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Let X1, . . . , Xn denote a random sample from a N (µ, σ2) distribution,where the mean µ and the variance σ2 are both unknown so that the param-eter vector is given byθ = (µ, σ2)T .(i) Show that the joint p.d.f. of X1, . . . , Xn belongs to the two-parameterregular exponential family, stating what the canonical parameter isequal to in terms of θ

Question

Let X1, . . . , Xn denote a random sample from a N (µ, σ2) distribution,where the mean µ and the variance σ2 are both unknown so that the param-eter vector is given byθ = (µ, σ2)T .(i) Show that the joint p.d.f. of X1, . . . , Xn belongs to the two-parameterregular exponential family, stating what the canonical parameter isequal to in terms of θ

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Solution

The probability density function (pdf) of a normal distribution N(µ, σ^2) is given by:

f(x; µ, σ^2) = (1/√(2πσ^2)) * exp{ -1/2σ^2 * (x - µ)^2 }

Let X1, ..., Xn be a random sample from this distribution. The joint pdf of X1, ..., Xn is the product of the individual pdfs, since the X's are independent:

f(x1, ..., xn; µ, σ^2) = ∏_{i=1}^{n} f(xi; µ, σ^2)

Substituting the expression for the normal pdf, we get:

f(x1, ..., xn; µ, σ^2) = (1/(2πσ^2)^(n/2)) * exp{ -1/2σ^2 * ∑_{i=1}^{n} (xi - µ)^2 }

This can be rewritten as:

f(x1, ..., xn; µ, σ^2) = exp{ -1/2σ^2 * ∑_{i=1}^{n} (xi - µ)^2 - n/2 * log(2πσ^2) }

This is in the form of a two-parameter regular exponential family distribution, with canonical parameters µ and -1/2σ^2. The functions a(τ), b(θ), and c(x, τ) in the general form of the exponential family distribution are given by:

a(τ) = τ = σ^2 b(θ) = θ^2/2 = µ^2/2 c(x, τ) = -x^2/2τ - log(√(2πτ)) = -x^2/2σ^2 - log(√(2πσ^2))

So, the canonical parameters in terms of θ = (µ, σ^2) are µ and -1/2σ^2.

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