Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $N(x)$ denote the cdf of standard normal and $n(x)$ denote the pdf of standard normal.

How to evaluate the integral $\int\limits_{-\infty}^\infty N(a+x) n(x) \mathrm{d} x$ ?

Thanks a lot!

share|cite|improve this question

1 Answer 1

Use $\partial_a N(a+x) = n(a+x)$ and $n(-x) = n(x)$: $$ \partial_a \int_{-\infty}^\infty N(a+x) n(x) \mathrm{d} x = \int_{-\infty}^\infty n(a+x) n(x) \mathrm{d} x = \int_{-\infty}^\infty n(-a-x) n(x) \mathrm{d} x = n\left(-\frac{a}{\sqrt{2}} \right)\frac{1}{\sqrt{2}} $$ The last equality reflects that the the sum $Z = X+Y$ of two standard normal random variates $X$ and $Y$ is another random variable with zero mean and variance $2$. The pdf of $Z$ is a convolution of pdfs of $X$ and $Y$: $$ f_Z(z) = \int_{-\infty}^\infty f_X(z-y) f_Y(y) \mathrm{d} y $$ Thus: $$ \int_{-\infty}^\infty N(a+x) n(x) \mathrm{d} x = \int_{-\infty}^a n\left(-\frac{b}{\sqrt{2}} \right) \frac{1}{\sqrt{2}} \mathrm{d} b = 1 - N\left(-\frac{a}{\sqrt{2}} \right) = N\left(\frac{a}{\sqrt{2}} \right) $$


In[15]:= With[{a = 2/3.},
   CDF[NormalDistribution[], a + x] PDF[NormalDistribution[], 
     x], {x, -Infinity, Infinity}], 
  CDF[NormalDistribution[], a/Sqrt[2]]}]

Out[15]= {0.681324, 0.681324}
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.