Probability Density Function without having the definite integral

Question

So we're given a random variable $X$ and an $f(x) =\begin{cases} 0, & \text{if $x<0$} \\ cxe^{-x}, & \text{if $x≥0 $} \end{cases}$, where $c$ is a constant.

How do I find the constant if I'm not given something to use as a definite integral?

Answer 1

Every probability density function $f(x)$ must satisfy $\int_{-\infty}^{\infty} f(x) \mathop{dx} = 1$. The reason for this is that $\int_{a}^{b} f(x) \mathop{dx}$ can be interpreted as the probability of the random variable $X$ lying in $[a, b]$. Thus, $\int_{-\infty}^{\infty} f(x) \mathop{dx}$ represents the probability of the random variable $X$ lying anywhere in $\mathbb{R}$, which occurs with probability one.

Hence, we have $$ \begin{align*} 1 &= c \cdot \int_{-\infty}^{\infty} f(x) \mathop{dx} \\[0.7em] &= c\left(\int_{-\infty}^{0} 0 \mathop{dx} + \int_{0}^{\infty} xe^{-x} \mathop{dx}\right) \\[0.7em] &= c \cdot \left(0 + 1\right), \end{align*} $$

which implies $\boxed{ c = 1}$