Question

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with unknown mean μ (−∞ < μ < ∞) and known precision \(\tau \) . Suppose also that the prior distribution of μ is the normal distribution with mean \({\mu _0}\) and precision \({\lambda _0}\) . Show that the posterior distribution of μ, given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the normal distribution with mean

\(\frac{{{\lambda _0}{\mu _0} + n\tau {{\bar x}_n}}}{{{\lambda _0} + n\tau }}\)with precision \({\lambda _0} + n\tau \)

Short answer

The posterior distribution of μ, given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the normal distribution with mean

\(\frac{{{\lambda _0}{\mu _0} + n\tau {{\bar x}_n}}}{{{\lambda _0} + n\tau }}\) with precision \({\lambda _0} + n\tau \)

Step-by-step solution

Given information

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with unknown mean μ (−∞ < μ < ∞) and known precision \(\tau \) .It is required to show that

The posterior distribution of μ, given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the normal distribution with mean

\(\frac{{{\lambda _0}{\mu _0} + n\tau {{\bar x}_n}}}{{{\lambda _0} + n\tau }}\) with precision \({\lambda _0} + n\tau \)

Proof: The posterior distribution of μ, given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the normal distribution with mean\(\frac{{{\lambda _0}{\mu _0} + n\tau {{\bar x}_n}}}{{{\lambda _0} + n\tau }}\) with precision \({\lambda _0} + n\tau \)

Precision of a Normal Distribution. The precision \(\tau \) of a normal distribution is defined as the reciprocal of the variance; that is, \(\tau = \frac{1}{{{\sigma ^2}}}\)

Suppose that \({X_1},{X_2},....,{X_n}\) form a random sample from the normal distribution with unknown mean μ and unknown precision \(\tau \) (−∞ <μ< ∞ and \(\tau \) > 0). Suppose also that the joint prior distribution of μ and \(\tau \) is as follows: The conditional distribution of μ given \(\tau \) is the normal distribution with mean μ0 and precision \({\lambda _0}\tau \) (−∞ < \({\mu _0}\) < ∞ and \({\lambda _0}\) > 0), and the marginal distribution of \(\tau \)is the gamma distribution with parameters \({\alpha _0}\) and \({\beta _0}\) (\({\alpha _0}\)> 0 and \({\beta _0}\) > 0).

Then the joint posterior distribution of μ and \(\tau \) , given that \({X_i} = {x_i}\) for i = 1,...,n, is as follows: The conditional distribution of μ given \(\tau \) is the normal distribution with mean \({\mu _1}\) and precision \({\lambda _1}\tau \) , where

\({\mu _1}\)=\(\frac{{{\lambda _0}{\mu _0} + n\tau {{\bar x}_n}}}{{{\lambda _0} + n\tau }}\)and \({\lambda _1} = {\lambda _0} + n\)

Hence by the theorem stated above the unknown parameter \(\theta \) is the mean \(\mu ,\) the variance \({\sigma ^{2}} = \frac{1}{\tau }\) . In the exercise we are given the values of the mean and the posterior distribution which are \({\mu _0}\) and \(\frac{1}{{{\lambda _0}}}\) respectively. Now by substituting this values ,the posterior distribution of \(\mu \) given \({X_1} = {x_1},...,{X_n} = {x_n}\) is normal distribution with mean \({\mu _1}\)=\(\frac{{{\lambda _0}{\mu _0} + n\tau {{\bar x}_n}}}{{{\lambda _0} + n\tau }}\)and variance \(v_1^2 = \frac{1}{{{\lambda _0} + n\tau }}\) with precision \({\lambda _0} + n\tau \)