Maximum Likelihood Estimation (MLE) for Discrete Distribution

This problem asks us to find the Maximum Likelihood Estimate (MLE) for the parameter $\theta$ of a discrete random variable $X$. We are given the probability mass function (PMF) of $X$ and the observed counts from a random sample.

Understanding the Problem Setup

We have a discrete random variable $X$ with the following probability mass function (PMF):

• $P(X = 0) = \frac{\theta}{3}$
• $P(X = 1) = 1 - \frac{\theta}{2}$
• $P(X = 2) = \frac{\theta}{6}$

The parameter $\theta$ is constrained to be in the interval $(0,1)$, i.e., $\theta \in (0,1)$.

We are given a random sample of size $n = 90$. The observed counts for each value of $X$ are:

First, let's check if the counts sum up to the total sample size: $20 + 60 + 10 = 90$. This matches the given sample size $n=90$.

Constructing the Likelihood Function

The Maximum Likelihood Estimation method involves finding the parameter value that maximizes the likelihood of observing the given data. The likelihood function, $L(\theta)$, is the product of the probabilities of observing each data point. For a sample with counts $n_0, n_1, n_2$, the likelihood function is:

$ L(\theta) = [P(X=0)]^{n_0} \times [P(X=1)]^{n_1} \times [P(X=2)]^{n_2} $

Substituting the given probabilities and observed counts:

$ L(\theta) = \left(\frac{\theta}{3}\right)^{20} \left(1 - \frac{\theta}{2}\right)^{60} \left(\frac{\theta}{6}\right)^{10} $

Maximizing the Log-Likelihood Function

It is usually easier to work with the natural logarithm of the likelihood function, called the log-likelihood function, denoted as $\ln L(\theta)$. Maximizing $\ln L(\theta)$ is equivalent to maximizing $L(\theta)$ because the logarithm is a monotonically increasing function.

Taking the natural logarithm:

$ \ln L(\theta) = \ln \left[ \left(\frac{\theta}{3}\right)^{20} \left(1 - \frac{\theta}{2}\right)^{60} \left(\frac{\theta}{6}\right)^{10} \right] $

Using the properties of logarithms ($\ln(ab) = \ln a + \ln b$ and $\ln(a^b) = b \ln a$):

$ \ln L(\theta) = 20 \ln\left(\frac{\theta}{3}\right) + 60 \ln\left(1 - \frac{\theta}{2}\right) + 10 \ln\left(\frac{\theta}{6}\right) $

$ \ln L(\theta) = 20 (\ln \theta - \ln 3) + 60 \ln\left(1 - \frac{\theta}{2}\right) + 10 (\ln \theta - \ln 6) $

Combine terms involving $\ln \theta$:

$ \ln L(\theta) = (20 + 10) \ln \theta - 20 \ln 3 - 10 \ln 6 + 60 \ln\left(1 - \frac{\theta}{2}\right) $

$ \ln L(\theta) = 30 \ln \theta + 60 \ln\left(1 - \frac{\theta}{2}\right) - (20 \ln 3 + 10 \ln 6) $

Finding the Derivative and Solving for $\theta$

To find the value of $\theta$ that maximizes the log-likelihood, we need to compute the derivative of $\ln L(\theta)$ with respect to $\theta$ and set it equal to zero.

$ \frac{d}{d\theta} \ln L(\theta) = \frac{d}{d\theta} \left[ 30 \ln \theta + 60 \ln\left(1 - \frac{\theta}{2}\right) - (20 \ln 3 + 10 \ln 6) \right] $

The derivative of $30 \ln \theta$ is $\frac{30}{\theta}$.

The derivative of $60 \ln\left(1 - \frac{\theta}{2}\right)$ requires the chain rule: $60 \times \frac{1}{1 - \frac{\theta}{2}} \times \frac{d}{d\theta}\left(1 - \frac{\theta}{2}\right) = 60 \times \frac{1}{1 - \frac{\theta}{2}} \times \left(-\frac{1}{2}\right) = \frac{-30}{1 - \frac{\theta}{2}}$.

The derivative of the constant terms $-(20 \ln 3 + 10 \ln 6)$ is $0$.

So, the derivative is:

$ \frac{d}{d\theta} \ln L(\theta) = \frac{30}{\theta} - \frac{30}{1 - \frac{\theta}{2}} $

Simplify the second term:

$ 1 - \frac{\theta}{2} = \frac{2 - \theta}{2} $

$ \frac{30}{1 - \frac{\theta}{2}} = \frac{30}{\frac{2 - \theta}{2}} = \frac{30 \times 2}{2 - \theta} = \frac{60}{2 - \theta} $

Therefore, the derivative becomes:

$ \frac{d}{d\theta} \ln L(\theta) = \frac{30}{\theta} - \frac{60}{2 - \theta} $

Now, set the derivative equal to zero to find the critical point(s):

$ \frac{30}{\theta} - \frac{60}{2 - \theta} = 0 $

$ \frac{30}{\theta} = \frac{60}{2 - \theta} $

Divide both sides by 30:

$ \frac{1}{\theta} = \frac{2}{2 - \theta} $

Cross-multiply:

$ 1 \times (2 - \theta) = 2 \times \theta $

$ 2 - \theta = 2\theta $

Add $\theta$ to both sides:

$ 2 = 3\theta $

Solve for $\theta$:

$ \theta = \frac{2}{3} $

Since $\theta = \frac{2}{3}$ is within the interval $(0,1)$, this is a potential MLE. The second derivative test confirms this is a maximum, but typically for these types of problems, the unique critical point found is the MLE.

Result

The maximum likelihood estimate of the parameter $\theta$ is $\frac{2}{3}$.