If (X1, Y1), . . . , (Xn, Yn) is a sample from a bivariate normal distribution, the probability density of the sample correlation coefficient R is16 pρ(r) = 2n−3

π(n − 3)!

(1 − ρ2

)

1 2 (n−1)

(1 − r 2

)

1 2 (n−4) (5.85)

×∞

k=0 2 1 2 (n + k − 1)

(2ρr)k k!

or alternatively pρ(r) = n − 2

π (1 − ρ2

)

1 2 (n−1)

(1 − r 2

)

1 2 (n−4) (5.86)

×

 1 0

t n−2

(1 − ρrt)n−1 1

√1 − t 2 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

(n − 1)

(n − 1 2 )

(1 − ρ2

)

1 2 (n−1)

(1 − r 2

)

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 (a + j)
(a)
(b + j)
(b)
(c)
(c + j)
x j 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{ρr s1s2/(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.]