z< z) = 0.5714 z= 0.180 x= 200 + 10(0.180) = 201.80
Solving P(Z< z) = 0.8238 z= 0.930 y= 201.80 + 0.930 = 202.73
t= 201.80 202.73 = 40911
Solving P(Z< z) = 0.3192 z= 0.470 x= 200 + 10( 0.470) = 195.3
Solving P(Z< z) = 0.6844 z= 0.480 y= 195.3 + 0.480 = 195.78
t= 195.3 195.78 = 38236
I am clear with the answer . However, in the exam I would have instead tried to find the f(x,y) = f(x) * (fy/x=x). Basically the joint distribution of the two normal distribution. And then use the random number as subject of formula and basically then get confused.
Pls help me to clarify the logic behind the simple answer which the examiner gave. Also what does the relevance of the uniform distribution
As a final point > see Sep 07/3 - very simple question where it gave u a CDF and three random numbers . The question ask you to simulate the observations of X. Can you help me to understand why we cannot use the same approach above.
April 06
Qu 6
One variable of interest, T, in the description of a physical process can be modelled as
T = XY where X and Y are random variables such that X ~ N(200, 100) and Y depends
on X in such a way that Y|X = x ~ N(x, 1).
Simulate two observations of T, using the following pairs of random numbers
(observations of a uniform (0, 1) random variable), explaining your method and
calculations clearly:
Random numbers
0.5714 , 0.8238
0.3192 , 0.6844
Ans6
Solving P(Z< z) = 0.5714 z= 0.180 x= 200 + 10(0.180) = 201.80
Solving P(Z< z) = 0.8238 z= 0.930 y= 201.80 + 0.930 = 202.73
t= 201.80 202.73 = 40911
Solving P(Z< z) = 0.3192 z= 0.470 x= 200 + 10( 0.470) = 195.3
Solving P(Z< z) = 0.6844 z= 0.480 y= 195.3 + 0.480 = 195.78
t= 195.3 195.78 = 38236