If (X1, Y1),..., (Xn, Yn) is a sample from a bivariate normal distribution, the probability density of the sample correlation coefficient R is15 pρ(r) = 2n−3

π(n − 3)! (1 − ρ2

)

1 2 (n−1)(1 − r2

)

1 2 (n−4) (5.85)

×∞

k=0

Γ2

#

1 2 (n + k − 1)$ (2ρr)

k k!

or alternatively pρ(r) = n − 2

π (1 − ρ2

)

1 2 (n−1)(1 − r2

)

1 2 (n−4) (5.86)

×

 1 0

t n−2

(1 − ρrt)n−1 1

√1 − t2 dt.

Another form is obtained by making the transformation t = (1 − v)/(1 − ρrv) in the integral on the right-hand side of (5.86). The integral then becomes 1

(1 − ρr)

1 2 (2n−3)  1 0

(1 − v)

n−2

√2v

#

1 − 1 2 v(1 + ρr)

$−1/2 dv. (5.87)

Expanding the last factor in powers of v, the density becomes n − 2

√2π

Γ(n − 1)

Γ(n − 1 2 )

(1 − ρ2

)

1 2 (n−1)(1 − r2

)

1 2 (n−4)(1 − ρr)

−n+ 3 2 (5.88)

×F 1

2 ; 1 2 ; n − 1 2 ;

1 + ρr 2



, where F

(a,

b, c, x) = ∞

j=0

Γ(a + j)

Γ(a)

Γ(b + j)

Γ(b)

Γ(c)

Γ(c + j)

xj j! (5.89)

is a hypergeometric function.

[To obtain the first expression make a transformation from (S2 1 , S2 2 , S12) with density (5.84) to (S2 1 , S2 2 , R) and expand the factor exp{ρs12/(1 − ρ2)στ} = exp{ρrs1s2/(1 − ρ2)στ} into a power series. The resulting series can be integrated term by term with respect to s2 1 and s2 2. The equivalence with the second expression is seen by expanding the factor (1 − ρrt)
−(n−1) under the integral in (5.86) and integrating term by term.]