Posterior distribution

Question

I read the following in some document:

Let $Y$ be a random variable with distribution $\mathcal{N}(\theta,\sigma^{2})$. The variance $\sigma^{2}$ is known. Let $p(\theta) = 1$ a flat prior on $\theta$. Then, the posterior distribution $p(\theta \vert Y)$ is $\mathcal{N}(Y,1)$.

From Bayes' theorem, I know that:

$$ \begin{eqnarray*} p(\theta \vert Y) & \propto & p(Y \vert \theta) p(\theta) \\ & \propto & \exp \Big( -\frac{1}{2\sigma^{2}} (\theta-Y)^{2} \Big) \\ \end{eqnarray*} $$

To me, the posterior distribution would be $\mathcal{N}(Y,\sigma^{2})$. I don't understand why the document says it is $\mathcal{N}(Y,1)$.

Answer 1

You are correct assuming we have only one observation. The true form of the posterior is $N(\bar{Y},\frac{\sigma^2}{n})$. To demonstrate:

$$ \begin{eqnarray*} p(\theta \vert Y) & \propto & p(Y \vert \theta) p(\theta) \\ & \propto & \prod_{i=1}^{n}\exp \Big( -\frac{1}{2\sigma^{2}} (\theta-Y_i)^{2} \Big) \\ & \propto & \exp \Big( -\frac{1}{2\sigma^{2}} \sum_{i=1}^{n}(\theta-Y_i)^{2} \Big) \\ & \propto & \exp \Big( -\frac{1}{2\sigma^{2}} (n\theta^2-2\theta\sum_{i=1}^{n}Y_i +Y_i^2) \Big) \\ & \propto & \exp \Big( -\frac{1}{2\sigma^{2}} (n\theta^2-2\theta\sum_{i=1}^{n}Y_i ) \Big) \\ & \propto & \exp \Big( -\frac{n}{2\sigma^{2}} (\theta-\bar{Y})^2 \Big) \\ \end{eqnarray*} $$

This also agrees with our intuition that the variance of the posterior should shrink with the number of observations we have.