Showing the pdfs of two normal distributions multiplied together is still the pdf of a normal distribution Given $p_1(x) = \cfrac{1}{\sqrt{2\pi}\sigma_1}e^{-x^2/2\sigma_1^2}$ and $p_2(x) = \cfrac{1}{\sqrt{2\pi}\sigma_2}e^{-x^2/2\sigma_2^2}$
I want to show that $p_1(x)p_2(x)$ is normal.
My thought
$$
\begin{align}
 p_1(x)p_2(x)
&= \cfrac{1}{\sqrt{2\pi}\sigma_1}e^{-x^2/2\sigma_1^2} \cdot
\cfrac{1}{\sqrt{2\pi}\sigma_2}e^{-x^2/2\sigma_2^2} \\
&=  \cfrac{1}{2\pi\sigma_1\sigma_2}
     e^{-x^2/2\sigma_1^2}\cdot e^{-x^2/2\sigma_2^2} \\
&= \cfrac{1}{2\pi\sigma_1\sigma_2}
     e^{-(x^2\sigma_1^2+x^2\sigma_2^2)/(2\sigma_1^2\sigma_2^2)} \\
&= \cfrac{1}{2\pi\sigma_1\sigma_2}
     e^{-\frac{x^2}{2}\cdot(\sigma_1^2+\sigma_2^2)/(\sigma_1^2\sigma_2^2)} \\
\end{align}
$$
I'm given that the variance of these two functions mutiplied together is $\sigma_1\sigma_2/(\sigma_1^2\sigma_2^2) = \sigma^2$ and has mean 0.
So
$$
\begin{align}
\cfrac{1}{2\pi\sigma_1\sigma_2}
     e^{-\frac{x^2}{2}\cdot(\sigma_1^2+\sigma_2^2)/(\sigma_1^2\sigma_2^2)} 
=
\cfrac{1}{2\pi\sigma_1\sigma_2}
     e^{-x^2/2\sigma^2}
\end{align}
$$
But I'm not sure how to show
$$
\cfrac{1}{2\pi\sigma_1\sigma_2}
=
\cfrac{1}{\sqrt{2\pi}\sigma}
$$

$$
\frac{x^2}{\sigma_1^2} + \frac{x^2}{\sigma_2^2} = \frac{x^2}{\left( \dfrac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \right)} = \frac{x^2}{\text{a positive number}}
$$
It is not generally true that
$$ \require{cancel}
\xcancel{\sigma_1 \sigma_2 = \frac{\sigma_1\sigma_2}{\sqrt{\sigma_2^2 + \sigma_2^2}},}
$$
as may be seen by looking at any case in which $\sigma_2^2 + \sigma_2^2 \ne 1.$
What you get is a Gaussian function, but not a probability density function.
