The Inverse of the Cumulative Standard Normal Probability Function

Home , Error function

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x 1 t2 N(x)= e− 2 dt √2π Z−∞ which we call S(x). In section 2 we derive an ODE satisfied by S(x), and solve it using a power series. We introduce a family of polynomials Pn related to the calculation of higher derivatives of S(x). In section 3 we study some properties of the Pn, such as relations between coefficients, recurrences, and generating functions. We also derive a general formula for Pn using the idea of “nested derivatives”, and we compare the Pn with the Hermite polynomials Hn. In section 4 we extend the definition of the Pn to n < 0 and use them to calculate the integrals of S(x). We also compute the integrals of powers of S(x) on the interval [0, 1]. Section 5 is dedicated to asymptotics of S(x) for x 0, x 1 using the function Lambert W. With the help of those formulas we→ derive→ an approximation to S(x) valid in the interval [0, 1] with error ε, ε < 0.0023. Finally, appendix A contains the first 20 non-zero coefficients in | | the series of S(x), and the first 10 polynomials Pn.

2 Derivatives

Deﬁnition 1 Let S(x) denote the inverse of

x 1 t2 N(x)= e− 2 dt √2π Z−∞ satisfying S N(x)= N S(x)= x (1) ◦ ◦ In terms of the error function erf(x),

1 x N(x)= erf +1 2 √ 2 Proposition 2 S(x) satisﬁes the IVP

2 S′′ = S(S′) (2) 1 1 √ S 2 =0, S′ 2 = 2π

2 Proof. Since N(0) = 1/2, in follows that S(1/2) = 0. From (2.1) we get

2 1 x2 S [N(x)] S′[N(x)] = = √2πe 2 = √2πe 2 N ′(x) Substituting N(x)= y we have

S2(y) 1 S′(y)= √2πe 2 , S′( )= √2π 2

Diﬀerentiating ln[S′(y)], we get (2.2). Proposition 3

n i (n+2) n i (n i) (i j+1) (j+1) S (x)= S − (x)S − (x)S (x), n 0. i j ≥ i=0 j=0 X X Proof. Taking the nth derivative of (2.2), and using Leibnitz’s Theorem, we have n (n+2) n (n i) (i) S = S − (S′S′) i i=0 X n i n (n i) i (i j+1) (j+1) = S − S − S i j i=0 j=0 X X n i n i (n i) (i j+1) (j+1) = S − S − S . i j i=0 j=0 X X

dnS 1 Corollary 4 If Dn = dxn ( 2 ) , then

D n =0, n 0. 2 ≥ n Putting Dn = (2π) 2 Cn , we can write

2n+1 C2n+1 1 2n+1 S(x)= (2π) 2 (x ) (2n + 1)! − 2 n 0 X≥ where C1 =1, C3 =1, C5 =7, C7 = 127, ... . Proposition 5 (n) n S = Pn 1(S)(S′) n 1 − ≥ where Pn(x) is a polynomial of degree n satisfying the recurrence

P0(x)=1, Pn(x)= Pn′ 1(x)+ nxPn 1(x), n 1 (3) − − ≥ so that

2 3 P1(x)= x, P2(x)=1+2x , P3(x)=7x +6x , ... .

3 Proof. We use induction on n. For n = 2 the result follows from (2.2). If we assume the result is true for n then

(n+1) n S = [Pn 1(S)(S′) ]′ − n n 1 = Pn′ 1(S)S′(S′) + Pn 1(S)n(S′) − S′′ − − n+1 n 1 2 = Pn′ 1(S)(S′) + Pn 1(S)n(S′) − S(S′) − − n+1 = [Pn′ 1(S)+ nSPn 1(S)](S′) − − n+1 = Pn(S)(S′)

Since Pn 1(x) is a polynomial of degree n 1 by hypothesis, is clear that − −

Pn(x)= Pn′ 1(x)+ nxPn 1(x) − − is a polynomial of degree n.

