Total probability theorem with normal probability density functions Let $X$ and $Y$ be two continuous random variables. Suppose that:


*

*$X$ has normal pdf with mean mu_1 and variance $\sigma_1^2$

*$Y|X$ has normal pdf with mean $mu_2=(a+b*X)$ and variance $\sigma_2^2$; where a and b are two constants from linear regression

*$f(Y)=\int  f(Y|X)f(X)  dx$; where $f()$ is the pdf


Question: How do I show that $f(Y)$ is a normal distribution? How do I determine its parameters based on the information I'm given?
Note: I have tried to brute force multiply the densities and integrate, which yields the answer to my question. However, is there a faster way to see this because I'm interested in extending to the case where $X$ is a random vector?
 A: What you should know is the well known property of normal distributions: "The sum of two independent normal random variables is normal."
More precisely, given two independent normal random variables, X and Y, the pdf of $aX+bY$ is
$$
\begin{align*} 
aX + bY &\sim N(a\mu_X + b\mu_Y, a^2\sigma_X^2 + b^2\sigma_X^2),
\end{align*}
$$ 
where $\mu_X, \mu_Y$ are means of $X, Y$ and $\sigma_X^2, \sigma_Y^2$ are the variance of $X,Y$.
You can find the proof in any textbook for introductory statistics/probability. The same things holds for vector valued case. (Modify the scalar products to vector/matrix products.)
Now you can easily verify that Y is normal and get its mean and variance. 
The first condition of your question means
$$
\begin{align*} 
X &\sim N(\mu_1, \sigma_1^2).
\end{align*}
$$ 
The second condition of your question means that 
there is a random variable Z, which is independent from X, and
$$
\begin{align*} 
Z &\sim N(a,\sigma_2^2) \\
Y &= Z + bX.
\end{align*}
$$ 
Combining these observations, you get
$$
 Y \sim N(a+\mu_1, b^2 \sigma_1^2 + \sigma_2^2).
$$
