Suppose $$ f(x)= \begin{cases} cxe^{-2x}, & x\ge 0\\ 0, & \text{otherwise} \end{cases} $$ is the density function. Determine $c$ and find the distribution function.
Can anyone help me solve this problem? My teacher has already given me the solution, but I don't think it is correct. Thank you.
By the definition of a probability density, we must have $\int_{\Omega} f = 1$, where $\Omega$ is the support of the random variable.
That is, we must have $$\int_0^{\infty} ce^{-2x} \; dx = 1.$$ Solving for $c$, we get: \begin{align*} \int_0^{\infty} e^{-2x} \; dx & = \frac{1}{c} \\ \frac{1}{2} & = c \\ c & = 2. \end{align*} Alternatively, we could recognize the density function as the form of an exponential distribution, which has density function $\lambda e^{-\lambda x}$ with parameter $\lambda > 0$, which takes support on the non-negative real line.