Corollary 6 Cn = Pn 1(0) −

3 The polynomials Pn(x) Lemma 7 If we write n n k Pn(x)= Qk x k X=0 we have

n n 1 Q0 = Q1 − n n 1 n 1 Qk = nQk−1 + (k + 1)Qk+1− k =1,...,n 2 (4) − − n n 1 Q = nQ − k = n 1,n k k − Proof.

n n k Qk x = Pn k X=0 d = Pn 1 + nxPn 1 dx − − n 1 n 1 − n 1 k 1 − n 1 k+1 = Qk− kx − + nQk− x k k X=0 X=0 n 2 n − n 1 k n 1 k = Qk+1− (k + 1)x + nQk−1 x − k k X=0 X=1

4 (n) n 1 n (n) (n+1) n Corollary 8 In matrix form (3.1) reads A Q − = Q where A × is given by ∈ℜ i, j = i +1, i =1,...,n 1 A(n) = n, j = i 1, i =2,...,n−+1 i,j  −  0, ow (n) In other words, A is a rectangular matrix with zeros in the diagonal, the numbers 1, 2, 3,...,n 1 above it, the number n below it and zeros everywhere else − 0 1000 0 0 n 0 200 · · · 0 0  0 n 0 3 0 · · · 0 0  · · · (n)  ......  A =  ......     0 0000 0 n 1   · · · −   0 0000 n 0   · · ·   0 0000 0 n   · · ·  and   n Q0 n Q1 n  n  Q = Q2  .   .     Qn   n  With the help of these matrices, we have an expression for the coeﬃcients of Pn(x) n n (n k+1) (n) (n 1) (1) Q = A − = A A − A · · · k Y=1 Proposition 9 The polynomials Pn(x) satisfy the recurrence relation

n i n i Pn+1(x)= Pn i 1(x)Pi j (x)Pj (x) i j − − − i=0 j=0 X X Proof.

n+2 Pn+1(S)(S′) n i (n+2) n i (n i) (i j+1) (j+1) =S = S − S − S i j i=0 j=0 X X n i n i n i i j+1 j+1 = Pn i 1(S)(S′) − Pi j (S)(S′) − Pj (S)(S′) i j − − − i=0 j=0 X X n i n+2 n i = (S′) Pn i 1(S)Pi j (S)Pj (S). i j − − − i=0 j=0 X X

5 Proposition 10 The exponential generating function of the polynomials Pn(x) is k 1 2 x2 S [N(x)+tN ′(x)] t e 2 − 2 = F (x, t)= P (x) k k! k 0 X≥ Proof. Since tk F (x, t)= P (x) k k! k 0 X≥ d tk tk =1+ Pk 1(x) + kxPk 1(x) dx − k! − k! k 1 k 1 X≥ X≥ d tk+1 tk+1 =1+ P (x) + xP (x) dx k (k + 1)k! k k! k 0 k 0 X≥ X≥ t ∂ =1+ F (x, s)ds + xtF (x, t) ∂x Z0 it follows that F (x, t) satisﬁes the diﬀerential-integral equation

∂ t 1 + (xt 1)F (x, t)+ F (x, s)ds = 0 (5) − ∂x Z0 Diﬀerentiating (3.2) with respect to t we get

∂ ∂ xF (x, t) + (xt 1) F (x, t)+ F (x, t)=0 − ∂t ∂x whose general solution is of the form

x2 x2 1 F (x, t)= e− 2 G te− 2 + √2π N(x) − 2 for some function G(z). From (3.2) we know that F (x, 0) = 1, and hence

1 x2 G √2π N(x) = e 2 , − 2 which implies that 1 S2 z + 1 G(z)= e 2 √2π 2 . Therefore, 1 2 x2 S [N(x)+tN ′(x)] F (x, t)= e 2 − 2

