Continuous Probability

Question

The Cumulative Density Function of a random variable $X$ is $CDF(x)=c*e^{-2*x}$ if $0 \leq X < \infty$ and is $0$ otherwise. What is $P(X \geq 2)$?

Answer

$PDF(x)=\frac{\mathrm{d}}{\mathrm{d}x}CDF(x)$


First we must normalize the PDF:


$1=\int_{0}^{\infty}PDF(x)\ \mathrm{d}x$
$1=(c*e^{-2*\infty})-(c*e^{-2*0})$
$c=-1$


$PDF(x)=-\frac{1}{2}*e^{-2*x}$


Now we can calculate the probability:


$P(X \geq 2)=\int_{2}^{\infty}-\frac{1}{2}*e^{-2*x}\ \mathrm{d}x$
$P(X \geq 2)=(-e^{-2*\infty})-(-e^{-2*2})$
$P(X \geq 2)=e^{-4}$
$P(X \geq 2) \cong 0.018316$