Find $k$ such that $f(x) = kx^3 e^{-x/2}$ is a density function - shortcut for integration?

Question

Given the PDF $$f(x) = kx^3 e^{-x/2},\ \ \forall x > 0$$ And $0$ otherwise, find the value of $k$ that makes $f(x)$ a density function. I know this is solvable by integration by parts, but I recall there being a shortcut that you can use to skip integration since this is an exponential distribution, does anyone remember it?

Answer 1

No, it is not exponential. It resembles the Gamma distribution. If $X\sim \operatorname{Gamma}(r,\lambda)$, then $$f_X(x) = \frac{1}{\Gamma(r)}\lambda^r x^{r-1} e^{-\lambda x}.$$

Your pdf is $f(x) = kx^{4-1} e^{-\frac{1}{2}x}$, which suggests that $\lambda = \frac{1}{2}$ and $r = 4$. So we can deduce that $k = \frac{1}{\Gamma(4)}\left(\frac{1}{2}\right)^4$ and $$f(x) = \frac{1}{\Gamma(4)}\left(\frac{1}{2}\right)^4x^{4-1}e^{-\frac{1}{2}x}.$$ You can check that it does integrate to $1$.

Answer 2

Hint: If $n > 0$ $$I_n = \int_0^\infty t^ne^{-t}\ dt = n\int_0^\infty t^{n-1}e^{-t}\ dt = n\cdot I_{n-1}$$