# One-Sample Log-rank test statistic

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:

• That zero in #1 ('0') is meant to be an 'O' as in observed events, right? Jul 25, 2015 at 18:01
• yes, sorry my bad, edited Jul 26, 2015 at 17:16
• I think both are correct. Jul 27, 2015 at 1:13
• 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. Jul 27, 2015 at 1:15
• O and E are sums. Sorry my bad. Jul 27, 2015 at 1:59

• 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. Jul 28, 2015 at 14:01
• 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). Jul 29, 2015 at 0:01