# Total probability theorem with normal probability density functions

Let $X$ and $Y$ be two continuous random variables. Suppose that:

1. $X$ has normal pdf with mean mu_1 and variance $\sigma_1^2$
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
3. $f(Y)=\int f(Y|X)f(X) dx$; 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?

• There is no such thing as the random variable $Y|X$; what you are saying is that the conditional density of $Y$ given the value of $X$ is a normal density. Commented Nov 21, 2013 at 3:02

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.)
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 $$Y \sim N(a+\mu_1, b^2 \sigma_1^2 + \sigma_2^2).$$