Parameter Estimation for N(theta, theta)
This problem requires finding the Maximum Likelihood Estimate (MLE) of the parameter $\theta$ for a random sample $X_1, \ldots, X_n$ from a normal distribution $N(\theta, \theta)$. The parameter space is constrained to $\theta > 0$.
Deriving the Log-Likelihood Function
The probability density function (PDF) for a single observation from $N(\theta, \theta)$ is: $f(x; \theta) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{(x-\theta)^2}{2\theta}\right)$
The likelihood function $L(\theta)$ for the sample is the product of the PDFs: $L(\theta) = \prod_{i=1}^n f(X_i; \theta) = \left(\frac{1}{\sqrt{2\pi\theta}}\right)^n \exp\left(-\frac{1}{2\theta} \sum_{i=1}^n (X_i-\theta)^2\right)$
Working with the log-likelihood function $\ell(\theta) = \log L(\theta)$ simplifies differentiation: $\ell(\theta) = -\frac{n}{2} \log(2\pi) - \frac{n}{2} \log(\theta) - \frac{1}{2\theta} \sum_{i=1}^n (X_i-\theta)^2$
Maximizing the Log-Likelihood
To find the MLE, we compute the derivative of $\ell(\theta)$ with respect to $\theta$ and set it to zero. Expanding the sum term $\sum_{i=1}^n (X_i-\theta)^2 = \sum X_i^2 - 2\theta \sum X_i + n\theta^2$. The derivative $\frac{d\ell}{d\theta}$ is calculated as: $\frac{d\ell}{d\theta} = -\frac{n}{2\theta} - \frac{1}{2} \frac{d}{d\theta}\left(\frac{\sum X_i^2}{\theta} - 2\sum X_i + n\theta\right)$ $\frac{d\ell}{d\theta} = -\frac{n}{2\theta} - \frac{1}{2} \left(-\frac{\sum X_i^2}{\theta^2} + n\right)$ $\frac{d\ell}{d\theta} = -\frac{n}{2\theta} + \frac{\sum X_i^2}{2\theta^2} - \frac{n}{2}$
Setting the derivative to zero gives the equation for critical points: $-\frac{n}{2\theta} + \frac{\sum X_i^2}{2\theta^2} - \frac{n}{2} = 0$
Solving the Quadratic Equation for $\theta$
Multiply the equation by $2\theta^2$ to eliminate denominators: $-n\theta + \sum X_i^2 - n\theta^2 = 0$ Rearrange into the standard quadratic form $n\theta^2 + n\theta - \sum X_i^2 = 0$. Divide by $n$: $\theta^2 + \theta - \frac{1}{n} \sum X_i^2 = 0$
Using the quadratic formula $\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1, b=1, c = -\frac{1}{n} \sum X_i^2$: $\theta = \frac{-1 \pm \sqrt{1^2 - 4(1)(-\frac{1}{n} \sum X_i^2)}}{2(1)}$ $\theta = \frac{-1 \pm \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$
Selecting the Valid MLE Solution
This yields two potential solutions for $\theta$: $\theta_1 = \frac{-1 - \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$ $\theta_2 = \frac{-1 + \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$
The problem specifies that $\theta > 0$. Since $\sqrt{1 + \frac{4}{n} \sum X_i^2}$ is always greater than 1 (because $n>0$ and $\sum X_i^2 \ge 0$), the first solution $\theta_1$ is always negative. The second solution $\theta_2$ is always positive.
Therefore, the Maximum Likelihood Estimate (MLE) of $\theta$ is the positive root: $\hat{\theta} = \frac{-1 + \sqrt{1 + \frac{4}{n} \sum X_i^2}}{2}$
Important Questions from Random Variables
Let $X$, $Y$, and $Z$ be independent Normal random variables with means $-1$, $0$, and $1$, respectively, and variances $1$, $1$, and $3$, respectively. Which of the following random variables has a Cauchy distribution with location parameter $0$ and scale parameter $1$?
- Consider a finite population of size $N = 100$. Let $T_1$ be the sample mean of a study variable based on a sample of size $n$ ($1 < n < N$) under simple random sampling with replacement scheme. Let $T_2$ be the sample mean of the same study variable based on a sample of size $n$ under simple random sampling without replacement scheme. If $Var(T_1) = 9Var(T_2)$, then the sample size $n$ equals
Let $X_1$ and $X_2$ be a random sample from Uniform$[0, \theta]$ distribution, where $\theta > 0$. For testing the hypothesis
$H_0: \theta = 1$ against $H_1: \theta = 2$,
consider a test which rejects $H_0$ if $X_1 + X_2 > \frac{4}{5}$. Then, the probability of type-I error is- Let $\{Y_n: n \ge 1\}$ be a sequence of independent and identically distributed random variables, where $Y_1 \sim \text{Bernoulli}(\frac{1}{2})$. Define $Z = \sum_{n=1}^\infty \frac{4Y_n}{5^n}$. Then, which of the following statements is true?
Let $X$ be a single sample from an absolutely continuous distribution with probability density function
$f(x|\theta) = \begin{cases} \frac{2}{\theta^2}(\theta - x), & \text{if } 0 < x < \theta \\ 0, & \text{otherwise,} \end{cases}$
where $\theta > 0$ is unknown. Which of the following intervals is a $95\%$ confidence interval for $\theta$?
