2
$\begingroup$

I came across a sentence like this: "A is independently associated with B". What does it means? (They have a lot of factors to compare)

Exact quotation: In the subgroup of patients without MetS PNPLA3 (OR: 2.002 [CI95%: 1.062–3.772], P = 0.032) and presence of type 2 diabetes (OR: 5.317 [CI95%: 1.009–28.014], P = 0.049) but not hepatic coppercontent (OR: 0.991 [CI95%: 0.963–1.020], P = 0.539) were significantly and independently associated with NASH (Table 4). The paper is:

Stättermayer, A. F., Traussnigg, S., Aigner, E., Kienbacher, C., Huber-Schönauer, U., Steindl-Munda, P., … Ferenci, P. (2017). Low hepatic copper content and PNPLA3 polymorphism in non-alcoholic fatty liver disease in patients without metabolic syndrome. Journal of Trace Elements in Medicine and Biology, 39, 100–107. doi:10.1016/j.jtemb.2016.08.006. Retrieved from http://www.sciencedirect.com/science/article/pii/S0946672X16302218

My background is biomedical (without any statistic/maths) so simple explanation about the relation will be much appreciated.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ You will probably get better answers if you quote the exact passage and cite it. $\endgroup$
    – Kodiologist
    Commented Mar 9, 2017 at 19:58
  • $\begingroup$ The answers to you question are all mumbo jumbo. Their just using technical language to answer the non-technical question. They still haven’t answered the plain language question. $\endgroup$
    – Peter Williamson
    Commented Jul 6, 2022 at 2:15

3 Answers 3

Reset to default
2
$\begingroup$

The paper in question is clearly not statistically sophisticated. For example, the authors only included predictors in the multiple-regression models if they had been significant in a model with only the given predictor; this method of variable selection is popular, but has no statistical motivation.

It seems that by "independently associated", the authors mean that a variable had a significantly nonzero coefficient in a multiple-regression model. This doesn't have anything to do with independence in the sense relating to joint probability distributions. The way one would usually describe this situation is that a predictor is significant while "controlling for" the other predictors in the model.

Improve this answer
$\endgroup$
1
$\begingroup$

Unless someone can tell me otherwise, the only "independent association" I can think of occurs in spurious correlation. Spurious correlation is a non-causal association between variables. Spurious correlation occurs when two independent (i.e., orthogonal and unrelated) variables are related through a modifying third variable that has a causal link to both otherwise independent variables. For example, consider the famous nonsense statement, "Ice-cream consumption causes drowning." That is, if we plot a time series of monthly ice-cream sales, it is highly correlated to monthly drowning case occurrences. In this case the modifying third variable is temperature. When it's hot, people eat more ice-cream and drown more frequently.

Spurious correlation has many websites, here is one. It can be hilarious. So a completed sentence from the partial quote might read as "A is independently associated with B... through modifying variable C."

Also, the phrase "independently associated" appears in the medical literature with some frequency. It sounds nicer than "spurious correlation via some modifying variable the nature of which, we may not have even thought about." Why don't you cough up the rest of the sentence? We could all do with a good belly laugh if it is indeed spurious.

Update: Having the paper now in hand, it is, indeed medical. The evidence for independence is soft, and the evidence for correlations with a modifying variable is indirect. Unfortunately, "We are not amused."

Improve this answer
$\endgroup$
2
  • $\begingroup$ Does it mean that the author is confident that there is no causal link between 2 variables or there is still probability about the causative relationship? $\endgroup$
    – java
    Commented Mar 10, 2017 at 13:56
  • $\begingroup$ The author states that A and B are independent (no causal linkage) but covary as C. The implication is that the probability of a causal relationship between A and B is not significant when the correlation with C is removed. So yes, there is still a probability. This implies that the correlation of A and B with C removed is not significantly different from zero. It does prove that the correlation of A and B with C removed is zero. In other words, the hypothesis of independence of A and B with C removed was not rejected. $\endgroup$
    – Carl
    Commented Mar 10, 2017 at 19:33
0
$\begingroup$

It is not preferred language, but if B and A have a significant bivariate association which is not found to be so after control of several possible factors, we might say "A and B are independently associated".

The problem is that further adjustment for C or D or... may introduce effects which change the relation between B and A. For instance, confounders may change the association so that their correlation is no longer significant. Adjustment for mediators will eliminate the indirect effect between B and A so that if there is no direct effect, the association "goes away". Another possibility is simply a loss of degrees of freedom so that significance tests become under powered after overadjustment for unrelated factors.

Causal modeling illustrates several relationships where adjustment for other factors will cause a "significant bivariate relationship" to not be so after further control of factors.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.