# Bounding higher moments of truncated normal

I'm looking for a convenient upper bound on the integral

\begin{equation*} \int_y^\infty x^k \exp(-(x-\mu)^2/2) dx \end{equation*}

for (possibly large) positive integer $k.$ This is equivalent to finding higher moments of a truncated normal distribution. A bound that works for non-integer $k$ as well would be even better.

Of course "convenient" is in the eye of the beholder, but I'd like some sort of fairly simple expression that I can use in further calculations. For example, an upper bound of the form $f(x) \exp( -g(x))$ where where $f$ and $g$ are low-degree polynomials would be great. I'm more interested in simplicity of form than in obtaining the tightest possible bound.

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I assume $y \gt 0$ and $y \gg \mu$. If you replace $x^k$ by $y^k \exp {\left( (k/y)x - k \right) }$ you will overestimate the $x^k$ term (because this is the exponential of the first two terms of the MacLaurin series of $\log(x^k)$ expanded around $x=y$, the series is alternating, and the remainder term is negative). Completing the square yields a closed-form formula for an upper bound, one of whose factors is a Gaussian integral:

$$\sqrt{2\pi }y^k \exp \left( {\frac{k (k+2 y (-y+\mu ))}{2 y^2}} \right) \Phi \left(\frac{k}{y}-y+\mu \right)$$

This will work extremely well when $k$ is large compared to $y$ and $\mu$ because then most of the mass of the integral is concentrated at its lower limit where the exponential upper bound to $x^k$ is a good approximation. To avoid exponential overflow, use logarithms to compute the product.

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Thanks! That's clever and useful. – John D. Cook Oct 4 '10 at 11:22

Let $I_k=\int_y^\infty x^k\mathrm{exp}(-(x-\mu)^2/2)dx$ and $a_k=y^{k}\mathrm{exp}(-(y-\mu)^2/2)$.

Integration by parts gives $I_k=a_{k-1}+(k-1)I_{k-2}$.

Therefore, for even $n$, \begin{multline} I_n=a_{n-1}+(n-1)a_{n-3}+(n-1)(n-3)a_{n-5}+\cdots\newline+(n-1)(n-3)\cdots 1 a_1+(n-1)(n-3)\cdots 1 I_0. \end{multline}

Feel free to expand this answer and deduce an estimate from the above.

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Thanks. I've updated the question to be more specific regarding what I'm looking for. – John D. Cook Sep 30 '10 at 14:50
What happened to $\mu$? $I_n$ clearly must include a polynomial in $\mu$ times the exponential, so the recursion coefficients have to depend on $\mu$ as well as on $n$. – whuber Oct 1 '10 at 18:18
@whuber: You are right. What I wrote only holds for $\mu=0$. – Rasmus Oct 2 '10 at 15:43