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 $(5, \frac{3}{4})$.
Important Questions from Discrete Probability
- Suppose two fair dice are thrown independently at random. Let $X$ and $Y$ be the numbers on the upper face of the first die and that of the second die, respectively. Then which of the following statements are true?
- A box contains 40 numbered red balls and 60 numbered black balls. From the box, balls are drawn one by one at random without replacement till all the balls are drawn. The probability that the last ball drawn is black equals
Consider the problem of testing $H_0 : \theta = 1$ vs $H_1 : \theta = \frac{1}{2}$ where $\theta$ is the mean of a Poisson random variable. Let $X$ and $Y$ be a random sample from Poisson ($\theta$) distribution. Consider the following test procedure:
Reject $H_0$ if either $X = 0$ or $(X = 1 \text{ and } X + Y \leq 2)$; otherwise accept $H_0$.
Which of the following are true?
In a football league, the goals scored by home teams over 380 matches have the following frequency distribution.
Number of goals 0 1 2 3 4 5 Frequency 92 121 91 50 19 7 The average goals scored by home teams is 1.49. We want to test $H_0$: Goal distribution is Poisson. Based on observations the value of the $\chi^2$-statistic for goodness of fit is 1.27. Given $\chi^2_{0.05, 6} = 1.64, \chi^2_{0.05, 5} = 1.15, \chi^2_{0.95, 6} = 12.59$ and $\chi^2_{0.95, 5} = 11.07$, which of the following are true?
Ten balls are put in 6 slots at random. Then the expected total number of balls in the two extreme slots is
