Binomial Parameters MLE Calculation

The objective is to determine the Maximum Likelihood Estimate (MLE) for the parameters $(n, p)$ of a Binomial distribution, given that an observation $X=3$ was made. The parameter space is restricted to $n \in \{5, 6\}$ and $p \in \{\frac{1}{4}, \frac{3}{4}\}$. We need to find the pair $(n, p)$ that maximizes the likelihood function $L(n, p)$ for the observed $X=3$.

Likelihood Function Formulation

The probability mass function (PMF) of a Binomial$(n, p)$ random variable $X$ is:

$P(X=k | n, p) = \binom{n}{k} p^k (1-p)^{n-k}$

The likelihood function for the observed value $X=3$ is:

$L(n, p | X=3) = \binom{n}{3} p^3 (1-p)^{n-3}$

Evaluating Likelihood for Each Parameter Pair

We compute the likelihood value for each of the four possible combinations of $(n, p)$:

• Case 1: $(n=5, p=\frac{1}{4})$ $L(5, \frac{1}{4}) = \binom{5}{3} (\frac{1}{4})^3 (1 - \frac{1}{4})^{5-3} = 10 \times (\frac{1}{64}) \times (\frac{3}{4})^2 = 10 \times \frac{1}{64} \times \frac{9}{16} = \frac{90}{1024}$
• Case 2: $(n=5, p=\frac{3}{4})$ $L(5, \frac{3}{4}) = \binom{5}{3} (\frac{3}{4})^3 (1 - \frac{3}{4})^{5-3} = 10 \times (\frac{27}{64}) \times (\frac{1}{4})^2 = 10 \times \frac{27}{64} \times \frac{1}{16} = \frac{270}{1024}$
• Case 3: $(n=6, p=\frac{1}{4})$ $L(6, \frac{1}{4}) = \binom{6}{3} (\frac{1}{4})^3 (1 - \frac{1}{4})^{6-3} = 20 \times (\frac{1}{64}) \times (\frac{3}{4})^3 = 20 \times \frac{1}{64} \times \frac{27}{64} = \frac{540}{4096} = \frac{135}{1024}$
• Case 4: $(n=6, p=\frac{3}{4})$ $L(6, \frac{3}{4}) = \binom{6}{3} (\frac{3}{4})^3 (1 - \frac{3}{4})^{6-3} = 20 \times (\frac{27}{64}) \times (\frac{1}{4})^3 = 20 \times \frac{27}{64} \times \frac{1}{64} = \frac{540}{4096} = \frac{135}{1024}$

Identifying the Maximum Likelihood Estimate

Compare the computed likelihood values:

• $L(5, \frac{1}{4}) = \frac{90}{1024}$
• $L(5, \frac{3}{4}) = \frac{270}{1024}$
• $L(6, \frac{1}{4}) = \frac{135}{1024}$
• $L(6, \frac{3}{4}) = \frac{135}{1024}$

The largest likelihood value is $\frac{270}{1024}$, obtained when $(n=5, p=\frac{3}{4})$.

Thus, the maximum likelihood estimate for $(n, p)$ is $(5, \frac{3}{4})$.