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Box Muller Method

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(1)

Box Muller Method

Takashi Shinzato([email protected]) January 27, 2007

1 Motivation

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

-4 -2 0 2 4

p(x)

x Box Muller method

Box Muller method p(x)

Figure 1: Box Muller method; random variable x is asymptotically distributed onN(0,1).

In this informal note, we discuss normal gaussian ran- dom variable x N(0,1)1 using stochastic variables on uniform distribution in [0,1]. At first step, we indicate then the following relation as

Z

−∞

dxex

2

2 =

2π. (2)

In order to prove eq.(2), we present via µZ

−∞

dxex

2 2

2

= Z

−∞

dxdyex2 +y

2

2 , (3)

and replacexandy torandθ like2

x = rcosθ, (4)

y = rsinθ. (5)

Using eq.(4) and eq.(5), they are also held as

x2+y2 = r2 (6)

tan1y

x = θ (7)

¯¯¯¯∂(x, y)

∂(r, θ)

¯¯¯¯ = r (8)

1Normal distribution is distributed on

p(x) : = ex

2

2

=N(0,1) (1)

with the mean 0 and the variance 1.

2r >0 and 0θ

Eq.(8) is Jaccobian. Hence, we show so Z

−∞

dxdyex2 +y

2

2 =

Z 0

Z

0

drrer

2 2

= 2π Z

0

dueu µ

u= r2 2

= 2π£

−eu¤

0 = 2π. (9)

Namely, eq.(2) is true. Therefore, we give then

1 = 1

2π Z

−∞

dxdyex2 +y

2

2 , (10)

and in the case thatxis distributed on probability mea- sure (or mass probability) p(x) = e

x2

2

, stochastic vari- ablexis called gaussian random variable.

2 Box Muller Method

In this section, we present and introduce Box Muller method. Providing with the law of large numbers or cen- ter limit theorem, gaussian random variable has played an important role and one makes usefull for hypothesis and test in Data science and so on. We define a new function as

U(R) : = 1 2π

Z

x2+y2R2

dxdyex2 +y

2

2 . (11)

This integral interval is in x2+y2 R2. We calcurate then briefly

U(R) = 1 2π

Z 0

Z R

0

drrer

2 2

= Z R22

0

dueu= 1−eR

2

2 (12)

U(R) is nondecreasing function satisfied as lim

R0U(R) = 0 (13)

Rlim→∞U(R) = 1 (14) Moreover, by eq.(11), the integral interval is replaced using 0≤r≤Rand eq.(4) and eq.(5). We represent that using p∈ [0,1] and eq.(12), variable R of U(R) =pis decided as

U(R) =p = R = p

2 log(1−p). (15) In settings:= 1−p∈[0,1] andt∈[0,1], it is given as 1

(2)

x =

½ pp2 log(s) cos (2πt)

2 log(s) sin (2πt). (16) That is to say, we give then the scheme that we gener- ate gaussian random variable distributed on N(0,1) us- ing uniform distribution on [0,1]; its scheme is called Box Muller method. More being generalized, stochastic vari- ablez∼N(µ, σ2), with the meanµand the variance σ2, is given as

z = µ+σp

2 log(s) cos (2πt), (17) or

z = µ+σp

2 log(s) sin (2πt). (18)

3 Conclusion

We introduce and discuss, in this informal note, the scheme of normal distribution N(0,1) using unique dis- tribution on [0,1], Box Muller method.

0 0.1 0.2 0.3 0.4 0.5

-4 -2 0 2 4

p(x)

x Box Muller method

Frequency on Box Muller method(106) N(0,1)

Figure 2: Frequency on Box Muller method is asymptoti- cally distributed onN(0,1).

2

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