Question

If the density function of \(X\) equals $$ f(x)=\left\\{\begin{array}{ll} c e^{-2 x}, & 02\\} ?\)

Short answer

The constant 'c' is 2, and the probability P(X > 2) is \(e^{-4}\).

Step-by-step solution

Find the constant 'c'

To find the constant 'c', we will integrate the density function over its range and set the result equal to 1. Since random variable X has a range of (0, ∞), we will integrate the function with respect to x in this range. \[ \int_{0}^{\infty} ce^{-2x} dx = 1 \]

Integrate the density function

To solve for c, integrate the density function with respect to x. \[ c \int_{0}^{\infty} e^{-2x} dx = 1 \] To integrate the exponential function, we will use: \[ \int e^{-2x} dx = -\frac{1}{2}e^{-2x} \] Therefore, \[ c \left[-\frac{1}{2}e^{-2x} \right]_0^\infty = 1 \]

Evaluate the integral and solve for 'c'

Next, we will evaluate the integral and solve for c. \[ c \left[ -\frac{1}{2}e^{-2(\infty)} - \left(-\frac{1}{2}e^{-2(0)}\right) \right] = 1 \] The first term is zero since e^(-2∞) = 0: \[ c \left[ 0 - \left(-\frac{1}{2}e^{0}\right) \right] = 1 \] \[ c \left[\frac{1}{2} \right] = 1 \] \[ c = 2 \] The value of c is 2. Now we update the density function: \[ f(x) = \left\{\begin{array}{ll} 2e^{-2x}, & 0<x<\infty \\ 0, & x<0 \end{array}\right. \]

Finding the probability P(X > 2)

Now we need to find the probability P(X > 2). Integrate the updated density function over the range (2, ∞). \[ P(X > 2) = \int_{2}^{\infty} 2e^{-2x} dx \]

Integrate the updated density function

Integrate the updated density function with respect to x. \[ 2 \int_{2}^{\infty} e^{-2x} dx \] We already know that \(\int e^{-2x} dx = -\frac{1}{2}e^{-2x}\): \[ 2 \left[-\frac{1}{2}e^{-2x} \right]_2^\infty \]

Evaluate the integral

Evaluate the integral to find the probability P(X > 2). \[ 2 \left[ -\frac{1}{2}e^{-2(\infty)} - \left(-\frac{1}{2}e^{-2(2)}\right) \right] \] \[ 2 \left[ 0 - \left(-\frac{1}{2}e^{-4}\right) \right] \] \[ P(X > 2) = e^{-4} \] The probability P(X > 2) is \(e^{-4}\).