6 Deﬁnition 11 We deﬁne the “nested derivative” D(n) by

D(0)[f](x) 1 ≡ (n) d (n 1) D [f] (x)= f(x) D − [f] (x) , n 1 dx × ≥ n o Example 12 1. D(n) [eax]= n!anenax

2. D(n) [x]=1

3. D(n) x2 = (n + 1)!xn

Proposition 13 2 n x2 (n) x Pn(x)= e− 2 D e 2 h i Proof. We use induction on n. For n = 0 the result follows from the deﬁnition of D(n). Assuming the result is true for n 1 −

Pn(x)= Pn′ 1(x)+ nxPn 1(x) − − (n 1) 2 x2 (n 1) 2 x2 d − x (n 1) − x (n 1) = [e− 2 D − (e 2 )] + nxe− 2 D − (e 2 ) dx (n 1) 2 x2 (n 1) 2 x2 − x (n 1) − x d (n) = (n 1)xe− 2 D − (e 2 )+ e− 2 [D (e 2 )]+ − − dx (n 1) 2 x2 − x (n 1) + nxe− 2 D − (e 2 )

(n 1) 2 x2 x2 − x (n 1) d (n 1) = e− 2 xD − (e 2 )+ D − (e 2 ) dx (n 1) 2 1 2 1 2 x2 − x x d x (n 1) = e− 2 e− 2 e 2 D − (e 2 ) dx 2 n x2 (n) x h i = e− 2 D e 2 h i

Summary 14 We conclude this section by comparing the properties of Pn(x) with the well known formulas for the Hermite polynomials Hn(x) [7]. Since the Hn are deeply related with the function N(x), we would expect to see some similarities between the Hn and the Pn.

7 Pn(x) Hn(x)

2 n x2 (n) x n x2 dn x2 Pn(x)= e− 2 D (e 2 ) Hn(x) = ( 1) e n (e− ) − dx

2 tk 1 2 x tk 2 2 S [N(x)+tN ′(x)] 2 2xt t Pk(x) k! = e − Hk(x) k! = e − k 0 k 0 P≥ P≥

Pn(x)= Pn′ 1(x)+ nxPn 1(x) Hn(x)= Hn′ 1(x)+2xHn 1(x) − − − − − n 1 (n) n (n) 1 − x S = Pn 1(S)(S′) N = − Hn 1 N ′ − √2 − √2

4 Integrals of S(x)

Deﬁnition 15

x x1 xn 1 − ( n) S − (x)= S(xn) dxndxn 1 . . . dx1, n 1. · · · − ≥ Z0 Z0 Z0 Lemma 16

x n x2 n t2 Pn 1(x)= e− 2 Pn 1(0) + e 2 Pn(t)dt (6) −  −  Z0   Proof. It follows immediately from solving the ODE for Pn 1 in terms of Pn. −

Proposition 17 Using (4.1 ) to deﬁne Pn(x) for n< 0 yields

P 1(x)= x − P 2(x)= 1 (7) − − x2 P 3(x)= √πe N √2x − − and the relation (n) n S = Pn 1(S)(S′) − still holds.

Proof. For n = 0 we have

P 1(x)= P 1(0) + x − − (0) S = S = P 1(S) −

8 so P 1(0) = 0. For n = 1 − − x2 P 2(x)= e 2 [P 2(0) + 1] 1 − − − ( 1) We can calculate S − explicitly by

x S(x) 2 ( 1) 1 z S − (x)= S(t)dt = S[N(z)]e− 2 dz √2π Z0 Z −∞ S(x) 1 z2 = ze− 2 dz √2π Z −∞ 2 1 S(x) 1 = e− 2 = (S′)− −√2π −

Hence, P 2(0) = 1. Finally, for− n = −2 − x2 √π P 3(x)= e P 3(0) √πN(√2x)+ − − − 2 A similar calculation as the one above, making a change of variables t = N(z) ( 1) in the integral of S − (x) yields

