To analyse the distribution properly you cannot start with the assumption that the result will be Gaussian. If $$X|\theta,\tau\sim\mathcal N(\theta,\tau^2)$$ and $$\theta\sim\mathcal N(\mu,\sigma^2_\theta)\qquad\tau\sim\mathcal N(\sigma,\omega^2_\tau)$$ then one can write $$X=\theta+\tau\xi=\mu+\sigma_\theta\epsilon_\theta+(\sigma+\omega_{\tau}\epsilon_\tau)\xi$$ where $$\epsilon_\theta,\epsilon_\tau,\xi\sim\mathcal N(0,1)$$independently. From this representation, one can see that the resulting distribution of $X$ is not Normal due to the product $\epsilon_\tau\xi$.

Note that the Normal assumption on $\tau$ is rather unusual because usually $\tau$ is assumed to be positive. A more standard approach (in a Bayesian perspective) is to assume that $\tau^{-2}$ is distributed from a Gamma distribution$$\tau^{-2}\sim\mathcal Ga(\alpha,\beta)$$ Then$$X|\tau\sim\mathcal N(\mu,\sigma^2_\theta+\tau^2)$$and the marginal density of $X$ is given by $$ f(x)=\int_0^\infty \frac{(\sigma^2_\theta+\iota^{-1})^{-1/2}}{\sqrt{2\pi}} \exp\left\{-(x-\mu)^2\iota/2\right\}\iota^{\alpha-1}\exp\{-\beta\iota\}\,\text d\iota$$ for which there is no closed-form expression (unless using special functions).

To analyse the distribution properly you cannot start with the assumption that the result will be Gaussian. If $$X|\theta,\tau\sim\mathcal N(\theta,\tau^2)$$ and $$\theta\sim\mathcal N(\mu,\sigma^2_\theta)\qquad\tau\sim\mathcal N(\sigma,\omega^2_\tau)$$ then one can write $$X=\theta+\tau\xi=\mu+\sigma_\theta\epsilon_\theta+(\sigma+\omega_{\tau}\epsilon_\tau)\xi$$ where $$\epsilon_\theta,\epsilon_\tau,\xi\sim\mathcal N(0,1)$$independently. From this representation, one can see that the resulting distribution of $X$ is not Normal due to the product $\epsilon_\tau\times\xi$ of standard Normal variates. Since by integrating out $\theta$ $$X|\tau\sim\mathcal N(\mu,\sigma^2_\theta+\tau^2)$$the marginal density of $X$ is given by $$ f(x)=\int_{-\infty}^\infty \frac{(\sigma^2_\theta+\tau^2)^{-1/2}}{\sqrt{2\pi}} \exp\left\{-(x-\mu)^2/2\tau^2\right\}\frac{1}{\sqrt{2\pi}\omega_\tau}\exp\{-(\tau-\sigma)^2/2\omega^2_\tau\}\,\text d\tau$$ which is not particularly manageable (i.e., does not return a closed form expression).

Note that the Normal assumption on $\tau$ is rather unusual because usually $\tau$ is assumed to be positive. A more standard approach (in a Bayesian perspective) is to assume that $\tau^{-2}$ is distributed from a Gamma distribution$$\tau^{-2}\sim\mathcal Ga(\alpha,\beta)$$ Then$$X|\tau\sim\mathcal N(\mu,\sigma^2_\theta+\tau^2)$$and the marginal density of $X$ is given by $$ f(x)=\int_0^\infty \frac{(\sigma^2_\theta+\iota^{-1})^{-1/2}}{\sqrt{2\pi}} \exp\left\{-(x-\mu)^2\iota/2\right\}\iota^{\alpha-1}\exp\{-\beta\iota\}\,\text d\iota$$ for which there is no closed-form expression (unless using special functions).

Another Bayesian modelling is to assume that $\theta$ and $\tau$ are independent, with $$\theta|\tau\sim\mathcal N(\mu,\rho^2\tau^2)\qquad\tau^{-2}\sim\mathcal Ga(\alpha,\beta)$$ in which case the marginal distribution of $X$ is a Student's $t$ distribution (cf. Bayesian textbooks).

To analyse the distribution properly you cannot start with the assumption that the result will be Gaussian. If $$X|\theta,\tau\sim\mathcal N(\theta,\tau^2)$$ and $$\theta\sim\mathcal N(\mu,\sigma^2_\theta)\qquad\tau\sim\mathcal N(\sigma,\omega^2_\tau)$$ then one can write $$X=\theta+\tau\xi=\mu+\sigma_\theta\epsilon_\theta+(\sigma+\omega_{\tau}\epsilon_\tau)\xi$$ where $$\epsilon_\theta,\epsilon_\tau,\xi\sim\mathcal N(0,1)$$independently. From this representation, one can see that the resulting distribution of $X$ is not Normal due to the product $\epsilon_\tau\xi$.

To analyse the distribution properly you cannot start with the assumption that the result will be Gaussian. If $$X|\theta,\tau\sim\mathcal N(\theta,\tau^2)$$ and $$\theta\sim\mathcal N(\mu,\sigma^2_\theta)\qquad\tau\sim\mathcal N(\sigma,\omega^2_\tau)$$ then one can write $$X=\theta+\tau\xi=\mu+\sigma_\theta\epsilon_\theta+(\sigma+\omega_{\tau}\epsilon_\tau)\xi$$ where $$\epsilon_\theta,\epsilon_\tau,\xi\sim\mathcal N(0,1)$$independently. From this representation, one can see that the resulting distribution of $X$ is not Normal due to the product $\epsilon_\tau\xi$.

Note that the Normal assumption on $\tau$ is rather unusual because usually $\tau$ is assumed to be positive. A more standard approach (in a Bayesian perspective) is to assume that $\tau^{-2}$ is distributed from a Gamma distribution$$\tau^{-2}\sim\mathcal Ga(\alpha,\beta)$$ Then$$X|\tau\sim\mathcal N(\mu,\sigma^2_\theta+\tau^2)$$and the marginal density of $X$ is given by $$ f(x)=\int_0^\infty \frac{(\sigma^2_\theta+\iota^{-1})^{-1/2}}{\sqrt{2\pi}} \exp\left\{-(x-\mu)^2\iota/2\right\}\iota^{\alpha-1}\exp\{-\beta\iota\}\,\text d\iota$$ for which there is no closed-form expression (unless using special functions).