Using the Standard Deviation to Calculate a Return Tail Probability

Summary

The document compares two calculations for the probability that a normally distributed return falls below five percent. With a stated mean of ten percent and standard deviation of 1.5 percent, direct integration of the normal density gives a small lower-tail probability. The question then compares this result with a z-score calculation that divides by the standard error for a sample of ten, producing a much smaller probability.

The answer identifies the source of the discrepancy: the question concerns one return drawn from the distribution, so its standard deviation is used in standardizing the threshold. Dividing by the standard error would apply to a sample mean, not an individual observation. The exchange offers a focused correction rather than a broader treatment of return distributions, estimation uncertainty, or whether the assumed normal model is appropriate for real financial returns.

Key ideas

• For a single return, standardize a threshold using the distribution’s standard deviation.
• The standard error of the mean applies when calculating probabilities for a sample average.
• Using the sample-size adjustment changes the question and can greatly alter a tail probability.
• The result depends on the assumption that returns follow the stated normal distribution.

Tags

# calculating probability of a return below a specific value


# calculating probability of a return below a specific value












assume a probability distribution with a mean of %10 and standard deviation of %1.5. In wanting to solve the probability being lower than %5, the normal distribution is written down and integrated as follows:

$\int_{-∞}^{0.05} e^{(-\frac{1}{2} ((x - 0.1)/0.015)^2)}/(\sqrt{2 π} 0.015) dx = 0.00042906$

which gives %0.04

if you calculate it through the z score(assuming a sample size of 10) it would be:

Z=(0.05-0.10)/(0.015/sqrt(10))=-10.5409

looking at a z score table would then give a probability of 2.798*10^-26

why do the two approaches give different answeres?

## Answer by Quant In Spe (score 1)

https://quant.stackexchange.com/a/71994

Seems like you calculated the z-score incorrectly. There’s no need for the sqrt(10) in the denominator since we just want the standard deviation there.

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