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Let $CDF$ be the cumulative distribution function for the standard normal distribution. Let $Z$ be a standard normal random variable.

Then $CDF(Z)$ is uniformly distributed on the unit interval, so by integration we can show that $E(ln(CDF(Z))) = -1$.

My question: there an easy way to compute $E[\ln(CDF(Z + c))]$ for constant $c$?

What I'm really looking for is $E[\ln(CDF(Z + c))]$ on the $(-\infty, 0]$ interval and the $(0, \infty)$ interval.

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    $\begingroup$ Perhaps you can tell us in just a tad more detail what integration you are doing to come up with $\ln(\operatorname{cdf}(Z)) = -1$? $\operatorname{cdf}(Z)$ is a random variable taking on values in $(0,1)$, and its natural logarithm $\ln$ is also a random variable with values in $(-\infty,0)$. So, what does it mean when you say that $\ln(\operatorname{cdf}(Z)) = -1$?? $\endgroup$ Feb 26, 2013 at 21:14
  • $\begingroup$ Dilip, sorry for the conclusion, I meant E(ln(CDF(Z)))=−1. Or equivalently, if U is a uniformly distributed RV on the unit interval, E(ln(U)) = -1. $\endgroup$
    – garyrob
    Feb 26, 2013 at 21:19
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    $\begingroup$ E(ln(CDF(Z))) is equivalent to E(ln(U)) where U is uniformly distributed on (0, 1]. That can be found by integrating ln(), which is x(ln(x))-x + C, which is -1 if C is 0 x is 1. $\endgroup$
    – garyrob
    Feb 26, 2013 at 21:31

2 Answers 2

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No,you'll have to do the integration numerically for each $c$ value.

Let $\Phi(x)$ be the Gaussian CDF function.

You want to evaluate $$ I=\int \Phi'(x-c) \ln \Phi(x) dx $$

Idea 1: this looks like a convolution; the Fourier transform of $\Phi'$ is easy enough, but that of $\ln \Phi$ is not; and even if you could take the transform of the second factor, inverting the resulting expression is unlikely to be feasible.

Idea 2: do it by parts, this works out to $$ \Phi(x-c) \ln \Phi(x) \vert^{\infty}_{\infty} - \int_{\infty}^{\infty} \frac{ \Phi(x-c)}{\Phi(x)} \Phi'(x) dx $$ the first term is conveneiently zero, but the second is no better off.

Idea 3: expand the $\ln \Phi$ term. Define $U=1-\Phi$ so that : $$ I=-\int U'(x-c) \ln [ 1-U(x)] dx $$ and then expand the logarithm as though $U$ were small. You end up with terms like $U'(x-c) U^n(x)$, which are still not easily integrable due to the shift in the argument.

At this point, it seems that the simplifications that arise in $E[ \ln \Phi(x)]$ aren't panning out, so, numerical integration is a feasible solution.

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  • $\begingroup$ plugging the relevant integrand into Wolfram Alpha yielded no analytic results. $\endgroup$
    – Dave
    Feb 27, 2013 at 0:23
  • $\begingroup$ Wolfram Alpha is only a crude test--there is plenty it cannot do. Your exposition here is much more interesting and convincing than that! (+1) $\endgroup$
    – whuber
    Feb 27, 2013 at 14:12
  • $\begingroup$ Thank you for your thoughts. I had never tried numeric integration before, but I found scipy.integrate (docs.scipy.org/doc/scipy/reference/tutorial/integrate.html) and it did the job. And now I have another great took in my arsenal. $\endgroup$
    – garyrob
    Feb 27, 2013 at 17:44
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In short there is no "easy" way to do this, and Dave has tried a few sensible approaches. One approach perhaps worth a try is to taylor expand since your function is smooth and analytic, and we have a simple form for the moments of the standard normal:

Notice that

$L^* = \ln(\Phi(X))$

$L = \frac{\phi(x)}{\Phi(x)}$

$L' = \frac{\phi(x)^2}{\Phi(x)^2} - x \frac{\phi(x)}{\Phi(x)} = L(x)^2 - xL(x)$

$L'' = 2LL'-L-xL' = 2L^3 - 3xL^2 +(x^2-1) L$

$L''' = 6L^2L' - 6xLL'-3L^2+2xL+(x^2-1)L' = 6L^4-6xL^3-6xL^3+6x^2L^2-3L^2 + 2xL+(x^2-1)(L^2-xL)$

$ =6L^4-12xL^3+(7x^2-4)L^2 + (3x-x^3)L$

I had hoped we could spot a pattern there incolving Hermite polynomials or some-such, but I can't see one... however, because of the recurrence it is basic calculus and algebra that a computer could chunk through to arbitrary depth.

Armed with these we can now notice that we can define $X = Z + c$ so that $X \sim N(c,1)$ and then Taylor expand $L^*$ around the point c

$E[\ln(\Phi(X))] = E[\ln(\Phi(c)) + L(c) Z + L'(c)\frac{Z^2}{2} + L''(c)\frac{Z^3}{3!}+\cdots]$

Now remember that $E[Z^p] = (p-1)!!$ for even $p$ and $0$ otherwise, so we only keep every other term:

$E[\ln(\Phi(X))] = \ln(\Phi(c)) + L'(c)\frac{1}{2} + L'''(c)\frac{3!!}{4!}+ L^{(5)}(c)\frac{5!!}{6!} \cdots$

$E[\ln(\Phi(X))] = \ln(\Phi(c)) + L'(c)\frac{1}{2} + L'''(c)\frac{1}{8}+ L^{(5)}(c)\frac{1}{48} + L^{(7)}(c) \frac{1}{384} + \cdots + L^{(2k-1)}(c) \frac{1}{2^k k!} + \cdots$

(where we use that $(2k-1)!! = \frac{(2k)!}{2^k k!}$ )

So all that remains is to plug in the derivatives above and work out the total. If anyone spots a pattern then this should work exactly.


Incidentally, if you hadn't asked about $\ln(\Phi(x))$ but simply $\Phi(x)$ then we have a much simpler expression.

Carrying on from the expression above but using $\Phi$ in place of $L^*$ we have:

$E[\Phi(X)]=\Phi(c) + \phi(c) \sum_{k=1}^\infty \frac{H_{2k-1}(c)}{2^k k!}$

Where $H_n$ is the nth Hermite polynomial, and we have used the fact that $\frac{d^n \phi(x)}{dx^n}=H_n(x)\phi(x)$

Notice that inside the sum we have a polynomial over an exponential, so this is going to converge nicely.

Notice also that since $H_{2k-1}(0) = 0$ for all k then in the case where $c=0$ this colapses back to 0.5 as expected.

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