Integrating the step size function under a Gaussian

Question

I am interesting in the following integral:

$\int \Theta(f_i -f_c)\mathcal{N}(f_c|m_c,\sigma^2_c)df_c$

where $\Theta()$ represent the heavised function https://en.wikipedia.org/wiki/Heaviside_step_function. I have seen in many works that the result of this integral is given by:

$\phi(\frac{f_i-m_c}{\sigma_c})$

where $\phi()$ is the cumulative density function of a Gaussian.

What I have seen so far is that smooth approximations to the step-side function can be shown to be CDF functions in the limit. For instance:

$\theta(x)=\underset{k\rightarrow\infty}{\lim}\frac{1}{2}[1+\text{erf}[kx]]=\phi(x)$

where $\text{erf}$ is the error function.

However I am not able to use this to solve the above integral. Can someone help me?. One more point, is this result also valid for a multivariate Gaussian?

Thank you.

Answer 1

As Calvin Khor suggested, your integral is just a fancy way to write this: $$ \int_{-\infty}^{f_i}\mathcal{N}_{m_c,\sigma_c^2}(f_c)df_c=\int_{-\infty}^{\frac{f_i-m_c}{\sigma_c}}\mathcal{N}_{0,1}(u)du=\phi\left(\frac{f_i-m_c}{\sigma_c}\right) $$

If the first transition is still problematic, write the $\mathcal{N}_{m_c,\sigma_c^2}(f_c)$ as $\frac{1}{\sigma_c\sqrt{2\pi}}\exp\left(-\frac{(f_c-m_c)^2}{\sigma_c^2}\right)$ and substitute $u=(f_c-m_c)/\sigma_c$.

Answer 2

Okei, I want to add where my confusion came from, which is posted as an answer to Vasily.

I was considering that the value that changes is $f_i$, but it is $f_c$. We should consider that for values of $f_c$ above $f_i$ the integral is evaluated to 0, thus the part we are integrating from the Gaussian, is the left part, starting from $-\infty$ until $f_i$, which is the starting value from where the integral evaluates to 0.