806 CHAPTER 39. STATISTICAL TESTS
Now it is necessary to invert the transformations and solve for v,w in terms of t,u.
t =w√v/r
,u = v
So w = t√ u
r ,v = u. Thus
f (t,u) =1√2π
e−t2u2r
1Γ(r/2)2r/2
√1r
uu(r/2)−1e−u/2
=1√2π
e−12r t2ue−
12 u 1
Γ(r/2)2r/2 u12 r− 1
2
√1r
Now this is the density for a random vector (T,U) and it is desired to find the density forT. This means U can be anywhere in (0,∞) and so to get this density we do the followingintegral.
1√2π
√1r
1Γ(r/2)2r/2
∫∞
0e−
12r t2ue−
12 uu
12 r− 1
2 du
Consider the integral. It is ∫∞
0e−u
(t22r +
12
)u
12 (r−1)du
Change variables letting x = u(
t2
2r +12
),dx =
(t2
2r +12
)du. Then it equals
∫∞
0e−x
(x
t2
2r +12
) 12 (r−1)
1t2
2r +12
dx
=
(1
t2
2r +12
) 12 r+ 1
2 ∫ ∞
0e−xx
12 (r−1)dx
Let α−1 = 12 (r−1) . Then the above equals(
1t2
2r +12
) 12 r+ 1
2
Γ(α) =
(1
t2
2r +12
) 12 r+ 1
2
Γ
(12
r+12
)Therefore, the density function for T is
=1√2π
√1r
1Γ(r/2)2r/2
(2
(t2/r+1)
) 12 r+ 1
2Γ
(12
r+12
)=
1√π
√1r
Γ( 1
2 r+ 12
)Γ(r/2)
(1
(t2/r+1)
) 12 r+ 1
2
Then
f (t)≡ 1√π
√1r
Γ( 1
2 r+ 12
)Γ(r/2)
(1
(t2/r+1)
) 12 r+ 1
2
is the density for the T distribution. Here t ∈R. Here is a graph of F (x) = P(X ≤ x) for Xdistributed as a T distribution in which r = 10.