By following the distribution I provided here on stackmathematics

Q. A biased die favours the number 2. It's rolled 4 times and 2 comes up twice. It's rolled again 4 times and 2 comes up once. Calculate the Likelihood and find the MLE.

The likelihood is: $L(\theta;3) = \binom{8}{3}\theta^3(1-\theta)^5$, whereas we have the local maximum at $\theta = \frac{3}{8}$.

I want to verify that, approximately $(0.0718, 0.6026)$ is a 2-likelihood region for $\theta$ by making a plot to indicate this.

From what I understand, the 2-likelihood region is given as the following:

The region $R_k = \{\theta \in \mathbb{R}: L(\theta;x) \ge e^{-k}L(\hat{\theta};x)\}$ for the k-unit likelihood region for parameters $\theta$ based on data $x$.

We know that $x=3$ and $\theta = \frac{3}{8}$, then we apply the following:

$$\binom{8}{3}\theta^3(1-\theta)^5 \ge e^{-2}L\left(\frac{3}{8};3\right)\approx0.035$$

So I think the next step is to plug $0.0718, 0.6026$ into $\theta$ to compare the inequality.

Something like:

$$\binom{8}{3}(0.0718)^3(1-0.0718)^5 \approx 0.014$$

However, $0.014 < 0.035$ and so $0.0718$ does not fall in the 2-likelihood region?

If this is the case, is there an easier way to check for values that fall within the confidence interval of the likelihood region on R?

## Best Answer

Let $a=3$ and $b=5,$ so that $\hat\theta = a/(a+b).$ Let the likelihood function be $L(\theta;a,b).$

Taking logarithms,express the region as$$R_k = \{\theta\in[0,1]\mid \log L(\theta;a,b) - \log L(\hat\theta;a,b) + k \ge 0\}.$$

Because for $k \gt 0$ there will be a lower limit $l$ between $0$ and $\hat\theta$ and an upper limit $u$ between $\hat\theta$ and $1,$ use any decent root finder within each of these intervals. Here is a sketch of the situation, showing the graph of the difference in log likelihoods as a function of $\theta$ on the interval $[0,1].$

This is a general picture of how maximum likelihood confidence intervals are foundin any one-parameter situation where the likelihood is smooth and has a unique global maximum in the interior of the parameter space. (Sometimes one or both limits don't exist, though, depending on how that parameter space is defined.)For an interval with confidence $1-\alpha,$ $2k$ is the upper $1-\alpha$ quantile of the chi-squared distribution with one degree of freedom. For instance, for $95\%$ confidence $k= 3.84\ldots/2 \approx 1.92.$

For $k=2,$ the built in

`uniroot`

function in`R`

computes $l=0.1060\ldots$ and $u=0.7157\ldots.$ The check of any putative limit, such as those mentioned in the question, is trivial: just compare it to these endpoints.Notice it was unnecessary to compute the constant factors $\binom{8}{3}$ in

`Lambda`

(the log likelihood function) because they cancel in taking the difference of log likelihoods. This, too, is a general phenomenon.