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Box –Muller Muller (1958) Gaussian distribution: distribution: y y2 1 − e dy σ z e σ dy ∫ 0 πσ P( y) 1 − y e σ = πσ = ∫ cannot be solved in closed <strong>for</strong>m. trick: r = y + y y tan tanϕϕ = y 2 2 2 1 2 1 2 1 2 dyy dyy = rdrdϕϕ y 2 y 2 y y 1 1 y y 2 2 1 − σ 1 − σ e e ∫ ∫ z ⋅ z = dy ⋅ dy 1 2 1 2 0 πσ 0 πσ y y y2 2 2 1 + y2 ϕ r r − − σ σ 1 2 1 1 ∫∫ ∫∫ e e = dy dy = rdrdϕ πσ πσ 1 2 0 0 0 0 2 ϕ σ 1 ⎛ y ⎞ = = ⎜ ⎟ πσ 2 2 2ππ ⎝ y2 ⎠ Box –Muller Muller trick 1 ⎛ y ⎞ 1 z1⋅ z2 = ⎜ ⎟⋅ 2π ⎝ y2 ⎠ ( 1 ) y + y =−σln −z 2 2 1 2 2 y sin 2π z = tan 2π z = y z 1 1 2 1 cos 2 2π 1 1 ( 1−e σ ) arctan ( 1−e σ ) 2 2 2 1 + 2 r y y − − 2 2 1+ 2 y y − e σ arctan ( 1− ) ⇒ ( ) y = −σσ ln 1 1− z sin 2 2ππ z 1 2 1 = − ( − π ( ) y σ ln 1 z cos 2πz 2 2 1 From two homogeneously g y distributed random numbers z1 and z 2 one gets two Gaussian distributed random numbers y1 and y2 . 39 40
Rejection method Generate random numbers y ∈ [0, [, [0,B] ] sampled p according to a distribution P(y) ) with P(y) < AA. . Sample two homogeneously distributed random numbers z 1 , , z 2 ∈ [0,1) [0,1). . If the point (B (Bz Bz1, , A Az Az2) )li ) lies above b the th curve P(y), ), i.e. P(Bz Bz1) < Az Az2 then reject the attempt, otherwise y = Bz Bz1 is retained as a random number which which is is distributed according to P(y). ). P 41
Page 1 and 2: Autumn term 2008 I Introduction t d
Page 3 and 4: Plan of this course • 19.09. Intr
Page 5 and 6: Sedimentation Sedimentation compari
Page 7 and 8: Book Series • „Monte Carlo Meth
Page 9 and 10: Computer tools • Object oriented
Page 11 and 12: Distribution of random numbers +∞
Page 13 and 14: Mersenne prime p numbers n M = 2
Page 15 and 16: n-cube cube-test test One One can c
Page 17 and 18: Making 64 64-bit bit integers ..…
Page 19: Transformation method We want rando

y - Institute for Building Materials - ETH Zürich

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