# I know $Y|Z \sim\mathbb{N}(Z,\sigma^2)$, but what is $Z|Y$?

Say I know the conditional distribution:

$Y|Z \sim\mathbb{N}(Z,\sigma^2)$

Now, what if I reversed this though and wanted to find the conditional distribution $Z|Y$?

From intuition I would expect the distribution to still be normal with mean of $Y$ (i.e., $Y$ is an unbiased predictor of $Z$?), but I would also think the variance is larger.

• You can't really say anything about the distribution of $Z|Y$ without making some assumptions on $P(Z)$ and $P(Y)$. $Z$ could be a constant, or a die roll, or Poisson or anything you can imagine. And once you state something about $P(Z)$ and $P(Y)$ then Bayes theorem will give the answer. Commented Feb 19, 2016 at 17:24
• This is the answer to the question, you might as well paste the same thing as an answer
– Bach
Commented Feb 19, 2016 at 17:30
• @Corone Hmm, so are you saying I would need to know something like Z ~ $\Bbb N$($\mu$,$\sigma_z^2$) to be able to fully specify the answer?
– JPJ
Commented Feb 19, 2016 at 18:39
• @user3496060 More precisely, you need to know both the marginal distributions of Z and of Y.
– Sycorax
Commented Feb 19, 2016 at 18:42
• OK, so like $f_y$(Y|Z) = $f_z$(Z|Y)$f_y(Y)\frac1{f_z(Z)}$
– JPJ
Commented Feb 19, 2016 at 18:52

You can't really say anything about the distribution of $Z|Y$ without making some assumptions on $P(Z)$ and $P(Y)$. $Z$ could be a constant, or a die roll, or Poisson or anything you can imagine. And once you state something about $P(Z)$ and $P(Y)$ then Bayes theorem will give the answer:
$P(Y|Z) = \frac{P(Z|Y)P(Y)}{P(Z)}$
In the special case where $Y$ and $Z$ are both normally distribution, then Y|Z being normal with mean proportional to $Z$ fixed variance is sufficient to conclude that $Y$ and $Z$ are jointly normal. In this case:
$Z|Y \sim N(\mu_Z + \frac{\sigma_Z \rho}{\sigma_Y} (Y-\mu_Y),(1-\rho^2)\sigma_z^2 )$
Notice that the distribution is NOT centred around $Y$ as you supposed - this means that if the Y you observe is very high above the mean of Z, then it is likely that that was due to a quite high value of Y combined with a quite high value of Z, rather than an extremely high value of Z. This effect is the source of regression toward the mean