( 2) 1 S − (x)= N[√2S(x)] −2√π

√π and we conclude that P 3(0) = . − − 2 Corollary 18 ( 2) S 2√πS − = √2S − Proposition 19 h i

1 k (2i + 1), n =2k, k 1 Sn(x)dx = ≥  i=1 Z Q0, n =2k +1, k 0 0  ≥ Proof.  1 ∞ n n S (x)dx = z N ′(z)dz

Z0 Z −∞ ∞ 1 n 1 z2 = z e− 2 dz √2π Z −∞ k = (2i + 1), n =2k, k 1 ≥ i=1 Y

9 5 Asymptotics

Deﬁnition 20 We’ll denote by LW (x) the function Lambert W [4], LW (x)eLW (x) = x (8) This function has the series representation [5]

n 1 ( n) − LW (x)= − xn, n! n 1 X≥ the derivative d LW (x) LW = if x =0, dx x[1 + LW (x)] 6 and it has the asymptotic behavior LW (x) ln(x) ln[ln(x)] x . ∼ − → ∞ Proposition 21

1 S(x) g (x)= LW , x 0 ∼ 0 − 2πx2 → s 1 S(x) g (x)= LW , x 1 ∼ 1 2π(x 1)2 → s − Both functions g0(x) and g1(x) satisfy the ODE

2 2 2 g′′ = g(g′) 1+ g(g′) , for g g2(1 + g2) ∼ | | → ∞ Proof.

1 x2 1 N(x) e− 2 , x ∼ √2π x → −∞ 1 S(t)2 1 t e− 2 , t 0 ∼ √2π S(t) → S(t)2 1 S(t)e 2 , t 0 ∼ √2πt → 2 1 S2(t)eS (t) , t 0 ∼ 2πt2 → Using the deﬁnition of LW (x) we have 1 S2(t) LW , t 0 ∼ 2πt2 → or 1 S(t) LW , t 0 ∼− 2πt2 → s The case x 1 is completely analogous. →

10 Corollary 22 Combining the above expressions, we can get the approximation

1 S(x) g (x)=(2x 1) LW (9) ≃ 2 − 2πx2(x 1)2 s − good through the interval (0, 1). We can reﬁne it even more by putting

1 S(x) g (x)= Q(x) LW (10) ≃ 3 2πx2(x 1)2 s − Q(x)= 1+(6 2√2π)x + ( 12+6√2π)x2 + (8 4√2π)x3 − − − − where Q(x) has been chosen such that

Q(0) = 1,Q(1) = 0,Q(1/2)=0,Q′(1/2) = √2π, Q′′(1/2)=0 − 6 Appendix

The ﬁrst 10 Pn(x) are

P0(x)=1

P1(x)= x 2 P2(x)=1+2x 3 P3(x)=7x +6x 2 4 P4(x)=7+46x + 24x 3 5 P5(x) = 127x + 326x + 120x 2 4 6 P6(x) = 127+1740x + 2556x + 720x 3 5 7 P7(x) = 4369x + 22404x + 22212x + 5040x 2 4 6 8 P8(x) = 4369+102164x + 290292x + 212976x + 40320x 3 5 7 9 P9(x) = 243649x + 2080644x + 3890484x + 2239344x + 362880x 2 4 6 P10(x) = 243649+ 8678422x + 40258860x + 54580248x + + 25659360x8 + 3628800x10

The ﬁrst few odd Cn are

11 n Cn 1 1 3 1 5 7 7 127 9 4369 11 243649 13 20036983 15 2280356863 17 343141433761 19 65967241200001 21 15773461423793767 23 4591227123230945407 25 1598351733247609852849 27 655782249799531714375489 29 313160404864973852338669783 31 172201668512657346455126457343 33 108026349476762041127839800617281 35 76683701969726780307420968904733441 37 61154674195324330125295778531172438727 39 54441029530574028687402753586278549396607 41 53789884101606550209324949796685518122943569

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