Exercise 11.1 - Student T as infinite mixture of Gaussian

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Recall that the 1d Student-t distribution takes the form:

St ( x | μ , σ 2 , v ) = Γ ( v 2 + 1 2 ) Γ ( v 2 ) ( 1 𝜋𝑣 σ 2 ) 1 2 ( 1 + ( x − μ ) 2 v σ 2 ) − v + 1 2 .

While the r.h.s. of (11.61) is:

∫ 𝒩 ( x | μ , σ 2 z ) ⋅ Ga ( z | v 2 , v 2 ) d z = ∫ z 2 π σ 2 ⋅ exp ⁡ { − z 2 σ 2 ( x − μ ) 2 } ⋅ ( v 2 ) v 2 Γ ( v 2 ) ⋅ z v 2 − 1 ⋅ exp ⁡ { − 𝑣𝑧 2 } d z = 1 2 π σ 2 ⋅ ( v 2 ) v 2 Γ ( v 2 ) ∫ z v − 1 2 ⋅ exp ⁡ { − ( v 2 + ( x − μ ) 2 2 σ 2 ) z } d z .

Instead of integrating analytically, we observe that the term being integrated takes the form of a Gamma density:

Ga ( z | v + 1 2 , v 2 + ( x − μ ) 2 2 σ 2 ) = ( v 2 + ( x − μ ) 2 2 σ 2 ) v + 1 2 Γ ( v + 1 2 ) ⋅ z v − 1 2 ⋅ exp ⁡ { − ( v 2 + ( x − μ ) 2 2 σ 2 ) z } .

Therefore the r.h.s. of (11.61) becomes:

1 2 π σ 2 ⋅ ( v 2 ) v 2 Γ ( v 2 ) ⋅ Γ ( v + 1 2 ) ( v 2 + ( x − μ ) 2 2 σ 2 ) v + 1 2 = Γ ( v + 1 2 ) Γ ( v 2 ) ⋅ 1 2 π σ 2 ⋅ ( v 2 ) v 2 ( v 2 + ( x − μ ) 2 2 σ 2 ) v + 1 2 = Γ ( v + 1 2 ) Γ ( v 2 ) ⋅ 1 2 π σ 2 ⋅ ( v 2 ) − 1 2 ( 1 + ( x − μ ) 2 2 v σ 2 ) − v + 1 2 .

This completes the proof.

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2021-03-24 13:42
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