Using L’Hôpital’s Rule to Evaluate a Normal Tail Limit

Summary

The document works through a limit involving the upper tail probability and density of a normal distribution, arising in a shortfall-to-quantile ratio. The key step is to evaluate a related tail expression as the threshold tends to infinity. The response applies L’Hôpital’s rule to a ratio whose numerator and denominator both approach zero, differentiating the normal tail and density terms.

Using the identities that the derivative of the cumulative normal distribution is its density and that the density derivative is minus the threshold times the density, the resulting expression tends to zero. This establishes the limit needed for the original simplification. The explanation is a narrow asymptotic calculation; it does not discuss the broader risk measure or conditions beyond the displayed normal-distribution setup.

Key ideas

• The upper tail probability of a standard normal distribution is one minus its cumulative distribution function.
• The derivative of the normal density is the negative threshold multiplied by the density.
• L’Hôpital’s rule can evaluate the displayed indeterminate tail ratio.
• The resulting ratio tends to zero as the threshold grows without bound.
• The argument addresses a specific normal-distribution limit rather than broader risk-measure properties.

Tags

# Showing that the shortfall-to-quantile ratio of a normal distribution goes to one


# Showing that the shortfall-to-quantile ratio of a normal distribution goes to one












I dont get why $$\lim_{x \to \infty} \frac{\mu \{1 - \Phi(x)\} + \sigma \phi(x)}{(\mu + \sigma x) \{1 - \Phi(x)\} } = \lim_{x \to \infty} \frac{1}{1 - \sigma \frac{1 - \Phi(x)}{(\mu + \sigma x)\phi(x)}}, $$. This is the only part that I'm stuck with, otherwise I don't see how to evaluate the limit. Thanks for reading

## Answer by ir7 (score 3, accepted)

https://quant.stackexchange.com/a/65991

$$ \lim_{x \to \infty} \frac{\sigma - \sigma \Phi(x)}{(\mu + \sigma x)\phi(x)} $$

$$\stackrel{0/0}{=} \lim_{x \to \infty} \frac{-\sigma \phi(x)}{\sigma\phi(x) + (\mu + \sigma x)\phi'(x) } $$

$$ =\lim_{x \to \infty}\frac{-\sigma \phi(x)}{\sigma\phi(x) - (\mu + \sigma x)x\phi(x) } = 0$$

(Used $\phi'(x) = -x\phi(x) $ and $\Phi'(x) =\phi(x)$, just like in your first L'Hospital application to get your equality. And L'Hospital works the second time too as $\lim_{x \to \infty}\phi(x) = 0$, $\lim_{x \to \infty}x \phi(x) = 0$.)

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.