Another random variable $Y$ follows $t_1$ distribution with probability density function $h(y)$ and cumulative distribution function $H(y)$. Let $c$ be the positive real number for which $g(c) = h(c)$.
Which of the following statements is/are correct?
Let's analyze the question and determine the correct statements by understanding the characteristics of the standard normal distribution and the t-distribution with 1 degree of freedom.
Conceptual Background:
- The probability density function (PDF) for a standard normal distribution, \(g(z)\), is given by: \(g(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}}\)
- The cumulative distribution function (CDF) for a standard normal distribution, \(G(z)\), is: \(G(z) = P(Z \leq z)\)
- The t-distribution with 1 degree of freedom is equivalent to the standard Cauchy distribution. Its PDF, \(h(y)\), is: \(h(y) = \frac{1}{\pi} \frac{1}{1 + y^2}\)
- The corresponding CDF, \(H(y)\), is: \(H(y) = \frac{1}{2} + \frac{1}{\pi}\arctan(y)\)
Now, we need to determine which of the provided statements are correct:
Statement Analysis:
- $G(0) = H(0)$:
The CDF \(G(0)\) for a standard normal distribution at \(z = 0\) is: \(G(0) = 0.5\) (since the standard normal is symmetric about 0).
The CDF \(H(0)\) for the t-distribution at \(y = 0\) is: \(H(0) = \frac{1}{2} = 0.5\).
Thus, this statement is correct. - $G(c) < H(c)$:
The condition \(g(c) = h(c)\) is used to find \(c\).
Given \(g(c) = h(c)\), the CDFs are compared for their values: \(G(c)\) and \(H(c)\) will differ based on distribution's tail properties.
Typically, for a given \(c\), the tail of the t-distribution is heavier compared to the normal distribution.
Therefore, generally \(G(c) > H(c)\), making this statement incorrect. - $G(-c) < H(-c)$:
Similar to \(c\), \(-c\) needs to be analyzed in context of symmetrical CDF properties: With the t-distribution having heavier tails, \(H(-c) > G(-c)\) for positive \(c\).
This statement is correct. - $g(0) = h(0)$:
Evaluating the PDFs at \(x = 0\): \(g(0) = \frac{1}{\sqrt{2\pi}}\) and \(h(0) = \frac{1}{\pi}\), indicating that \(g(0) \ne h(0)\).
This statement is incorrect.
Conclusion: The correct statements are $G(0) = H(0)$ and $G(-c) < H(-c)$, corresponding to the logic and evaluation steps above.
Important Questions from Continuous Distributions
Suppose X is a continuous random variable with probability density function
\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), -∞ < x < ∞.
Define
\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)
Then which of the following statements are true?
Let X1, X2, ..., Xn be a random sample from an absolutely continuous distribution with the probability density function
\(f(x \mid \theta)=\left\{\begin{array}{cl} e^{\theta-x}, & \text { if } x \geq \theta \\ 0, & \text { if } x<\theta \end{array},\right.\)
where θ ∈ ℝ is unknown. Define \(\bar{X}=\frac{1}{n} \sum_{i=1}^n X_i\) and X(1) = min{X1, ..., Xn}. Then
which of the following statements are true?
Suppose that X is a continuous random variable with probability density function given by:
f(x) = \(\left\{ {\begin{array}{c} {\frac{x}{8},}&{x \in \left[ {0,2} \right)}\\ {\frac{1}{4},}&{x \in \left[ {2,4} \right)}\\ { - \frac{x}{8} + \frac{3}{4},}&{x \in \left[ {4,6} \right)} \end{array}}\right.\)
Find the mean of X.
The variable x takes a value between 0 and 10 with uniform probability distribution. The variable y takes a value between 0 and 20 with uniform probability distribution. The probability of the sum of variables (x + y) being greater than 20 is _________
Probability density function of a random variable X is given below
\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{if\;1 \le x \le 5}\\ 0&{otherwise} \end{array}} \right.\)
P (X ≤ 4) is
