# Read e-book online A Simple Gamma Random Number Generator for Arbitrary Shape PDF

By Tanizaki H.

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Additional info for A Simple Gamma Random Number Generator for Arbitrary Shape Parameters

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Combined with the fact that I ( 0 )is completely decomposable, which will be shown in the next section, it will imply that I ( 0 ) is irreducible. 3: Let L ( s ) = 1/(1- q-'). In particular, c(s) = 0 i f and only i f Then s =fl. Proof: In order to calculate A, ( - s ) o A, ( s ) , we need to make the formula for a,(f) more explicit. Clearly, f is Ki-invariant for some i, and we shall show that f(1) determines f ( w n ) for all n in N \ N-i+l. Indeed, note the following relation in SLZ(F): Let j 5 -2, and assume that x is in wjC3 \ wj+lO.

Suvin 40 for all k and k' in K . As usual, the multiplication in H , ( G o / / K )is defined by convolution of functions (fl * f2)(g) = / f l ( h ) f i ( h - l d dg. G O Note that H , ( G o / / K )is contained in V , and equal to V K + the , subspace consisting of all functions f in V such that 7r(k)f = ~ ( k ) ffor all k in K . The element e K defined above is the identity element in H , ( G o / / K ) . 2: There i s a natural isomorphism Homco(V,V) HE(GO//K). Since V is generated by e K , the operator A is completely determined by its value A ( e K ) .

Proof: The right exactness is clear. We only need to check that Vh --+ V, is injective. This is equivalent to V‘ n V ( N ) = V ’ ( N ) ,which will follow from the following important characterization of V ( N ) . 2: Let V be a smooth N-module. A vector v is in V ( N ) i f and only i f k3 n(n)v d n = 0 f o r a sufficiently large open compact subgroup N j . I n particular, V ( N ) = U j e z V ( N j ) , where V ( N j ) is the subspace of all v in V such that the above integral vanishes. Proof: If v is in V ( N ) ,then v is a finite linear combination of terms 7r(n)u- u.