IAI Actuarial Core Principles · Actuarial Statistics · Basic univariate distributions and generating samples
A random variable X has density f(x) = 3x^2 for 0 < x < 1. Using the inverse transform method with u = 0.512, what is the simulated value of X?
The simulated value is 0.800. The distribution function is F(x) = x^3 on (0,1), so the inverse is x = u^(1/3). Since 0.8 cubed equals 0.512, the generated value is 0.8.
- A0.800Correct
- B0.717
- C0.512
- D0.171
- 0.640
Explanation
F(x)=x^3, so x = u^(1/3). With u=0.512, 0.8^3 = 0.512, hence x=0.8. Using the square root gives 0.717, a wrong power. Using u itself gives 0.512.
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