Question

Probability Density function $f(x)$ of a continuous random variable is given by $f(x) = ce^{-|x|}, -\infty<x<\infty$. If the total probability $\int_{-\infty}^{\infty} f(x) dx = 1$ then $c =$

• A. \( 0.6 \)
• B. \( 0.5 \)
• C. \( 0.4 \)
• D. \( 0.3 \)

Hint:

To find the constant in a PDF, set the integral of the PDF over its entire domain equal to 1 and solve for the constant. Remember the definition of the absolute value function when integrating.

Answer 1

The Correct Option is B

For a probability density function, the integral over its entire range must equal 1. Given $f(x) = ce^{-|x|}$ for $-\infty<x<\infty$, we have: $$\int_{-\infty}^{\infty} ce^{-|x|} dx = 1$$ Since $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x<0$, we can split the integral into two parts: $$\int_{-\infty}^{0} ce^{-(-x)} dx + \int_{0}^{\infty} ce^{-(x)} dx = 1$$ $$c \int_{-\infty}^{0} e^{x} dx + c \int_{0}^{\infty} e^{-x} dx = 1$$ $$c [e^x]_{-\infty}^{0} + c [-e^{-x}]_{0}^{\infty} = 1$$ $$c (e^0 - \lim_{x \to -\infty} e^x) + c (-\lim_{x \to \infty} e^{-x} - (-e^{-0})) = 1$$ $$c (1 - 0) + c (-0 - (-1)) = 1$$ $$c(1) + c(1) = 1$$ $$2c = 1$$ $$c = \frac{1}{2} = 0.5$$ Thus, the value of $c$ is $0.5$.