Answer to Question #343635 in Statistics and Probability for Rohit

Question

f(x) = kx * e ^ (- x / 2); k is a constant, x > 0 a) Find the value of k for f(x) to be a valid probability function

Answer 1

$$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{\infin}kxe^{-x/2}dx=1$$

$$\int xe^{-x/2}dx$$

$$u=x, du=dx$$

$$dv=e^{-x/2}dx, v=-2e^{-x/2}$$

$$\int xe^{-x/2}dx=-2xe^{-x/2}+2\int e^{-x/2}dx$$

$$=-2xe^{-x/2}-4e^{-x/2}+C$$

$$\displaystyle\int_{0}^{\infin}kxe^{-x/2}dx=k\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t}xe^{-x/2}dx$$

$$=k\lim\limits_{t\to \infin}[-2xe^{-x/2}-4e^{-x/2}]\begin{matrix} t \\ 0 \end{matrix}$$

$$=k(-0-0-(-0-4))=4k=1$$

$$k=\dfrac{1}{4}$$