Meta-analysis help: how to convert hazard ratio to risk ratio

[Al Bothwell]

I am working on a meta-analysis using Stata. All the primary studies report risk ratio except one study that reported hazard ratio. Is there a formula to convert a hazard ratio to a risk ratio?

Thanks,
Al Bothwell

[Leonardo Guizzetti]

In short, probably not. Under very specific conditions, the HR and RR will coincide when everyone has complete followup (no dropouts or censoring). This may be unrealistic in your specific situation. It may be possible to compute an approximate RR by inspection of the Kaplan-Meier curves by finding the group-specific failure (or event) rates at a fixed time-point, since these are probabilities. Where you will likely run into problems, is how to compute the variance of the survival probabilities. Greenwood's formula (or more modern methods used in Stata) require you to know how many censoring events there were up to the fixed time-point. (It's not justified to use the binomial variance formula because of the censoring.) If somehow you have this information, then you can proceed by the usual methods to get your RR, variance and CI.

An alternative is to contact the authors, explain your research question and request they perform a reanalysis to obtain the RR estimates you need. They would be in the best position to provide this.

[Clyde Schechter]

Leonardo Guizzetti

Under very specific conditions, the HR and RR will coincide when everyone has complete followup (no dropouts or censoring).

I don't think that's right.

Suppose we have two exponential survival functions, say one with constant hazard 1 and the other with constant hazard 2. Then the hazard ratio is 2:1 = 2. But at any given time t, the surviving proportions will be exp(-t) and exp(-2t), respectively. But the relative risk of death at time t will be (1-exp(-t))1-exp(-2t)). While that approaches 2 in the limit as t -> 0, it also approaches 1 in the limit as t -> infinity, and takes on all possible values in between at realistic values of t.

As far as I am aware, there is no systematic relationship between RR and hr over time periods that are not, in effect, infinitesimal.

[Leonardo Guizzetti]

Thanks for the counterexample Clyde. I was imagining maybe a different condition where one estimated a regression model using survival data under those conditions I noted but I hadn't really thought the idea all the way through.

[Clyde Schechter]

I've been playing around with this for the past hour or so. It is not a coincidence that in the limit as t->0, hr(t) and RR(t) tend to be equal. I can prove it, subject to mild regularity conditions on the survival functions:

Theorem:
Let S*₁(t) and S₂(t) be two survival functions, both continuously differentiable in a neighborhood of t = 0.
Let h₁ and h₂ be the corresponding hazard functions, and hr(t) the hazard ratio h₁(t)/h₂(t). Define the relative risk of failure up to time t to be
RR(t) = (1-S₁(t))/(1-S₂(t)).
Then lim_t->0 hr(t) exists if and only if if lim_t->0 S₁’(t)/S₂’(t) exists (‘ denotes d/dt), and, in that case, lim_t->0 RR(t) also exists and all three of these limits are equal.

Proof:
By definition, for any survival function S(t), its hazard function h(t) = -S’(t)/S(t). Consequently,
lim_t->0 hr(t) = lim_t->0 = lim_t->0 [-S₁’(t)/S₁(t)]/[ -S₂’(t)/S₂(t)] = lim_t->0 [S₁’(t)/S₂’(t)] * [S₂(t)/S₁(t)] .
Now, by definition of a survival function we have S(0) = 1, and by continuity, therefore, lim_t->0 S(t) = 1.
So, we get lim_t->0 hr(t) = lim_t->0 -S₁’(t)/S₂’(t) * 1 = lim_t->0 -S₁’(t)/S₂’(t).
Therefore lim_t->0 hr(t) exists if and only if lim_t->0 -S₁’(t)/S₂’(t) exists, and if so, they are equal.

Now let’s consider lim_t->0 RR(t).
Since RR(t) = (1-S₁(t))/(1-S₂(t)), and both S₁ and S₂ approach 1 as t -> 0, we can apply L’Hopital’s rule.
This gives us: lim_t->0 RR(t) = lim_t->0 -S₁’(t)/(-S₂’(t)) = lim_t->0 S₁’(t)/(S₂’(t)) if the last of these exists.
So lim_t->0 RR(t) exists if lim_t->0 S₁’(t)/(S₂’(t)) exists, and in that case they are equal.

But now we have established that lim_t->0 RR(t) and lim_t->0 hr(t) both exist if lim_t->0 S₁’(t)/(S₂’(t)) does, and in that case they are both equal to it, and, therefore, equal each other.

Comment:
Note that if lim_t->0 S₁’(t)/(S₂’(t)) does not exist, then neither does lim_t->0 hr(t). However, lim_t->0 RR(t) might exist anyway, in which case it is definitely not approximated by the hazard ratio.

How important is the condition that lim_t->0 S₁’(t)/(S₂’(t)) exist? For the commonly used survival distributions, I think this usually holds, if only because S_i’(0) is always finite, and is usually non-zero (i = 1, 2). One can, however, illustrate survival functions for which this will fail. For example, if S₁(t) = exp(-t) and S₂ = exp(-t²), then the derivative ratio in question tends toward infinity as t approaches 0.

In the end, I'm not sure how useful this will be to O.P. If he is interested in the RR over sufficiently short survival times, it will be approximately equal to the hazard ratio. But sufficiently short may be doing more work there than he can accommodate in his situation.

[Al Bothwell]

Thank you for the great comments. I will contact the authors to request for RR. As a backup option, I will try setting a 2x2 table using the data reported in the article and then calculate RR.