24 Inverse Transforms for Continous Distributions PDF

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Mersenne Twister...

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Inverse-transform Technique !

The concept generalized to continuous: !

For cdf function: r = F(x)

!

Generate r from uniform(0,1)

!

Find x by solving above equation for x !

 

i.e. apply the cdf inverse to the uniform random numbers r = F(x)



r1

x = F-1(r) x1

Exponential Distribution   [Inverse-transform] !

Exponential Distribution: !

Exponential pdf:

f(x) = λe-λx !

for x ≥ 0

Exponential cdf:

F(x) = 1 – e-λx for x ≥ 0 !

Derive CDF-1

r = F(x) r = 1 – e-λx

solve for x

(isolate x, take log of both sides, then solve) !

To generate X1, X2, X3 …

Xi = F-1(Ri) = -(1/λ) ln(1 – Ri)

Figure: Inverse-transform technique for exp(λ = 1)

Other Distributions   [Inverse-transform] !

Examples of other distributions for which inverse cdf works are: ! ! !

Uniform distribution Weibull distribution Triangular distribution

Empirical Continuous Dist’n  ! !

[Inverse-transform] When theoretical distribution is not applicable To collect empirical data: ! !

!

Resample the observed data Interpolate between observed data points to fill in the gaps

For a small sample set (size n): !

Arrange the data from smallest to largest

!

The slope for each line segment is

!

Assign the probability 1/n to each interval

x(i ) − x(i−1) x(i ) − x(i−1) = n(x(i ) − x(i−1) ) = ai = i / n − (i − 1) / n 1/ n i⎞ ⎛ X = Fˆ −1 (R) = x(i ) + ai+1 ⎜ R − ⎟ ⎝ n⎠

(

)

= x(i ) + x(i+1) − x(i ) ( nR − i )

i =floor(R/n)

(i − 1) / n ≤ R ≤ i / n x(i-1) ≤ x ≤ x(i )

Empirical Continuous Dist’n  !

[Inverse-transform] Example: Let the data collected for 100 broken-widget repair times be:

  R1 = 0.83: Consider  c3 = 0.66 < R1 < c4 = 1.00  

X1 = x(4-1) + a4(R1 – c(4-1)) = 1.5 + 1.47(0.83-0.66) = 1.7

Empirical Continuous Dist’n  !

[Inverse-transform] Example: Let the data collected for 100 broken-widget repair times be:

Textbook says accuracy increases with increasing samples.   R1 = 0.83: Consider  c3 = 0.66 < R1 < c4 = 1.00  

X1 = x(4-1) + a4(R1 – c(4-1)) = 1.5 + 1.47(0.83-0.66) = 1.7

Not true!!!

Empirical Continuous Dist’n  !

[Inverse-transform] Example: Let the data collected for 100 broken-widget repair times be:

Textbook says accuracy increases with increasing samples.   R1 = 0.83: Consider 

Smoothing needed!

c3 = 0.66 < R1 < c4 = 1.00  

X1 = x(4-1) + a4(R1 – c(4-1)) = 1.5 + 1.47(0.83-0.66) = 1.7...