Suppose that the data below are remission times (in weeks) of two treatment groups and that data follow an exponential distribution.

Group 1: (6,6,7,8,10,13,16,22,23,6+,9+,10+,11+,17+,19+,20+,25+,32+,32+,34+,35+)

Group 2: (1,1,2,2,3,4,4,5,5,7,8,8,9,11,11,12,12,15,17,22,23)

where $+$ mean censored data.

How I can use the likelihood ratio test in this case to test the hypothesis that the remission risk in Group 2 is 5 times the risk of Group 1?

Let $T\sim \exp(\alpha)$ then $$f(t)=\alpha\exp(-\alpha t)\qquad t>0$$

$$S(t)=P(T\geq t)=1-F_T(t)=\exp(-\alpha t)$$


$$\lambda(t)=\frac{f(t)}{S(t)}=\frac{\alpha\exp(-\alpha t)}{\exp(-\alpha t)}=\alpha$$

So I think if the $\lambda_1(t)=\alpha_1$ and $\lambda_2(t)=\alpha_2$ then the hypothesis is $$H_0:\lambda_2(t)=5\lambda_1(t)$$ in the particular case of exponential $$H_0:\alpha_2=5\alpha_1$$

What I think to do is find the likelihood estimate for two groups and compare

For Group 1

$$\alpha_1=\frac{\sum \delta_i}{\sum t_i}$$

where $\delta=1$ if are not censoring and $\delta=0$ if a particular time is censoring, and $t_i$ are the remission times.

I'm a litle lost. How the right way to test it?

This question is from an old exam, and need to be done at hand. I don't understand how I can use the likelihood test in this case, because if two groups follow a exponential distribution, the degrees of freedom in the likelihood test will be 0, what no make sense.


1 Answer 1


If you would write the likelihood, then the answer would be easy. I use subscript $i$ for the first group and $j$ for the second group,

The likelihood for the model is $$ \prod_i \lambda_1 ^{\delta_i} \exp(-\lambda_1t_i) \prod_j \lambda_2^{\delta_j} \exp(-\lambda_2 t_j) $$ so the log-likelihood is $$ \ell = \log \lambda_1 \sum_i \delta_i - \lambda_1 \sum_i t_i + \log\lambda_2 \sum_j \delta_j - \lambda_2\sum_j t_j $$

Then you replace $\lambda_2 = 5 \lambda_1$ to obtain the (log)likelihood under the null $$ \ell_0 = \log \lambda_1 \sum_i \delta_i - \lambda_1 \sum_i t_i + (\log(5) + \log \lambda_1) \sum_j \delta_j - (5 + \lambda_1)\sum_j t_j $$

The difference in degrees of freedom is 1 since there is a difference of one parameter between the two models.

Then calculate the test statistic \begin{align*} 2(\ell - \ell_0) &= 2 * ((\log \lambda_2 - \log \lambda_1 - \log(5))\sum_j \delta_j + (-\lambda_2 + 5 + \lambda_1) \sum_j t_j ) \end{align*}

If you calculate $\lambda_1$ and $\lambda_2$, they are just the number of events / total time at risk (MLE of exponential distribution). You have the estimates $\lambda_1 = 9/361 = 0.024$ and $\lambda_2 = 21 / 182 = 0.11$ (maybe I made a mistake here, the most difficult part of the exercise).

Your test statistic is then $$ 2 * (21 * (-2.20 + 3.72 -1.6) + 182 * (-0.11 + 5 + 0.024)) = 1785$$ (again, calculations by hand might be wrong so you should check that).

Now if you vaguely know how the chi square distribution with 1 degree of freedom looks like, you'll have no trouble concluding that $1785 >> 1$ and that the null hypothesis is rejected.


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