Answer:
[tex](1)p(x)\geq 0\\(2)\int_{0}^{\infty} p(x) dx=0[/tex]
Explanation:
A function f(x) is a Probability Density Function if it satisfies the following conditions:
[tex](1)f(x)\geq 0\\(2)\int_{0}^{\infty} f(x) dx=0[/tex]
Given the function:
[tex]p(x)=\dfrac{1}{r}e^{-x/r} \: on\: [0,\infty), where\:r=\dfrac{20}{ln(2)}[/tex]
(1)p(x) is greater than zero since the range of exponents of the Euler's number will lie in [tex][0,\infty).[/tex]
(2)
[tex]\int_{0}^{\infty} p(x)=\int_{0}^{\infty} \dfrac{1}{r}e^{-x/r}\\=\dfrac{1}{r} \int_{0}^{\infty} e^{-x/r}\\=-\dfrac{r}{r}\left[e^{-x/r}\right]_{0}^{\infty}\\=-\left[e^{-\infty/r}-e^{-0/r}\right]\\=-e^{-\infty}+e^{-0}\\SInce \: e^{-\infty} \rightarrow 0\\e^{-0}=1\\\int_{0}^{\infty} p(x)=1[/tex]
The function p(x) satisfies the conditions for a probability density function.
