I have a question about calculation of conditional density of two normal distributions. I have random variables $X|M \sim \text{N}(M,\sigma^2)$ and $M \sim \text{N}(\theta, s^2)$, with conditional and marginal densities given by:
$$\begin{equation} \begin{aligned}
f(x|m) &= \frac{1}{\sigma \sqrt{2\pi}} \cdot \exp \Big( -\frac{1}{2} \Big( \frac{x-m}{\sigma} \Big)^2 \Big), \\[10pt]
f(m) &= \frac{1}{s \sqrt{2\pi}} \cdot \exp \Big( – \frac{1}{2} \Big( \frac{m-\theta}{s} \Big)^2 \Big).
\end{aligned} \end{equation}$$
I would like to know the marginal distribution of $X$. I have multiplied the above densities to form the joint density, but I cannot successfully integrate the result to get the marginal density of interest. My intuition tells me that this is a normal distribution with different parameters, but I can't prove it.
Best Answer
Your intuition is correct - the marginal distribution of a normal random variable with a normal mean is indeed normal. To see this, we first re-frame the joint distribution as a product of normal densities by completing the square:
$$\begin{equation} \begin{aligned} f(x,m) &= f(x|m) f(m) \\[10pt] &= \frac{1}{2\pi \sigma s} \cdot \exp \Big( -\frac{1}{2} \Big[ \Big( \frac{x-m}{\sigma} \Big)^2 + \Big( \frac{m-\theta}{s} \Big)^2 \Big] \Big) \\[10pt] &= \frac{1}{2\pi \sigma s} \cdot \exp \Big( -\frac{1}{2} \Big[ \Big( \frac{1}{\sigma^2}+\frac{1}{s^2} \Big) m^2 -2 \Big( \frac{x}{\sigma^2} + \frac{\theta}{s^2} \Big) m + \Big( \frac{x^2}{\sigma^2} + \frac{\theta^2}{s^2} \Big) \Big] \Big) \\[10pt] &= \frac{1}{2\pi \sigma s} \cdot \exp \Big( -\frac{1}{2 \sigma^2 s^2} \Big[ (s^2+\sigma^2) m^2 -2 (x s^2+ \theta \sigma^2) m + (x^2 s^2+ \theta^2 \sigma^2) \Big] \Big) \\[10pt] &= \frac{1}{2\pi \sigma s} \cdot \exp \Big( - \frac{s^2+\sigma^2}{2 \sigma^2 s^2} \Big[ m^2 -2 \cdot \frac{x s^2 + \theta \sigma^2}{s^2+\sigma^2} \cdot m + \frac{x^2 s^2 + \theta^2 \sigma^2}{s^2+\sigma^2} \Big] \Big) \\[10pt] &= \frac{1}{2\pi \sigma s} \cdot \exp \Big( - \frac{s^2+\sigma^2}{2 \sigma^2 s^2} \Big( m - \frac{x s^2 + \theta \sigma^2}{s^2+\sigma^2} \Big)^2 \Big) \\[6pt] &\quad \quad \quad \text{ } \times \exp \Big( \frac{(x s^2 + \theta \sigma^2)^2}{2 \sigma^2 s^2 (s^2+\sigma^2)} - \frac{x^2 s^2 + \theta^2 \sigma^2}{2 \sigma^2 s^2} \Big) \\[10pt] &= \frac{1}{2\pi \sigma s} \cdot \exp \Big( - \frac{s^2+\sigma^2}{2 \sigma^2 s^2} \Big( m - \frac{x s^2 + \theta \sigma^2}{s^2+\sigma^2} \Big)^2 \Big) \cdot \exp \Big( -\frac{1}{2} \frac{(x-\theta)^2}{s^2+\sigma^2} \Big) \\[10pt] &= \sqrt{\frac{s^2+\sigma^2}{2\pi \sigma^2 s^2}} \cdot \exp \Big( - \frac{s^2+\sigma^2}{2 \sigma^2 s^2} \Big( m - \frac{x s^2 + \theta \sigma^2}{s^2+\sigma^2} \Big)^2 \Big) \\[6pt] &\quad \times \sqrt{\frac{1}{2\pi (s^2+\sigma^2)}} \cdot \exp \Big( -\frac{1}{2} \frac{(x-\theta)^2}{s^2+\sigma^2} \Big) \\[10pt] &= \text{N} \Big( m \Big| \frac{xs^2+\theta\sigma^2}{s^2+\sigma^2}, \frac{s^2 \sigma^2}{s^2+\sigma^2} \Big) \cdot \text{N}(x|\theta, s^2+\sigma^2). \end{aligned} \end{equation}$$
We then integrate out $m$ to obtain the marginal density $f(x) = \text{N}(x|\theta, s^2+\sigma^2)$. From this exercise we see that $X \sim \text{N}(\theta, s^2+\sigma^2)$.