# Product Theorem for Gaussian Functions

By | March 27, 2016

It is a well known fact in mathematics that the product of two Gaussian functions is also a Gaussian function. In this post I want to present this result for future reference for me and for anyone who might find it useful.

The product theorem for Gaussian functions states that the product of two overlapping Gaussian functions is also a Gaussian function and determines the center and width of the resulting function in terms of the parameters of the two original functions.

To illustrate the identity, consider a Gaussian function of the form $G_i(x)=A_i\exp[-(x-\mu_i)^2/2\sigma_i^2]$, where $\mu_i$, $\sigma_i$ and $A_i$, correspond to its center, width and amplitude, respectively. The result of the product of two Gaussian functions $G_1(x)$ and $G_2(x)$ with different amplitudes, widths and central positions, $G_3(x) = G_1(x)G_2(x)$, is equal to

where the centroid is given by

and the width by

The theorem shows that the result of the product of two Gaussian functions is a new Gaussian function of width $\tilde{\sigma}$, centered in the position $\tilde{\mu}$, whose amplitude strongly depends on the factor $\exp\left(-(\mu_1-\mu_2)^2/2(\sigma_1^2+\sigma_2^2)\right)$. In addition, the resulting function is narrower than either of the two original Gaussian functions, and its center lies within the interval $(\mu_1,\mu_2)$. Notice that the above result can be further simplified if we consider the scenario where the Gaussian functions are centered in different positions but have the same width. If we define $\sigma_1 = \sigma_2 = \sigma$, the centroid simplies to $\tilde{\mu}= (\mu_1+\mu_2)/2$, whereas the width is given by $\tilde{\sigma}^2 = \sigma^2/2$.

# Examples

The next figures show two representative cases where two different Gaussian functions with different and equal widths are multiplied. In both cases the result is a Gaussian function narrower than the original functions centered in between the two original centroids, $\mu_1$ and $\mu_2$.