(New page: OK, so what we have initially is a uniform random variable on the interval [0,1]. In Homework 6 question 4, it is proven that an exponential random variable with λ=0.5 is made out of two...) |
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D = ln(1-U)/λ | D = ln(1-U)/λ | ||
| − | In this case you just have to replace U with drand48(). | + | In this case you just have to replace U with drand48() and λ by 0.5 . |
Now that we have our exponential RV let's see how we can obtain X from D. | Now that we have our exponential RV let's see how we can obtain X from D. | ||
Latest revision as of 10:47, 21 October 2008
OK, so what we have initially is a uniform random variable on the interval [0,1].
In Homework 6 question 4, it is proven that an exponential random variable with λ=0.5 is made out of two gaussian random variables with the relationship $ D=X^2+Y^2 $
So, what we need first is an exponential RV with λ=0.5.
Question 1.b on the same homework shows an exponential RV can be obtained from a uniform~[0,1] by
D = ln(1-U)/λ
In this case you just have to replace U with drand48() and λ by 0.5 .
Now that we have our exponential RV let's see how we can obtain X from D.
Thinking in polar form we can easily argue that X = D^(1/2)cos(ω)
So we need another uniform RV ω, which ranges from [0, 2π]
Thinking about how scaling a uniform RV also scales its range, dividing drand48() by 2π gives us ω. This is true since the height has been reduced to 1/2π, the base must expand to [0, 2π] so that the total area remains to be 1.
Now that we have defined D and ω in terms of drand48() it is just a matter of plugging the definitions in the equation.
