I am having quite some troubles with the following homework:

In a city it's measured for the whole year whether it rained or not. A distribution $\textrm{Bernoulli}(r_t|\rho)$ characterizes the observations $\{r_t\}_{t=1}^{365}$ where $r_t = 1$ if it rained that day and $r_t = 0$ if it did not rain.

My goal is finding the value of $\rho$ as a maximum likelihood.

One first step is to find the likelihood of the entire set of observations, which I believe is:

$p(r_1,r_2,\ldots,r_t|\rho) = \prod_{t=1}^{365}\rho^{r_t}(1 - \rho)^{1-r_t}$

Considering $n_1$ as the number of rainy days, and $n_0$ the number of not raining days the likelihood is simplified:

$p(r_1,r_2,\ldots,r_t|\rho) =\rho^{n_1}(1 - \rho)^{n_0}$

Hence, the log-likelihood calculated as follows:

$\mathrm{log}p(r_1,r_2,\ldots,r_t|\rho) = n_1\mathrm{log}(\rho) + n_0\mathrm{log}(1-\rho)$

Its derivative seems to be:

$\frac{\mathrm{d}}{\mathrm{d}\rho}\mathrm{log}p(r_1,r_2,\ldots,r_t|\rho) = \frac{n_1}{\rho} + \frac{n_0}{1- \rho}$

Setting the derivative to 0:

$\rho = \frac{n_1}{n_1 - n_0}$

Plugging in the yearly measurements $n_1 = 217$, $n_0 = 148$:

$\rho = \frac{217}{217-148} = 3.145$

Unfortunately this value is not allowed for $\rho$, which should be in the interval $[0,1]$.

What's wrong? I did the calculations several times with the same result.


1 Answer 1


$$\frac{d \log(p)}{d \rho} = \frac{n_1}{\rho} - \frac{n_0}{1-\rho} \Rightarrow \rho = \frac{n_1}{n_0+n_1}$$

  • $\begingroup$ hmm, why do you have a minus between $n_1/\rho$ and $n_0/(1-\rho)$ ? $\endgroup$
    – fstab
    Sep 17, 2014 at 11:30
  • 2
    $\begingroup$ I don't know why someone downvoted this answer since it correctly points out that the OP mistakenly wrote the derivative of $\log(1-\rho)$ as $\frac{1}{1-\rho}$ instead of as $-\frac{1}{1-\rho}$ as told to us by the chain rule of differentiation. $\endgroup$ Sep 17, 2014 at 14:05
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    $\begingroup$ francesco the minus is there because $\frac{d}{d\rho} \log(1-\rho)=\frac{1}{1-\rho}\cdot -1$ by the chain rule of differentiation. $\endgroup$
    – Glen_b
    Sep 17, 2014 at 15:21

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