Find the value of λ such that the function f (x) is a valid probability density function. _______ \(f\left( x \right)\begin{array}{*{20}{c}} { = \lambda \left( {x - 1} \right)\left( {2 - x} \right)}&{for1 \le x \le 2}\\ { = 0}&{otherwise} \end{array}\)
Concept:
We know for probability density function
\(\mathop \smallint \nolimits_{ - \infty }^\infty f\left( x \right)dx = 1\)
Calculation:
Given:
f(x) = λ (x – 1) × (2 – x) 1 ≤ x ≤ 2, 0 otherwise
\(\mathop \smallint \nolimits_1^2 \lambda \left( { - {x^2} + 3x - 2} \right)\;dx = 1\)
\(\Rightarrow \lambda \left[ {\dfrac{{ - {x^3}}}{3} + 3 \times \dfrac{{{x^2}}}{2} - 2x} \right]_1^2 = 1\)
\(\Rightarrow \lambda \left[ {\dfrac{{ - 14 + 27 - 12}}{6}} \right] = 1\)
λ = 6Suppose 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.
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?