Integrate CDF, PDF, part 2

We want:

$$ I = \int_{-\infty}^\infty \Phi(\lambda x) \frac{1}{\sigma} \phi\left( \frac{x - \mu}{\sigma} \right) \, dx $$

differentiate wrt $\mu$:

$$ \partial_\mu I = \int_{-\infty}^\infty \Phi(\lambda x) \frac{1}{\sigma} \left( \frac{-1}{\sigma} \right) (-\mu) \phi \left( \frac{x - \mu} {\sigma} \right) \, dx $$

Dont think that will work. Let's try differentiating wrt $\lambda$.

$$ \partial_\lambda I = \int_{-\infty}^\infty \lambda \phi(\lambda x) \frac{1}{\sigma} \phi \left( \frac{x - \mu} {\sigma} \right) \, dx $$

Let's define:

$$ E_1 = \frac{\lambda}{\sigma} \phi(\lambda x)\phi\left( \frac{x-\mu}{\sigma} \right) \, dx $$

Then:

$$ E_1 = \frac{\lambda}{\sigma} \frac{1}{2\pi} \exp \left( -\frac{1}{2} \left( \lambda^2 x^2 + \frac{(x - \mu)^2}{\sigma^2} \right) \right) $$

$$ = \frac{\lambda}{2\pi\sigma} \exp \left( -\frac{1}{2} \left( \frac{\lambda^2\sigma^2x^2 + x^2 - 2\mu x + \mu^2} {\sigma^2} \right) \right) $$

$$ =\frac{\lambda}{2\pi\sigma} \exp \left( - \frac{1}{2\sigma^2}\left( \left( \sqrt{(\lambda^2\sigma^2 + 1)}x - \frac{\mu} {\sqrt{(\lambda^2\sigma^2 + 1)}} \right)^2 - \frac{\mu^2}{\lambda^2\sigma^2 + 1} + \mu^2 \right) \right) $$

Looking at this term:

$$ - \frac{\mu^2}{\lambda^2\sigma^2 + 1} + \mu^2 $$

$$ = \frac{- \mu^2 + \mu^2\lambda^2\sigma^2 + \mu^2} {\lambda^2\sigma^2 + 1} $$

$$ = \frac{\mu^2\lambda^2\sigma^2} {\lambda^2\sigma^2 + 1} $$

Therefore $$ \partial_\lambda I = \frac{\lambda}{2\pi\sigma} \exp \left(

• \frac{\mu^2}{2\sigma^2} \left( \frac{\lambda^2\sigma^2}

   {\lambda^2\sigma^2 + 1}
  

  \right) \right) \int_{-\infty}^\infty \exp \left(
  • \frac{1}{2\sigma^2} \left( x\sqrt{\lambda^2\sigma^2+1}
    • \frac{\mu}{\sqrt{\lambda^2\sigma^2 + 1}} \right)^2 \right) \, dx $$

Substitute:

$$ y = \frac{1}{\sigma}\left( x\sqrt{\lambda^2\sigma^2 + 1} - \frac{\mu}{\sqrt{\lambda^2\sigma^2 + 1}} \right) $$

Therefore:

$$ dy = \frac{\sqrt{\lambda^2\sigma^2 + 1}} {\sigma} \, dx $$

Do substitution on the bounds, we have:

$$ x_1 = -\infty, x_2 = \infty $$

$\sigma$ is positive, and $\sqrt{\lambda^2\sigma^2 + 1}$ is positive. Therefore:

$$ y_1 = -\infty, y_2 = \infty $$

Therefore:

$$ \partial_\lambda I = \frac{\lambda}{\sigma\sqrt{2\pi}} \exp \left( -\frac{\mu^2\lambda^2} {2(\lambda^2\sigma^2+1)} \right) (1) \frac{\sqrt{\lambda^2\sigma^2+1}} {\sigma} $$

$$ = \frac{\lambda \sqrt{\lambda^2\sigma^2 + 1}} {\sigma^2 } \phi\left( \frac{\mu\lambda} {\sqrt{\lambda^2\sigma^2 + 1}} \right) $$

Therefore:

$$ I = \frac{\lambda\sqrt{\lambda^2\sigma^2 + 1}} {\sigma^2 } \frac{\sqrt{\lambda^2\sigma^2 + 1}} {\mu\lambda} \Phi \left( \frac{\mu\lambda} {\sqrt{\lambda^2\sigma^2 + 1}} \right) $$

not working....

Try:

Substitute:

$$ z = \frac{\mu}{\sqrt{\sigma^2 + \lambda^{-2}}} = \mu(\sigma^2 + \lambda^{-2})^{-1/2} $$

Therefore:

$$ dz/d\lambda = (-1/2)\mu(\sigma^2 + \lambda^{-2})^{-3/2}(-2\lambda^{-3}) $$

$$ = \frac{\mu} {(\lambda\sqrt{\sigma^2 + 1/\lambda^2})^3} $$

$$ = \frac{\mu} {(\sqrt{\sigma^2\lambda^2 + 1})^3} $$

Not working either...

How about, try substitution in the original integral first?

We have:

$$ I_1 = \int_{-\infty}^\infty \Phi(\lambda x) \frac{1}{\sigma} \phi\left( \frac{x - \mu}{\sigma} \right) \, dx $$

We know we can solve an integral in the form:

$$ I_2 = \int_{-\infty}^\infty \Phi\left( \frac{x - \mu}{\sigma} \right) \phi(x) \, dx $$

Let's try a variable substitution:

$$ y = \frac{x - \mu}{\sigma} $$

Therefore:

$$ dy/dx = 1/\sigma $$

And:

$$ dx = \sigma\,dy $$

For the limits, we have:

$$ x_1 = -\infty, x_2 = +\infty $$

Changing the variable, these become, bearing in mind \mu > 0, and $\sigma > 0$:

$$ y_1 = -\infty, y_2 = +\infty $$

Finally, $x$ in terms of $y$ is:

$$ x = y\sigma + \mu $$

Therefore we have:

$$ I_1 = \int_{-\infty}^\infty \Phi \left( \lambda\sigma y + \lambda\mu \right) \phi(y) \, dy $$

$$ = \int_{-\infty}^\infty \Phi \left( \frac{y + \frac{\mu}{\sigma}} {\frac{1}{\lambda\sigma}} \right) \phi(y) \, dy $$

... which is in the same form as $I_1$. Yay :)

And we have that the solution for I_2 is:

$$ I_2 = \Phi\left( \frac{\mu} {\sqrt{1 + \sigma^2}} \right) $$

But note that this was for negative $\mu$, whereas here we are constraining $\mu > 0$. The solution for $\mu > 0$ then will be:

$$ I_2[\mu > 0] = \Phi\left( \frac{-\mu} {\sqrt{1 + \sigma^2}} \right) $$

So, this suggests strongly that the solution for $I_1$ is:

$$ I_1 = \Phi \left( \frac{-\left(-\frac{\mu}{\sigma}\right)} {\sqrt{1 + \frac{1}{\lambda^2\sigma^2}}} \right) $$

$$ = \Phi \left( \frac{\mu} {\sqrt{\sigma^2 + \lambda^{-2}}} \right) $$

Yay! :)