# Calculate the confidence interval of a Hazard Ratio from the ones of the survival

I have 2 populations with their survival at n years and their 95% confidence interval.

Under the assumption of exponential distribution, the hazard ratio can be expressed as:

$$HR_{AB} = \frac{log(S_A)}{log(S_B)}$$

However, how to infer the confidence interval of the HR (or $$log(HR)$$) from these data?

If there is no analytical solution, could a bootstrap procedure be an alternative?

PS: this would be a very exploratory analysis, hence assumption on the distribution is not highly significant.

EDIT - Let's consider these notations:

• $$p_A [95\%CI :p_{A_{inf}} ; p_{A_{sup}}]$$ and $$p_B [95\%CI :p_{B_{inf}} ; p_{B_{sup}}]$$at time $$t$$
• measured on respectively $$n_A$$ and $$n_B$$ individuals
• with $$\Delta_A = p_{A_{sup}}-p_{A_{inf}}$$ and $$\Delta_B = p_{B_{sup}}-p_{B_{inf}}$$

With an exponential distribution, $$\log S_A(t)= -\lambda_A t$$ and $$\log S_B(t)=-\lambda_B t$$. The hazard ratio ($$\text{HR}$$) you request is:

$$\frac{\log S_A}{\log S_B} = \frac{\lambda_A}{\lambda_B},$$

and the $$\log \text{HR}$$ is:

$$\log \text{HR} = \log \lambda_A - \log \lambda_B.$$

The confidence interval (CI) for the $$\text{HR}$$ is related to the variance of the its estimate. The formula for the variance of a difference gives:

$$\text{Var} (\log \text{HR }) = \text{Var}(\log \lambda_A) +\text{Var} (\log \lambda_B) - 2 \text{ Cov}(\log \lambda_A,\log \lambda_B).$$

If the estimates of $$\lambda_A$$ and $$\lambda_B$$ were obtained independently, then the last covariance term is zero so all you need are the variances of the individual $$\lambda$$ estimates.

What you have are survival probabilities $$p_A$$ and $$p_B$$ and their 95% CI at some time $$t$$. A survival probability $$p$$ is reached at $$t= -\frac{\log p}{\lambda}$$, $$\lambda = - \frac{\log p}{t}$$ or $$\log \lambda = \log(-\log p) - \log t.$$

Thus $$\text{Var}(\log \lambda) = \text{Var}(\log(-\log p))$$.

That result suggests converting the CI on the survival probabilities $$p$$ to CI on $$\log(-\log p)$$ to get estimates for the CI of the individual $$\lambda$$ estimates. If those latter CI estimates are sufficiently symmetrical about the point estimates, you can then use a normal approximation, dividing the half-widths of the 95% CI by 1.96 to estimate the standard deviations of the $$\lambda$$ estimates and then squaring to get their estimated variances. Adding those variances, taking the square root, and multiplying by 1.96 then give estimated CI half-widths for $$\log \text{HR}$$.

• great, perfect answer as usual! Commented May 29, 2021 at 18:39
• Just a little question: when you say "dividing the half-widths of the 95% CI by 1.96", are your referring to $p_{A_{inf}}$ or to $log(-log(p_{A_{inf}}))$? Commented May 31, 2021 at 9:41
• Also, shouldn't the formula take $n_A$ and $n_B$ into account? Commented May 31, 2021 at 15:49
• @DanChaltiel $n_a$ and $n_b$ are implicitly taken into account in the original CI. More cases, narrower CI. The way I formulated the answer, I was referring to half-widths on the $\log (-\log p)$ scale, as the variance on that scale is what's most directly related to the variance of $\log \text{HR}$.
– EdM
Commented May 31, 2021 at 16:00