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2) Let Yi,., Yn be iid N(a,a2). Let a~ known Find the posterior distribution p(u|3y). This distribution will depend N (8, T2)

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Answer #1

This is the example of bayesian problem,

To calculate the p(u/y) which is posterior distribution we have formula,

p(u/y)=(p(y/u)*p(u))/p(y)

so first we need to calculate p(y/u)

here prior distribution is \mu~ N(\delta,\tau^2)

The probability density function of Yi is,

p(Yi/\mu) = (271\sigma^2)^(-1/2) exp(-(yi -\mu)^2/2\sigma^2)

since y1,y2...yn are independent,the likelihood is,

p(y/u)=\prod_{}^{}p(Yi/\mu)= (271\sigma^2)^(-n/2) exp(-_{}^{}\sum(yi -\mu)^2/2\sigma^2)

The prior is,

p(u) =(271\tau^2)^(-1/2) exp(-(\mu-\delta)^2/2\tau^2)

p(y)=(271\sigma^2)^(-1/2) exp(-(y-\mu)^2/2\sigma^2)

p(u/y)=(p(y/u)p(u)/p(y)

   (p(y/u)p(u)= [(271\sigma^2)^(-n/2)*(271\tau^2)^(-1/2) exp[-1/2(_{}^{}\sum(yi -\mu)^2/\sigma^2)+ exp(-(\mu-\delta)^2/\tau^2))]]

by solving square brackets in exponential and adding and subtracting by [(\delta/\tau^2)+(_{}^{}\sumyi/\sigma^2)]^2 / [(1/\tau^2)+(n/\sigma^2)]

here terms involving \delta^2 is ((1/\tau^2)+(n/\sigma^2))\delta^2

and another terms involving \delta is -2((\mu/\tau^2)+(_{}^{}\sumyi/\sigma^2))\delta

remaining terms are (\delta^2/\tau^2)+(_{}^{}\sumyi^2/\sigma^2)

after solving these we define some terms .,

\tau" =((1/\tau^2)+(n/\sigma^2)^(-1/2)) and \delta" =[(\delta/\tau^2)+(_{}^{}\sumyi/\sigma^2)]/[(1/\tau^2)+(n/\sigma^2)]

the expression which comes in bracket of exponent is ,

{((\mu-\delta")/\tau")^2 +[(\delta^2/\tau^2)+(_{}^{}\sumyi^2/\sigma^2)-((\delta/\tau^2)+(_{}^{}\sumyi/\sigma^2)]^2 / [(1/\tau^2)+(n/\sigma^2)])]}

posterior pdf for \mu is

p(u/y)=(271\tau"^2)^(-1/2) exp(-(\mu-\delta")^2/2\tau"^2)

posterior distribution is normal with mean \delta" and variance \tau"^2

NOTE: Solve the exponent brackets as per given instruction above,I've given all the possible terms which will come in exponent brackets.

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