The data $X$ is from $N(0, \cfrac{1}{\theta})$ distribution where $\theta$ is the model parameter.
Find $\theta$'s bayesian conjugate prior distribution and the appropriate posterior distribution.
Seems like conjugate distribution should be $Gamma$? What parameters with? Not sure with the constants.
For a sample $\boldsymbol x = (x_1, \ldots, x_n)$, the kernel of the likelihood is $$\mathcal L(\theta \mid \boldsymbol x) \propto \theta^{n/2} e^{-\frac{\theta}{2} \sum_{i=1}^n x_i^2},$$ where $\theta$ is the reciprocal variance, also called precision. For simplicity, let $\xi = \sum_{i=1}^n x_i^2$ be the sum of squares of observations. Then we note that the likelihood is proportional to a gamma density parametrized by shape and rate; i.e., if $$Y \sim \operatorname{Gamma}(\alpha,\beta), \\ f_Y(y) = \frac{\beta^\alpha y^{\alpha-1} e^{-\beta y}}{\Gamma(\alpha)},$$ then the kernel of $Y$ is $$y^{\alpha - 1} e^{-\beta y}$$ and the choice $$y = \theta, \quad \alpha = \frac{n}{2} + 1,\quad \beta = \frac{1}{2} \sum_{i=1}^n x_i^2 = \frac{\xi}{2},$$ matches the likelihood. Consequently, the conjugate prior for $\theta$ is gamma distributed, and the posterior distribution for $\theta$ for the prior shape and scale hyperparameters $\alpha, \beta$ will be $$\begin{align} f_\Theta(\theta \mid \boldsymbol x) &\propto \mathcal L(\theta \mid \boldsymbol x) p(\theta) \\ &= \theta^{n/2} e^{-\xi \theta} \theta^{\alpha - 1} e^{-\beta \theta} \\ &= \theta^{n/2 + \alpha - 1} e^{-(\xi + \beta)\theta} \end{align}$$ which is gamma with posterior hyperparameters $$\alpha^* = \alpha + \frac{n}{2}, \quad \beta^* = \beta + \xi.$$