# Meta analysis on studies with 0-frequency cells

I am familiar with meta analysis and meta regression techniques (using the R package metafor from Viechtbauer), but I recently stumbled on a problem I can't easily solve. Say we have a disease that can go from mother to the unborn child, and it has been studied already a number of times. Mother and child were tested for the virus right after birth. As an unborn child can impossibly get the virus other than from the mother, one would expect crosstabulations like :

           | neg kid | pos kid
mother neg |    A    |   C=0
-----------|---------|--------
mother pos |    B    |   D


Obviously using odds ratios (OR) gives errors as one would be dividing by 0. Same for relative risks :

$\frac{A/(A+B)}{0/(0+D)}$

Now the researchers want to test the (senseless) hypothesis whether infection of the child is related to the infection of the mother (which seems very, very obvious). I'm trying to reformulate the hypothesis and come up with something that makes sense, but I can't really find something.

To complicate things, some kids with negative moms actually are positive, probably due to infection in the first week. So I only have a number of studies where C = 0.

Anybody an idea on how to statistically summarize the data of different studies following such a pattern. Links to scientific papers are also more than welcome.

• I wouldn't call this data "malformed" - it just has a zero-frequency cell, which in large part is due to the effect being large. From the point of view of the application this is a "good thing". – Aniko Nov 17 '10 at 21:14
• @Aniko : I agree, malformed is a wrong word, but I didn't really know how to say it different. – Joris Meys Nov 17 '10 at 23:12

Seems to me this is one of the rare situations where it might well be better to meta-analyse risk differences rather than risk ratios or odds ratios. The risk difference $P(Kid_+ | Mum_+) - P(Kid_+|Mum_-)$ is estimated in each study by $D/(B+D) - C/(A+C)$. That should be finite in all studies even when $C=0$, so there should be no problem meta-analysing it.

I agree it seems pretty pointless to consider testing the hypothesis that this risk difference is zero. But it's meaningful to estimate how large it is, i.e. how much more likely a kid is to have the virus when their mum has it than when their mums doesn't.

• risk differences is indeed the way to go, as that one can be understood by non-statisticians as well. Accepted. – Joris Meys Nov 23 '10 at 21:15

Usually 0's imply that you have to use exact methods instead of relying on asymptotical methods such as meta-analysis with odds ratios. If you are willing to assume that the study effect is fixed, an exact Maentel-Hanszel test is the way to go. For an exact random effects analysis, you have to use a binomial regression model with a random study effect. I have done both in a recent applied paper, but the methods section there would not be more helpful to you, as it essentially conveys this information.

Edit

This paper is not applied, but this is where I got the idea from when confronted with the same issue:
 Hans C. van Houwelingen, Lidia R. Arends, and Theo Stijnen. Advanced methods in meta-analysis: multivariate approach and meta-regression. Statistics in Medicine, 2002; 21:589–624

Here is the paper where I used this approach (it is not apparent in the abstract, but is mentioned in the methods section):
 Trivedi H, Nadella R, Szabo A. Hydration with sodium bicarbonate for the prevention of contrast-induced nephropathy: a meta-analysis of randomized controlled trials. Clin Nephrol. 2010 Oct;74(4):288-96.

• +1 for using a binomial mixed effect models. Alas it was rejected as "not a standard method". If you can give me some links to papers where this approach is used in a meta-analysis setup, you would help me greatly. Thank you in advance. – Joris Meys Nov 23 '10 at 21:16
• I have edited my answer with some references. – Aniko Nov 23 '10 at 21:43

The metafor package's documentation says that "Adding a small constant to the cells of the 2x2 tables is a common solution to this problem." and also provides an option to do this within the call for rma().

• Common solutions are not always correct solutions. – Joris Meys Jul 29 '12 at 16:48