Completing the square: finding the conjugate normal analysis

[gabrielwong1991]

Basically I have this equation and trying to prove it from the middle bit to the LHS.

This derivation is actually when I do bayesian econometrics where we need to find the conjugate normal analysis with unknown mean and known variance for the prior.

I was told we can use completing the square to prove it but after many attempt I fail to recognise...

Perhaps anyone could help?

I also tried the completing the square like this notes but with no success.
http://www.cse.psu.edu/~rtc12/CSE598C/completingTheSquare.pdf

Many thanks

. . . . .\(\displaystyle \large \displaystyle \sum_{t = 1}^T\, (y_t\, -\, \theta_1)^2\, =\, \sum_{t = 1}^T\, (y_t\, -\, \bar{y})^2\, +\, T\, (\bar{y}\, -\, \theta_1)^2\, =\, vs^2\, +\, T\, (\bar{y}\, -\, \theta_1)^2\)

\(\displaystyle \large \mbox{For all }\, \theta_1,\, \mbox{ where}\)

. . . . .\(\displaystyle \large v\, =\, T\, -\, 1\)

\(\displaystyle \large \mbox{And}\)

. . . . .\(\displaystyle \large s^2\, =\, v^{-1}\, \)\(\displaystyle \displaystyle \sum_{t = 1}^T\, (y_t\, -\, \bar{y})^2\)

 

[Deleted member 4993]

Basically I have this equation and trying to prove it from the middle bit to the LHS.

This derivation is actually when I do bayesian econometrics where we need to find the conjugate normal analysis with unknown mean and known variance for the prior.

I was told we can use completing the square to prove it but after many attempt I fail to recognise...

Perhaps anyone could help?

Many thanks

. . . . .\(\displaystyle \large \displaystyle \sum_{t = 1}^T\, (y_t\, -\, \theta_1)^2\, =\, \sum_{t = 1}^T\, (y_t\, -\, \bar{y})^2\, +\, T\, (\bar{y}\, -\, \theta_1)^2\, =\, vs^2\, +\, T\, (\bar{y}\, -\, \theta_1)^2\)

\(\displaystyle \large \mbox{For all }\, \theta_1,\, \mbox{ where}\)

. . . . .\(\displaystyle \large v\, =\, T\, -\, 1\)

\(\displaystyle \large \mbox{And}\)

. . . . .\(\displaystyle \large s^2\, =\, v^{-1}\, \)\(\displaystyle \displaystyle \sum_{t = 1}^T\, (y_t\, -\, \bar{y})^2\)

Use the fact that:

\(\displaystyle \displaystyle{\sum_{t=1}^T \left (y_t - y_{mean}\right ) \ = \ 0}\)

 

[gabrielwong1991]

Use the fact that:

\(\displaystyle \displaystyle{\sum_{t=1}^T \left (y_t - y_{\mbox{me}\mbox{an}}\right ) \ = \ 0}\)

Hey mate thanks for your reply...

can you show me the steps on how to do it?

 

[stapel]

can you show me the steps on how to do it?

What did you get when you tried it? Please show your steps, even if you think they're wrong. Thank you!

 

[Deleted member 4993]

Basically I have this equation and trying to prove it from the middle bit to the LHS.

This derivation is actually when I do bayesian econometrics where we need to find the conjugate normal analysis with unknown mean and known variance for the prior.

I was told we can use completing the square to prove it but after many attempt I fail to recognise...

Perhaps anyone could help?

I also tried the completing the square like this notes but with no success.
http://www.cse.psu.edu/~rtc12/CSE598C/completingTheSquare.pdf

Many thanks

. . . . .\(\displaystyle \large \displaystyle \sum_{t = 1}^T\, (y_t\, -\, \theta_1)^2\, =\, \sum_{t = 1}^T\, (y_t\, -\, \bar{y})^2\, +\, T\, (\bar{y}\, -\, \theta_1)^2\, =\, vs^2\, +\, T\, (\bar{y}\, -\, \theta_1)^2\)

\(\displaystyle \large \mbox{For all }\, \theta_1,\, \mbox{ where}\)

. . . . .\(\displaystyle \large v\, =\, T\, -\, 1\)

\(\displaystyle \large \mbox{And}\)

. . . . .\(\displaystyle \large s^2\, =\, v^{-1}\, \)\(\displaystyle \displaystyle \sum_{t = 1}^T\, (y_t\, -\, \bar{y})^2\)

.\(\displaystyle \large \displaystyle \sum_{t = 1}^T\, (y_t\, -\, \theta_1)^2\, =\, \sum_{t = 1}^T\, \left [(y_t\, -\, \bar{y}) + (\bar{y} -\, \theta_1)\right ]^2 \)

Continue....

 

[gabrielwong1991]

Thanks Khan,

Here is what I thought, I am out of clue when i do this until khan told I could be written that way