Question

In: Math

X1,...,Xn are i.i.d Γ(α1,β),...,Γ(αn,β) respectively. Show S = X1 +···+Xn ∼ Γ(α1 +···,αn,β).

X₁,...,X_n are i.i.d Γ(α₁,β),...,Γ(α_n,β) respectively. Show S = X₁ +···+X_n ∼ Γ(α₁ +···,α_n,β).

Expert Solution

are identical and independently distributed (i.i.d) gamma variables as , then the probability density function of is given by the convolution of the probability density functions of respectively. Now, let and be two identical and independently distributed (i.i.d) gamma variables as and , then the probability density function of is given by the convolution of the probability density functions and of and respectively. Thus,

The above integral can be simplified to:

Now, substituting we get,

where,

That is

Thus, the additive property for two independent gamma variables is true,

Hence, let the property be true for variables, i.e. , let

where, are identical and independently distributed (i.i.d) gamma variables as

Then, and is independently distributed.

Now, given, and . Thus,

The above integral can be simplified to:

Now, substituting we get,

where,

Thus,

Hence, proved.

milcah answered 3 months ago

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