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Suppose that $\varphi$ is the density of the standard Gaussian distribution given by $$ \varphi(x)=\frac1{\sqrt{2\pi}}\exp(-x^2/2) $$ for each $x\in\mathbb R$. The derivatives of $\varphi$ are given by the formula $$ \varphi^{(n)}(x)=(-1)^n\mathit{He}_n(x)\varphi(x) $$ for each $x\in\mathbb R$ and $n\ge0$, where $\mathit{He}_n$ is the $n$-th probabilists' Hermite polynomial.

It is straightforward to see that $\varphi$ and its first two derivatives are bounded. We have that \begin{align*} |\phi(x)|&\le (2\pi)^{-1/2}\approx0.399\\ |\varphi'(x)|&\le\varphi(1)\approx0.242\\ |\varphi''(x)|&\le (2\pi)^{-1/2}\approx0.399 \end{align*} for each $x\in\mathbb R$.

Does there exist $c>0$ such that $|\varphi^{(n)}(x)|\le c$ for each $x\in\mathbb R$ and $n\ge0$? If not, are all derivatives of $\varphi$ bounded, i.e. does there exist a positive sequence $c_n$ such that $|\varphi^{(n)}(x)|\le c_n$ for each $x\in\mathbb R$ and $n\ge0$?

Any help is much appreciated!

Cm7F7Bb
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    Note that the Hermite-polynomials form an orthogonal basis w.r.t. the weight $\exp(-x^2/2)$. This means their square is integrable, which (assuming no strange oscillations) will tell you something about its behaviour towards infinity. – mlk May 19 '17 at 15:04

1 Answers1

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Each individual $\varphi^{(n)}(x)$ is bounded, simply because it is a continuous function tending to zero at infinity.

However, we can't find a $c$ that uniformly bounds the entire sequence. To see this, just write down the Taylor series for $\varphi$: $$ \varphi (x) = \exp(-x^2)= 1 - x^2 + \frac 1 {2!} x^4 - \frac 1 {3!} x^6 + \dots$$ [I dropped some constant factors to keep the notation simple.]

Differentiating, and evaluating at $x = 0$ for even $n$ (say $n = 2m$), we see that $$ |\varphi^{(2m)}( 0)| = \frac{(2m)!}{m!},$$ which tends to infinity as $m \to \infty$.

I plotted up to four derivatives of the function on Mathematica. You can see that the peaks of these functions grow pretty fast!

enter image description here

Kenny Wong
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