Assume that Yi|θ ∼ N(3θ,2), for i = 1,...,n, and Θ ∼ N(1,9). Find the conditional pdf of Θ given the sample Y = (Y1, . . . , Yn)′.
I'm not sure how to go about this but my thought process was the following: -We need to find f(Θ|Y)=f(Θ)/f(Y) and f(Y)=f(Yi|θ)f(Θ) and then we can inverse f(Y) and inverse it and find our answer. My problem tho is multiplying the two normal distributions together? It gets a messy answer and not sure how to fully continue it.
You only need to focus on the terms with $\theta$. \begin{align} f(y_1, \ldots, y_n, \theta) &= f(\theta) \prod_{i=1}^n f(y_i \mid \theta) \\ &= (2\pi \cdot 9)^{-1/2} e^{-(\theta - 1)^2/(2 \cdot 9)} \prod_{i=1}^n (2 \pi \cdot 4)^{-1/2} e^{-(y_i - 3 \theta)^2 / (2 \cdot 4)} \\ &= c_{y_1, \ldots, y_n} \exp\left(-\left(\frac{1}{18} + \frac{9}{8} n\right)\theta^2 + 2 \left(\frac{1}{18} + \frac{3}{8} \sum_{i=1}^n y_i\right)\theta\right) \end{align}
Dividing this by $f(y_1, \ldots, y_n)$ will give you the desired conditional density, but this term also does not involve $\theta$. So it suffices to focus on the big exponential term above. You can complete the square to find that it is a normal distribution with some mean and variance.