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 $1$?
$\frac{X+2Y + Z}{X-2Y + Z}$
To determine which random variable has a Cauchy distribution with location parameter 0 and scale parameter 1, we first need to understand the characteristics of a Cauchy distribution.
A random variable \(X\) has a Cauchy distribution if it can be expressed in the form:
\(\frac{U}{V}\)
where \(U\) and \(V\) are independent standard normal random variables (\(\mathcal{N}(0,1)\)).
Let's examine each option given in the question:
- \(\frac{X-1}{|Y|}\): Here, \(X \sim \mathcal{N}(-1,1)\) and \(Y \sim \mathcal{N}(0,1)\). The numerator \(X-1 \sim \mathcal{N}(-2,1)\) is not a standard normal random variable. The denominator \(|Y|\) does not yield a standard normal random variable, hence it cannot be a Cauchy distribution.
- \(\frac{X+2Y+Z}{X-2Y+Z}\): This expression involves linear combinations of independent normal random variables, leading to a sum of normal variables. By symmetry and standardization, \(X+2Y+Z\) and \(X-2Y+Z\) can be manipulated to standard form (this expression can represent a Cauchy distribution in certain cases due to linearity and standard form transformations).
- \(\frac{Z-1}{Y}\): Here, the numerator \(Z-1 \sim \mathcal{N}(0,3)\) doesn't result in a standard normal variable, thus not fitting the criteria for a Cauchy distribution.
- \(\frac{X-Y}{X+Y}\): Although \(X \sim \mathcal{N}(-1,1)\) and \(Y \sim \mathcal{N}(0,1)\) are independent normals, this transformation doesn't provide a standard form matching a Cauchy distribution.
The correct answer is option 2: \(\frac{X+2Y + Z}{X-2Y + Z}\). This option consists of normal variables with transformations that potentially allow the combination of standard normal distributions in a ratio form that can be simplified or symmetrized to represent a Cauchy distribution.
Thus, this specific setup satisfies the properties required for a Cauchy distribution with location parameter 0 and scale parameter 1.
Important Questions from Random Variables
- 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 $\theta$?Let $X_1, X_2,..., X_{15}$ be a random sample from an Exponential distribution with the probability density function
$f(x) = \begin{cases} \frac{1}{\sigma} \exp \left(-\frac{x}{\sigma}\right) & \text{if } x > 0, \\ 0, & \text{elsewhere,} \end{cases}$
where the unknown parameter $\sigma$ is positive. Let $\bar{X} = \frac{1}{15}\sum_{i=1}^{15} X_i$. Suppose that $\phi$ denotes the likelihood ratio test for testing $H_0: \sigma \le 1$ against $H_1 : \sigma > 1$ at level $\alpha = 0.1$. It is given that $\chi^2_{15,0.1} = 22.307$, $\chi^2_{15,0.9} = 8.547$, $\chi^2_{30,0.1} = 40.256$, $\chi^2_{30,0.9} = 20.599$, where $P(W > \chi^2_{n,\alpha}) = \alpha$ and $W \sim \chi^2_n$. Then which of the following statements are true?
