Normally distributed variable with normally distributed mean. What the idea behind the prove of following statement?
I am pretty sure the statement it is correct.
If $X \sim N(\text{mean}_x, \text{var}_x)$ and $Y \sim N(\text{mean}_y+X, \text{var}_y)$,
then $Y \sim N(\text{mean}_x+\text{mean}_y, \text{var}_x+\text{var}_y)$
 A: If you are specifying a hierarchical model, you need to use correct and precise notation:  $$\begin{align*} X &\sim \operatorname{Normal}(\mu_x, \sigma_x^2); \\ Y \mid X &\sim \operatorname{Normal}(\mu_y + X, \sigma_y^2). \end{align*}$$  Without the conditional on the LHS, you are failing to distinguish the difference between the normality of the unconditional or marginal distribution of $Y$, and the normality of the conditional distribution of $Y$ given $X$.
The marginal distribution is simply $$\begin{align*} f_Y(y) &= \int_{x=-\infty}^\infty f_{Y \mid X}(y \mid x) f_X(x) \, dx \\ &\propto \int_{x=-\infty}^\infty \exp\left(-\frac{(y - (\mu_y + x))^2}{2\sigma_y^2} - \frac{(x-\mu_x)^2}{2\sigma_x^2} \right) \, dx. \end{align*}$$  It is up to you to compute this integral and demonstrate your understanding.
That said, if you have trouble doing the computation, then note that your model can be standardized:  $$Y \mid X = Y \mid Z \sim \operatorname{Normal}(\mu_x + \mu_y + \sigma_x Z, \sigma_y^2),$$ where $Z \sim \operatorname{Normal}(0,1)$; thus with another location-scale transformation we can look at the marginal distribution of the variable $$W = \frac{Y - (\mu_x + \mu_y)}{\sigma_y},$$ whose conditional is $$W \mid Z \sim \operatorname{Normal}(\sigma_x Z, 1).$$  Now this should be tractable to anyone with a knowledge of basic calculus.
