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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 :

$\displaystyle \int_{-\infty}^{\infty} f_{\alpha}(t)dt= 2\int_{-\infty}^{\infty}\phi(t)\Phi(\alpha t)\,dt=2\left[\Phi(\alpha t)\int_{-\infty}^{\infty}\phi(t)\,dt\right]_{-\infty}^{\infty}-2\int_{-\infty}^{\infty}\left[\alpha\Phi'(\alpha t).\int_{-\infty}^{\infty}\phi(t)\,dt\right]\,dt=2[\Phi(\infty)-\Phi(-\infty)]-2\int_{-\infty}^{\infty}\alpha\phi(\alpha t)\,dt=\cdots=0$

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    Can you show your steps? Like, the integration by parts. – 2017-02-08
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    @Anna See edit please... – 2017-02-08
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    See [Integration by polar coordinates](https://en.wikipedia.org/wiki/Gaussian_integral). This is the method I have seen most often in probability books. – 2017-02-08
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    The integration by parts formula in your Edit simply does not make sense, please review this topic seriously. – 2017-02-08

2 Answers 2

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A probabilistic interpretation: consider $(X,Y)$ i.i.d. standard normal, then $\Phi(at)=P(X

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    Why is $\int_{-\infty}^\infty \varphi(t)P(X – 2017-02-08
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    Because $\varphi$ is the PDF of $Y$ and because $(X,Y)$ is independent hence $$\int_{-\infty}^\infty\varphi(t)P(X – 2017-02-08
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    Thank you, it can be so easy if you know what to do... And in the last equality you used independence. – 2017-02-08
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$ \begin{align*} \frac{d}{d\alpha}\int_{-\infty}^{\infty}f_{\alpha}(t)\,dt&=\int_{-\infty}^{\infty}\frac{d}{d\alpha}f_{\alpha}(t)\,dt \\ &=\int_{-\infty}^{\infty}\frac{d}{d\alpha}\left(2\phi(t)\int_{-\infty}^{\alpha t}\phi(s)\,ds\right)dt \\ &=\int_{-\infty}^{\infty}2t\phi(t)\phi(\alpha t)\,dt \\ &=0 \end{align*} $

since the integrand is odd. Therefore $\int_{-\infty}^{\infty}f_{\alpha}(t)\,dt$ is equal to its initial value at $\alpha=0$ which is $1$.

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    At $\alpha=0$ , the value of the integral becomes $\sqrt{2/\pi}$. How it is $1$ ? – 2017-02-08
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    @S717717 $f_0(t)=2\phi(t)\phi(0)=2\phi(t)(1/2)=\phi(t)$ which integrates to $1$. – 2017-02-09
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    How $\phi(0)=1/2$ ? – 2017-02-09
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    @S717717 My mistake, that should have been $\Phi(0)$ so it's $f_0(t)=2\phi(t)\Phi(0)$. – 2017-02-09