A tight upper bound for Gaussian integral.

Question

Consider two positive real number $\mu$ and $\sigma$. Let $m = 1, 2, \ldots$ be the natural number, I want to find a tight upper bound for the following part Gaussian integral: $$\int_0^{\infty} \exp \left[ - \frac{(x - \mu)^2}{2 \sigma^2} \right] x^m \, \mathrm{d} x.$$ A naive way for $m$ is even, we can use $$ \begin{aligned} \int_0^{\infty} \exp \left[ - \frac{(x - \mu)^2}{2 \sigma^2} \right] x^m \, \mathrm{d} x & \leq \int_{-{\infty}}^{\infty} \exp \left[ - \frac{(x - \mu)^2}{2 \sigma^2} \right] x^m \, \mathrm{d} x \\ & = \sqrt{2 \pi \sigma^2} \int_{-{\infty}}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2}}\exp \left[ - \frac{(x - \mu)^2}{2 \sigma^2} \right] x^m \, \mathrm{d} x \\ & = \sqrt{2 \pi \sigma^2}\mathrm{E} \left[ \mathcal{N}(\mu, \sigma^2) \right]^m \\ & = \sqrt{2 \pi \sigma^2} \sum_{j=0}^{\left\lfloor\frac{m}{2}\right\rfloor}\left(\begin{array}{c}m \\ 2 j\end{array}\right) \frac{\mu^{m-2 j} \sigma^{2 j}(2 j) !}{j ! 2^j}. \end{aligned} $$ However, this method will lose a lot of information containing in the integral interval $(0, \infty)$. Besides, we cannot apply this method when $m$ is odd. I wounder whether there exists an improved approach to deal with this integral? Thanks in advance!

Answer 1

If $$I_m=\int_0^{\infty} \exp \left[ - \frac{(x - \mu)^2}{2 \sigma^2} \right] x^m \, dx$$ $$\frac{I_m}{2^{\frac{m-1}{2}} \sigma ^m }=\sqrt{2} \mu \Gamma \left(\frac{m+2}{2}\right) \, _1F_1\left(\frac{1-m}{2};\frac{3}{2};-\frac{\mu ^2}{2 \sigma ^2}\right)+$$ $$\sigma \Gamma \left(\frac{m+1}{2}\right) \, _1F_1\left(-\frac{m}{2};\frac{1}{2};-\frac{\mu ^2}{2 \sigma ^2}\right)$$ where appears Kummer confluent hypergeometric function.

Let $\mu=\sqrt{2} \sigma t$

$$\frac{I_m}{2^{\frac{m-1}{2}} \sigma ^{m+1} }=2 \,t\, \Gamma \left(\frac{m}{2}+1\right) \, _1F_1\left(\frac{1-m}{2};\frac{3}{2};-t^2\right)+$$ $$\Gamma \left(\frac{m+1}{2}\right) \, _1F_1\left(-\frac{m}{2};\frac{1}{2};-t^2\right)$$

Not proved, it seems that the second term is slightly larger than the first term.

$$\frac{I_m}{2^{\frac{m-1}{2}} \sigma ^{m+1} }< 2\,\Gamma \left(\frac{m+1}{2}\right) \, _1F_1\left(-\frac{m}{2};\frac{1}{2};-t^2\right)$$

For example, using $\mu=3$, $\sigma=1$ and $m=5$, the exact value is $1398.699006$ while the bound is $1398.699435$.

At this point, I am stuck.

Answer 2

The thing is, there's a readily available formula for the moments of the truncated normal distribution, so any 'tight' bound will only be a corruption of the exact formula. For reference, let $J_m(\mu, \sigma^2)$ be your integral. Then \begin{align*} J_m(\mu, \sigma^2) &= \left(1 - \Phi\left(-\frac{\mu}{\sigma}\right)\right)\mathbb{E}[X^m|X>0] \\ &=\sum_{r=0}^{m}\binom{m}{r}\mu^{m-r}\sigma^rI_r \end{align*} where \begin{align*} I_r = \int_{-\mu/\sigma}^{\infty} t^r \phi(t) dt = \left(-\frac{\mu}{\sigma}\right)^{r-1}\phi\left(-\frac{\mu}{\sigma}\right) + (r-1)I_{r-2} \end{align*}