Likelihood for model with two data sources
$begingroup$
$theta$ is parameter for model and $D_{1} ,D_{2}$ are two i.i.d data sources. The parameter for model are jointly optimised by MLE. My goal is to just confirm likelihood for the model is right.
$ P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} ,theta)}{P(D_{1},D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} | theta)P(theta)}{P(D_{1})P(D_{2})}$
Since, $D_{1},D_{2}$ are independent given $theta$. Can I write this directly as?
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$ (1)
OR
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(theta | D_{2})}{P(D_{1})}$ (2)
What would be the right way of writing this? Also, is everything correct? I strongly feel that the first equation is wrong because even though the data sources don't depend on each other the optimisation of parameter does. So, the second equation would be more appropriate which is intuitive that the parameters are optimised with respect to let's say $D_{2}$ and then optimised further given $D_{1}, theta$.
So, the overall likelihood term for model would be $P(D_{1}| D_{2}, theta)P(D_{2}| theta)$ ?
probability proof-verification maximum-likelihood
asked Jan 22 at 0:03
gamabuntagamabunta
12
$endgroup$
add a comment |
$begingroup$
$theta$ is parameter for model and $D_{1} ,D_{2}$ are two i.i.d data sources. The parameter for model are jointly optimised by MLE. My goal is to just confirm likelihood for the model is right.
$ P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} ,theta)}{P(D_{1},D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} | theta)P(theta)}{P(D_{1})P(D_{2})}$
Since, $D_{1},D_{2}$ are independent given $theta$. Can I write this directly as?
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$ (1)
OR
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(theta | D_{2})}{P(D_{1})}$ (2)
What would be the right way of writing this? Also, is everything correct? I strongly feel that the first equation is wrong because even though the data sources don't depend on each other the optimisation of parameter does. So, the second equation would be more appropriate which is intuitive that the parameters are optimised with respect to let's say $D_{2}$ and then optimised further given $D_{1}, theta$.
So, the overall likelihood term for model would be $P(D_{1}| D_{2}, theta)P(D_{2}| theta)$ ?
probability proof-verification maximum-likelihood
asked Jan 22 at 0:03
gamabuntagamabunta
12
$endgroup$
add a comment |
$begingroup$
$theta$ is parameter for model and $D_{1} ,D_{2}$ are two i.i.d data sources. The parameter for model are jointly optimised by MLE. My goal is to just confirm likelihood for the model is right.
$ P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} ,theta)}{P(D_{1},D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} | theta)P(theta)}{P(D_{1})P(D_{2})}$
Since, $D_{1},D_{2}$ are independent given $theta$. Can I write this directly as?
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$ (1)
OR
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(theta | D_{2})}{P(D_{1})}$ (2)
What would be the right way of writing this? Also, is everything correct? I strongly feel that the first equation is wrong because even though the data sources don't depend on each other the optimisation of parameter does. So, the second equation would be more appropriate which is intuitive that the parameters are optimised with respect to let's say $D_{2}$ and then optimised further given $D_{1}, theta$.
So, the overall likelihood term for model would be $P(D_{1}| D_{2}, theta)P(D_{2}| theta)$ ?
probability proof-verification maximum-likelihood
asked Jan 22 at 0:03
gamabuntagamabunta
12
$endgroup$
$theta$ is parameter for model and $D_{1} ,D_{2}$ are two i.i.d data sources. The parameter for model are jointly optimised by MLE. My goal is to just confirm likelihood for the model is right.
$ P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} ,theta)}{P(D_{1},D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1} ,D_{2} | theta)P(theta)}{P(D_{1})P(D_{2})}$
Since, $D_{1},D_{2}$ are independent given $theta$. Can I write this directly as?
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$ (1)
OR
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(D_{2}| theta)P(theta)}{P(D_{1})P(D_{2})}$
$P (theta | D_{1},D_{2}) = frac{P(D_{1}| D_{2}, theta)P(theta | D_{2})}{P(D_{1})}$ (2)
What would be the right way of writing this? Also, is everything correct? I strongly feel that the first equation is wrong because even though the data sources don't depend on each other the optimisation of parameter does. So, the second equation would be more appropriate which is intuitive that the parameters are optimised with respect to let's say $D_{2}$ and then optimised further given $D_{1}, theta$.
So, the overall likelihood term for model would be $P(D_{1}| D_{2}, theta)P(D_{2}| theta)$ ?
probability proof-verification maximum-likelihood
probability proof-verification maximum-likelihood
asked Jan 22 at 0:03
gamabuntagamabunta
12
asked Jan 22 at 0:03
gamabuntagamabunta
12
edited Jan 22 at 0:40
gamabunta
asked Jan 22 at 0:03
gamabuntagamabunta
12
asked Jan 22 at 0:03
gamabuntagamabunta
12
asked Jan 22 at 0:03
gamabuntagamabunta
12
12
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