help on CP2 April 2020 P1 - Q1 (viii)

[Alexa]

Hi all. I have a question on the method used in the exam paper and question in the title. In the audit word doc, the steps used in this question is:

Min[100,100+ln(U_(i,n) )/λ_n + 〖points dropped s.d.〗_n/b]

This is done to convert the random uniform distribution numbers to exponential. What I dont understand is why we need to divide the ln figure by lambda and then add the standard deviation (in pink in the equation above). Am I missing something?

Thank you!

 

[Lucy England]

The formula you've quoted is being used to simulate scores for each competitor.

The ln(U_(i,n) )/λ_n bit converts the uniform random variable into an exponential random variable - it's called rv_(i,n) in the question paper.

The exponential random variable is then used to produce a simulated score. The points dropped bit tailors the score to competitor n, based on their historical scores.

 

[Alexa]

The formula you've quoted is being used to simulate scores for each competitor.

The ln(U_(i,n) )/λ_n bit converts the uniform random variable into an exponential random variable - it's called rv_(i,n) in the question paper.

The exponential random variable is then used to produce a simulated score. The points dropped bit tailors the score to competitor n, based on their historical scores.

Could you help me make sense of the part of the formula in pink? I see the formula is provided in the question paper but I'm not able to understand the significance/ meaning of it.

I get that -ln (U i,n) will give me a negative number since U i,n is between 0 and 1 inclusive. So i can deduct this from 100 to get a simulated score. Does dividing it by lamda and adding s.d. give it a "weight"?

I hope i'm making sense. Thanks again for your help!

 

[Lucy England]

Lambda is the parameter for the exponential distribution that the rv_(i,n) values follow and 1/lambda is the mean of that exponential distribution.

Lambda varies based on the past scores of each competitor, so rv_(i,n) is competitor-specific. Lambda will be highest for competitors with low points dropped and a low variation of points dropped, because the mean and SD of points dropped will be low. A high value of lambda will reduce rv_(i,n) and therefore we'll be reducing the maximum of 100 points by less, getting a higher expected score.

The points dropped sd_n adjustment seems to adjust the expected score further, based on how much a competitor's past scores vary. The adjustment is bigger over longer distances.

I should note that you wouldn't need to spend a long time in the exam trying to understand exactly what's going on here. As long as you follow the instructions and apply the formulae accurately in your model, you'd be able to gain the marks you need.