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I got a dataset with abnormal stock returns for around 400 securities and i need to find out if they're statistically different from zero. They don't follow a normal distribution, so i would like to complement a regular student t with a non-parametric t test.

I'm wondering which wilcoxon test i should use. I have trouble identifying the correct test, either rank sum or signed, as i only have one dataset that i need to test against a hypothesized median of 0.

Thank you!

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  • $\begingroup$ To provide a simple answer to the question, it is the signed-rank test that is used to compare a sample to zero, or another default value. $\endgroup$ Commented Mar 30, 2019 at 1:57

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I have to apologize, I misread your post. I am editing it to correct that.

The answer is "yes," you can use it for paired data, but there should be significant reservations that you should take into account.

The first reservation is to note that excess returns are a model specific concept. If the model does not hold, then the construction becomes very suspect. The simplest example is the CAPM. Fama and MacBeth completely falsified the model in 1973, yet it won't go away.

However, let us go forward with the idea anyway.

Let, $r_i$ be the return for an equity security indexed in the set $i\in{1}\dots{I}\in\mathbb{Z}^+,$

$r_m$ will be some proxy for a market asset, and

$r_f$ will be some asset that pays out the same return in all states of nature.

The excess return is defined as $$r_i-r_f$$ and $$r_m-r_f.$$

The first fiction is that $r_f$ is fixed, however interest rates appear to be distributed by a Gamma distribution.

Equity returns, removing the effects of dividends, mergers, bankruptcies and liquidity costs follow a truncated Cauchy distribution. See references below as to why.

You do not want to make zero median effects because the median is shifted from the center of location and based on a population study, the median isn't zero; however, the location of the mode is implying that all risks are paid for with dividends for the marginal security.

The excess return would follow the distribution of the difference of a truncated Cauchy distribution and a Gamma distribution. I am not ambitious enough today to calculate the convolution to work out what it would be. Nonetheless, it will not have a mean and as a consequence a variance.

You have several potential tests that you could perform. The first is on $r_i-r_f$ versus $r_m-r_f$ to see if they are drawn from the same population. You could certainly used the Wilcoxon signed-rank test.

You need to be very cautious here. If you chose an index, such as the S&P 500 there is a very good chance that at least some of your 400 securities are a member of the index. If so, you need to recalculate the index, minus the one security or you are at least partially testing a security against itself. For a handful of securities whose returns drive the index and constitute most of the independent variability, they will move nearly in tandem because it really will be a near mapping of a security on itself.

You median won't be zero because it is known not to be. It would take 100 years of continuous losses to shift the empirical median to zero. Markets would reprice and learn long before that happened.

Since equity returns are truncated at -100%, the median is not the center of location the mode is.

If you remove liquidity, dividend, merger and bankruptcy effects, then the distribution of equity returns is $$\left[\frac{\pi}{2}+\tan^{-1}\left(\frac{\mu}{\sigma}\right)\right]^{-1}\frac{\sigma}{\sigma^2+(x-\mu)^2}.$$

This is due to the fact that equity returns are the product of the ratio of prices times the ratio of quantities. It is $$\frac{\text{future value}}{\text{present value}}.$$

This makes it a ratio distribution. While you can use rank methods and they will provide you valid inference, the median should not be zero. The mode should be zero.

If you wanted to test if the mode was near zero, you could perform a Bayesian test on $r_i$ alone, without subtracting $r_f$. If you wanted to subtract $r_f$, then you would need to perform the convolution mentioned above to find the likelihood.

Dependence of returns between securities is assured as movement of members of a truncated Cauchy distribution cannot be independent of each other, even though they also cannot be correlated with each other. As returns cannot have independent errors, you cannot use tests built on independence.

See:

See the following articles:

Curtiss, J. H. (1941). On the distribution of the quotient of two chance variables. Annals of Mathematical Statistics, 12:409-421.

Fama, E. (1965). The behavior of stock market prices. Journal of Business, 38:34-105.

Fama, E. F. (1963). Mandelbrot and the stable paretian hypothesis. Journal of Business, 36:420 - 429.

Fama, E. F. and MacBeth, J. D. (1973). Risk, return, and equilibrium: Empirical tests. The Journal of Political Economy, 81(3):607-636.

Fama, E. F. and Roll, R. (1968). Some properties of symmetric stable distributions. Journal of the American Statistical Association, 63(323):pp. 817-836.

Fama, E. F. and Roll, R. (1971). Parameter estimates for symmetric stable distributions. Journal of the American Statistical Association, 66(334):331 - 338.

Fisher, R. A. 1934. Two new properties of mathematical likelihood. Proc. Roy. Soc. Ser. A (144) 285-307.

Gull, S. F. (1988). Bayesian inductive inference and maximum entropy. In Erickson, G. J. and Smith, C. R., editors, Maximum-Entropy and Bayesian Methods in Science and Engineering: Foundations, volume 1 of Fundamental Theories of Physics, pages 53-74. Springer.

Gurland, J. (1948). Inversion formulae for the distribution of ratios. The Annals of Mathematical Statistics, 19(2):228-237.

Harris, D. E. (2017). The distribution of returns. The Journal of Mathematical Finance, 7(3):769-804.

Mandelbrot, B. (1963). The variation of certain speculative prices. The Journal of Business, 36(4):394-419.

Marsaglia, G. (1965). Ratios of normal variables and ratios of sums of uniform variables. Journal of the American Statistical Association, 60(309):193- 204.

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    $\begingroup$ "the median is not the center" ... and the signed rank test isn't really a test of the median. $\endgroup$
    – Glen_b
    Commented Mar 29, 2019 at 13:10
  • $\begingroup$ @Glen_b thanks for the post, I really misread the original posting and my answer was wrong. $\endgroup$ Commented Mar 30, 2019 at 23:02

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