# Does the following Integral have a Closed Form Solution?

Is there a solution to the following Integral that can get rid of the integral, but can leave the answer in terms of $\Phi()$ (the standard normal CDF) and $\phi$ (the standard normal PDF).?

$\int_{-\infty}^{\infty}\Phi(x)\phi(\frac{x-a}{b})dx$

where $a, b$ are constants with $b>0$,

I've tried several variants of integration by parts, all of which get me nowhere. I'm thinking the likely answer is no.

• Do you want the result in terms of $\Phi$, which does not exist in closed form? – Alex Apr 29 '13 at 20:13
• Yes, that's fine. I meant closed form as in get rid of the integral; i.e. $\int \Phi(x)dx = x\Phi(x)+\phi(x)$ – Greg Apr 29 '13 at 21:42
• It seems the answer should be $\frac{1}{2b}$ as the mean doesn't matter as $x \to \infty$ – Alex Apr 29 '13 at 22:17
• I'm not sure I follow. That also doesn't seem to coincide with some numerical integration examples I ran. – Greg Apr 30 '13 at 19:00

The integral in question can be given an probabilistic interpretation: $$\begin{split} \int_\mathbb{R} \Phi(x) \phi\left(\frac{x-a}{b}\right) \mathrm{d}x &= b \int_\mathbb{R} \Pr(Z \leqslant x) \phi\left(\frac{x-a}{b}\right) \frac{\mathrm{d}x}{b} \\ &= b \int_\mathbb{R} \Pr(Z \leqslant a+b y) \phi\left(y\right) \mathrm{d}y \\ &= b \Pr\left(Z \leqslant a + b Y \right) = b \Pr\left(Z - b Y \leqslant a \right) \end{split}$$ Since $Z$ and $Y$ are both independent standard normal random variables, $Z-b Y$ is also a normal random variables, with mean and variance: $$\mathbb{E}(Z-b Y) = 0, \quad \mathbb{Var}(Z-b Y) = \mathbb{Var}(Z) + b^2 \mathbb{Var}(Y) = 1+b^2$$ Thus $Z- b Y \stackrel{d}{=} X$, where $X \sim \mathcal{N}\left(0, \sqrt{1+b^2} \right)$ we have: $$\int_\mathbb{R} \Phi(x) \phi\left(\frac{x-a}{b}\right) \mathrm{d}x = b \Pr\left(Z - b Y \leqslant a \right) = b \Pr\left(X \leqslant a \right) = b \, \Phi\left(\frac{a}{\sqrt{1+b^2}}\right) \tag{1}$$ Numerical check of $(1)$:

In[25]:= With[{a = -3.,
b = 2.3}, {NIntegrate[
CDF[NormalDistribution[], x] PDF[
NormalDistribution[], (x - a)/b], {x, -Infinity, Infinity}],
b CDF[NormalDistribution[], a/Sqrt[1 + b^2]]}]

Out[25]= {0.266371, 0.266371}

• +1, obviously the way to go. (I seem to remember having explained this on the site relatively recently.) – Did May 1 '13 at 19:51
• Thanks very much for this. I must admit I'm having a little hard time following, especially where you have $b \int Pr(Z \leq a+by) \phi(y) dy=bPr(Z \leq a+bY)$. Could you elaborate on that step a bit? – Greg May 3 '13 at 15:54
• @Greg It makes use of the definition for the expectation $\int f(y) \phi(y) \mathrm{d}y = \mathbb{E}(f(Y))$ and the law of total expectation: $b \int \Pr(Z \leqslant a+b y) \phi(y) \mathrm{d}y = \mathbb{E}\left( \Pr(Z \leqslant a+b Y \mid Y) \right) = \Pr( Z \leqslant a+b Y )$. – Sasha May 3 '13 at 20:37