TITLE: Show that $f_{\alpha}(t)$ is a p.d.f.
QUESTION [6 upvotes]: Let $\displaystyle \phi(t)=\frac{1}{\sqrt{2\pi}}e^{-t^2/2}$,$t\in \Bbb R$ be the standard normal density function and $\displaystyle \Phi(x)=\int_{-\infty}^x\phi(t)\,dt$ be the standard normal distribution function. Let $f_{\alpha}(t)=2\phi(t)\Phi(\alpha t)$,$t\in \Bbb R$
where $\alpha \in \Bbb R$. Show that $f_{\alpha}$ is a probability density function.
we have $\Phi'(x)=\phi(x)$. We have to show that $\displaystyle\int_{-\infty}^{\infty}f_{\alpha}(t)\,dt=1$. I tried by integration by-parts but I got the value is $0$. Dose there any other process or where is my mistake.?
Edit : 
$\displaystyle \int_{-\infty}^{\infty} f_{\alpha}(t)dt= 2\int_{-\infty}^{\infty}\phi(t)\Phi(\alpha t)\,dt=2\left[\Phi(\alpha t)\int_{-\infty}^{\infty}\phi(t)\,dt\right]_{-\infty}^{\infty}-2\int_{-\infty}^{\infty}\left[\alpha\Phi'(\alpha t).\int_{-\infty}^{\infty}\phi(t)\,dt\right]\,dt=2[\Phi(\infty)-\Phi(-\infty)]-2\int_{-\infty}^{\infty}\alpha\phi(\alpha t)\,dt=\cdots=0$

REPLY [9 votes]: A probabilistic interpretation: consider $(X,Y)$ i.i.d. standard normal, then $\Phi(at)=P(X<at)$ and $\varphi$ is the PDF of $Y$ hence $$\int_{-\infty}^\infty 2\varphi(t)\Phi(at)dt=\int_{-\infty}^\infty 2\varphi(t)P(X<at)dt=2P(X<aY)=2P(Z<0)$$ where $Z=X-aY$ is centered normal with nonzero variance. Every centered normal distribution is symmetric hence $P(Z<0)=\frac12$. This shows that the integral on the LHS is $1$, as desired.