Question

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with known mean μ and unknown precision \(\tau \left( {\tau > 0} \right)\). Suppose also that the prior distribution of \(\tau \) is the gamma distribution with parameters\({\alpha _0}\,\,\,{\rm{and}}\,\,\,\,{\beta _0}\left( {{\alpha _0} > 0\,\,\,{\rm{and}}\,\,\,{\beta _0} > 0} \right)\) . Show that the posterior distribution of \(\tau \) given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the gamma distribution with parameters \({\alpha _0} + \frac{n}{2}\,\,\,\,{\rm{and}}\,\,\,\,\,{\beta _0} + \frac{1}{2}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \mu } \right)}^2}} \).

Short answer

posterior distribution of \(\tau \) given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the gamma distribution with parameters \({\alpha _0} + \frac{n}{2}\,\,\,\,{\rm{and}}\,\,\,\,\,{\beta _0} + \frac{1}{2}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \mu } \right)}^2}} \).

Step-by-step solution

Given information 

Suppose that \({X_1},...,{X_n}\) form a random sample from the normal distribution with known mean μ and unknown precision \(\tau \left( {\tau > 0} \right)\). Suppose also that the prior distribution of τ is the gamma distribution with parameters\({\alpha _0}\,\,\,{\rm{and}}\,\,\,\,{\beta _0}\left( {{\alpha _0} > 0\,\,\,{\rm{and}}\,\,\,{\beta _0} > 0} \right)\).It is required to show that the posterior distribution of \(\tau \) given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the gamma distribution with parameters \({\alpha _0} + \frac{n}{2}\,\,\,\,{\rm{and}}\,\,\,\,\,{\beta _0} + \frac{1}{2}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \mu } \right)}^{2}}} \)

Proof of posterior distribution of \(\tau \)  given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the gamma distribution with parameters \({\alpha _0} + \frac{n}{2}\,\,\,\,{\rm{and}}\,\,\,\,\,{\beta _0} + \frac{1}{2}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \mu } \right)}^2}} \) 

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

The joint distribution pdf of i.i.d. random variables \({X_1},...,{X_n}\)from the normal distribution with mean \(\mu \) and precision \(\tau \) is given by

\(\exp \left( { - \frac{1}{2}\tau \sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} } \right), - \infty < x < \infty \)

The posterior distribution of \(\tau \) given \({X_1} = {x_1},....,{X_n} = {x_n}\) is proportional to

\(\xi \left( {\tau ;x} \right) = f\left( {x;\mu ,\tau } \right)\xi \left( \tau \right)\) It is to be noted that the prior distribution\(\xi \left( \tau \right)\) is expressed as

posterior distribution of \(\tau \) given that \({X_i} = {x_i}\) (i = 1, . . . , n) is the gamma distribution with parameters \({\alpha _0} + \frac{n}{2}\,\,\,\,{\rm{and}}\,\,\,\,\,{\beta _0} + \frac{1}{2}\sum\limits_{i = 1}^N {{{\left( {{x_i} - \mu } \right)}^2}} \)

Hence the proof.