$$f ( x ) = \left\{ \begin{array} { l l }
4 e ^ { - 4 x } & \text { for } x \geq 0 \
0 & \text { for } x

Question

$$f ( x ) = \left\{ \begin{array} { l l } 
4 e ^ { - 4 x } & \text { for } x \geq 0 \
0 & \text { for } x < 0
\end{array} \right.$$ is a probability density function.

Quizplus · Accepted Answer

A function can be considered a probability density function (PDF) if it satisfies two conditions: (1) the function is non-negative for all values of $x$, and (2) the integral of the function over its entire range equals 1. In this case, $f(x) = 4e^{-4x}$ for $x \geq 0$ and $f(x) = 0$ for $x < 0$ meets both criteria. For the first condition, $f(x)$ is clearly non-negative for all $x$ since $e^{-4x}$ is always positive for $x \geq 0$, and $f(x) = 0$ for $x < 0$. For the second condition, integrating $f(x) = 4e^{-4x}$ from 0 to $\infty$ yields $\int_{0}^{\infty} 4e^{-4x} dx = [-e^{-4x}]_{0}^{\infty} = 1$, satisfying the requirement that the total area under the curve is 1.

f(x)={4e−4x for x≥00 for x<0f ( x ) = \left\{ \begin{array} { l l } 4 e ^ { - 4 x } & \text { for } x \geq 0 \\0 & \text { for } x < 0\end{array} \right.f(x)={4e−4x0​ for x≥0 for x<0​

$f ( x ) = \left\{ \begin{array} { l l } 4 e ^ { - 4 x } & \text { for } x \geq 0 \\0 & \text { for } x < 0\end{array} \right.$