Ok, I am totally confused, this is probably a pure math rather than a statistic question but please enlighten me.

I found the logrank statistic formula being reported in the literature in 2 ways, but I am not sure which one is correct. When I plug the same number in both formula, i get different results for the chi-square.

Which one should I use?

$O$ = total of observed events

$E$ = total of expected events

  1. $$\frac{(O-E)^2}{E} $$

  2. $$\frac{O-E}{\sqrt{E}}$$

Which one should be used? The main reason I am asking is because when I apply the weights (W) as in the formula below, it looks like the weights are being cancelled out:

one-sample logrank statistic

  • $\begingroup$ That zero in #1 ('0') is meant to be an 'O' as in observed events, right? $\endgroup$ Jul 25, 2015 at 18:01
  • $\begingroup$ yes, sorry my bad, edited $\endgroup$
    – Paris Char
    Jul 26, 2015 at 17:16
  • $\begingroup$ I think both are correct. $\endgroup$
    – Deep North
    Jul 27, 2015 at 1:13
  • 4
    $\begingroup$ The first thing you mention is simply the square of the second. Neither is exactly a log rank statistic, since that would involve some kind of sum of terms. Compare your second proposal with the statistic here for example and note several differences. $\endgroup$
    – Glen_b
    Jul 27, 2015 at 1:15
  • $\begingroup$ O and E are sums. Sorry my bad. $\endgroup$
    – Paris Char
    Jul 27, 2015 at 1:59

1 Answer 1


enter image description here

Just read this paragraph from the same page of your book and remember that square of a standard normal distribution is a chi-square distribution. Therefore the two tests are equivalent. You can use any of them.

  • $\begingroup$ I understand that. However I don't understand how the weights (W) don't cancel out in either of these 2 formulae. How can we apply weights when they cancel out in both numerator and denominator?Also another thing that confuses me (from the same page of the book): "a weight function is the weight W(t) = Y(t) which yields the one-sample log-rank test". So the one-sample logrank test is using Gehan weight by design? Am I missing something here? :S $\endgroup$
    – Paris Char
    Jul 28, 2015 at 13:22
  • $\begingroup$ Do you mean you try to cancel W for this ratio $\frac{\left \{ \sum_i^DW(t_i)\frac{d_i}{Y(t_i)}-\int_{0}^{r}W(s)b_0(s)ds \right \}^2}{\int_{0}^{r}W^2(s)\frac{b_0(s)}{y(s)}ds}$ ?W is a function or a random variable, I don't understand how you cancel them out? Now, i even don't know how to show it is a $\chi^2$ distribution. $\endgroup$
    – Deep North
    Jul 28, 2015 at 14:01
  • $\begingroup$ Doesn't the above formula simplify to W * (Σ(Observed) - Σ(Expected)) / (W * Var() ) ? $\endgroup$
    – Paris Char
    Jul 28, 2015 at 15:15
  • $\begingroup$ No, I think $\int_{0}^{r}W^2(s)\frac{b_0(s)}{Y(s)}ds$ itself is the variance when the Expected value $ E(W(s))$ is zero. Also, do you remember the definition of variance is $E(X^2)-[E(X)]^2$ and $E(X^2)=\int_{-\infty}^{\infty}x^2f(x)dx$ this is exactly the same when E(X)=0. I think you cannot factor out W here. It is not a constant, it is a random variable(function). $\endgroup$
    – Deep North
    Jul 29, 2015 at 0:01
  • $\begingroup$ Ok, so if I have observed deaths per year and Expected deaths per year, how do I calculate a weighted one-sample logrank test? I don't have the h0 so I cannot use the formula in the picture. All i know is that Var = sum of Expected. $\endgroup$
    – Paris Char
    Jul 29, 2015 at 1:11

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.