Consider the density of the Normal distribution given by

$$f(x; \mu, \sigma) = \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right)$$

It is obvious that, with a fixed $x$ and $\sigma$, the Normal density tends to $0$ as $\mu \to \infty$ or $\mu \to -\infty$. However, is there a way to characterize the rate of this convergence?

Similarly, what about the rate of convergence as $\mu \to -\infty$ or $\mu \to \infty$ for the distribution function

$$\int \limits_0^{\infty} \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right) \ d x,$$ where $\sigma > 0$?

It seems that knowing these rates of convergences would be helpful in answering the following question, which I would appreciate any insights on if people have them.

Does the product

$$\left[\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right)\right] \left[\dfrac{1}{\int \limits_0^{\infty} \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{1}{2}\left(\dfrac{x - \mu}{\sigma}\right)^2\right) \ d x}\right],$$ where $\sigma > 0$ is the same value in both factors and $x > 0$ in the first factor, tend to either $0$ or to $\infty$ as $\mu \to -\infty$? Intuitively, it seems that the answer would depend on the value of $\sigma$, but I'm not sure how to formally characterize this dependence.

Many thanks to all for any insights.


2 Answers 2


Let us introduce some notation: Let $Z \sim N(0,1)$ with density $\varphi(z) = f(z; 0, 1)$ and distribution function $\Phi$.

The denominator of the second term of the product you seek to study may be written as:

$$ \mathbb P[ Z/\sigma + \mu \geq 0] = \mathbb P[ Z \geq - \mu/\sigma] = 1 - \Phi(-\mu/\sigma).$$

Now by Mill's inequality (see e.g., this math.se post), it holds that:

$$ 1-\Phi(-\mu/\sigma) = \frac{\varphi(-\mu/\sigma)}{-\mu/\sigma}(1+o(1)) \text{ as } \mu/\sigma \to -\infty.$$

Hence the product you are interested in is equal to:

$$ \frac{\varphi((x-\mu)/\sigma)/\sigma}{1 - \Phi(-\mu/\sigma)} = -\mu\exp(-x^2/(2\sigma^2)) \exp(\mu x/\sigma)(1+o(1)) \text{ as } \mu/\sigma \to -\infty.$$

Since $x > 0$, we thus conclude that the ratio converges to $0$ as $\mu \to -\infty$.


With no loss of generality, choose units of measurement for which $\sigma=1.$

Notice that

$$\int_0^\infty \exp(-(x-\mu)^2/2)\,\mathrm{d}x = \int_{-\mu}^\infty \exp(-z^2/2)\,\mathrm{d}z.$$

Use this and the Fundamental Theorem of Calculus to compute the derivative of the denominator when applying L'Hopital's Rule to the quotient, which is a ratio of quantities both converging to $0:$

$$\begin{aligned} \lim_{\mu\to-\infty}\frac{\exp(-(x-\mu)^2/2)}{\int_0^\infty \exp(-(x-\mu)^2/2)\,\mathrm{d}x} &= \lim_{\mu\to-\infty}\frac{\frac{\mathrm{d}}{\mathrm{d}\mu}\exp(-(x-\mu)^2/2)}{\frac{\mathrm{d}}{\mathrm{d}\mu}\int_{-\mu}^\infty \exp(-z^2/2)\,\mathrm{d}z}\\ &= \lim_{\mu\to-\infty}\frac{-(x-\mu)\exp(-(x-\mu)^2/2)}{-\exp(-(-\mu)^2/2)}\\ &=\lim_{\mu\to-\infty}(x-\mu)\exp(-x^2/2)\exp(\mu x). \end{aligned}$$

This exhibits three asymptotic behaviors depending on the sign of $x:$

  1. When $x\lt 0,$ $\exp(\mu x)$ dominates and is has a positive argument, whence it diverges exponentially to $+\infty.$ See the left panel in the figure for an example.

  2. When $x=0,$ $\exp(\mu x)=1$ is constant, whence the limit (which is proportional to $x-\mu$) diverges linearly to $+\infty.$ See the middle panel for an example.

  3. When $x \gt 0,$ $\exp(\mu x)$ eventually has a negative argument that causes the limit to converge exponentially to $0.$ See the right panel for an example.


These are log-linear plots. Thus, exponential behavior appears linear. The question is concerned about what happens to the value as we move leftwards in each plot ($\mu\to-\infty$).

This is the R code to produce the plots. The first call to curve plots the log ratio. The second call overplots the log of the result obtained by L'Hopital's Rule (in dotted red). Although the black and red curves appear to coincide, that's only because the plots cover such a large range of $\mu:$ for $\mu \gt -4$ or so (and $x\in \{-1,0,1\}$ as shown here), the curves differ appreciably (by a few percent or more).

By computing the logarithms, the code is able to evaluate arguments $\mu$ and $x$ of considerable size (where computing the ratio itself will overflow or underflow the usual double precision arithmetic).

for (x in c(-1,0,1)) {
  curve(dnorm(x, mu, log = TRUE) - pnorm(0, mu, log = TRUE, lower.tail = FALSE),
        -100, 0, xname="mu", lwd=2,
        xlab=expression(mu), ylab="Log ratio", main = bquote(x==.(x)))
  curve(log(x-mu) + mu * x - x^2 / 2, xname="mu", add=TRUE, col="Red", lty=3)

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