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
We're trying to simulate T = XY ie X multiplied by Y - hence the need to simulate X and Y separately and then multiply them together. T is not the joint distribution of X and Y - that's why your approach will not work. The relevance of the uniform distribution is that the CDF F(x) takes values from 0 to 1 - so by choosing a value from 0 to 1 equally (ie the U(0,1)) we can then invert this to get a value from the distribution of X. However for the normal distribution we do not have a function for F(x) to invert - so we have to use the cumulative probabilities from page 162 of the Tables to find the values that correspond to the given CDF value.
