How to demonstrate that integral of N $\int$ N is equal to integral of N($\sigma$) incremented?

51 Views Asked by At

Recently, I have drawn the 2 following functions

$$ f(x) = \int_{t=-\infty}^{+\infty} \left(N(t\mid x,\frac{\sigma^2}{n}).\int_{y=Limit}^{+\infty} N(y\mid t,\sigma^2) \, dy \,\right)\, dt $$

and $$ g(x) = \int_{y=Limit}^{+\infty} N(y\mid x,\sigma^2(\frac{n+1}{n})) \, dy $$

where

$$ N(x\mid\mu,\sigma^2) = \frac{e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}}{\sigma \sqrt{2\pi}} $$ is the density function of a normal distribution.

On a graph, when $\mu$, $\sigma$, $n$ and $Limit$ are known, these 2 functions are full equal for all $x$ between $-\infty$ and $+\infty$

Question: how can I explain that these 2 functions are equivalent using pure mathematical expression ?

Just to help others, this formula is equivalent (but not equal) to integral described in How to compute normal integrals $\int_{-\infty}^\infty\Phi(x)N(x\mid\mu,\sigma^2)\,dx$ and $\int_{-\infty}^\infty\Phi(x)N(x\mid\mu,\sigma^2)x\,dx$