IAI Actuarial Core Principles · Actuarial Statistics · Basic univariate distributions and generating samples
The inverse transform method is used to simulate X with distribution function F(x) = 1 - (2/x)^3 for x >= 2. A Uniform(0,1) value u = 0.875 is generated and used as F(X) = u. What value of X results?
X equals 4. Setting F(x) = 0.875 gives (2/x)^3 = 0.125, so 2/x = 0.5 and x = 4. This is the inverse transform method: solve F(X) = u for X.
- A2.50
- B3.00
- C4.00Correct
- D8.00
- 16.00
Explanation
Solve 1 - (2/x)^3 = 0.875, so (2/x)^3 = 0.125 and 2/x = 0.5, giving x = 4. Check: 1 - 0.5^3 = 0.875. The value 8 arises from using 1-u incorrectly as 2/x = 0.125.
Did you get it right without looking?
One question tells you little. A timed set on Basic univariate distributions and generating samples shows your real accuracy, how long you take and where you lose marks.
More Basic univariate distributions and generating samples questions
- To simulate from a density f(x) = 6x(1 - x) on (0,1) by acceptance-rejection, a Uniform(0,1) proposal is used with the smallest constant M s…
- A simulation estimates the probability p that an annual loss exceeds Rs 50 crore. Out of 2,500 independent runs, 200 exceed Rs 50 crore. Wha…
- A simulation estimates the mean claim cost using n independent runs. The sample standard deviation of the outputs is Rs 400. What is the sta…
- X is uniform on (0,10). Given that X exceeds 4, what is the expected value of X?
- Claim counts N on a motor policy in a year are Poisson with mean 0.5. Independent claims are of two kinds, each claim being a theft claim wi…
- If U is Uniform(0,1) and F is a continuous, strictly increasing distribution function, which statement about the inverse transform method is…