To find the cumulative distribution function (CDF) of an exponential random variable \( X \) with parameter \( \lambda \), we start with the probability density function (PDF) of the exponential distribution. The PDF for an exponential random variable is given by:
\[ f_X(x) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0, \]
and \( f_X(x) = 0 \) for \( x < 0 \).
In this case, the parameter \( \lambda \) is given as 2. So, the PDF becomes:
\[ f_X(x) = 2 e^{-2x} \quad \text{for } x \geq 0, \]
and \( f_X(x) = 0 \quad \text{for } x < 0 \).
a) For \( x \leq 0 \):
For \( x \leq 0 \), the cumulative distribution function \( F_X(x) \) is simply the probability that \( X \) is less than or equal to \( x \). Since the exponential random variable only takes non-negative values, we have:
\[ F_X(x) = 0 \quad \text{for } x < 0. \]
b) For \( x > 0 \):
To compute the CDF for \( x > 0 \), we need to integrate the PDF from 0 to \( x \):
\[ F_X(x) = \int_0^x f_X(t) , dt = \int_0^x 2 e^{-2t} , dt. \]
Now we compute the integral:
\[ F_X(x) = 2 \int_0^x e^{-2t} , dt. \]
Using the antiderivative of \( e^{-2t} \):
\[ \int e^{-2t} , dt = -\frac{1}{2} e^{-2t} + C. \]
Evaluating the definite integral gives:
\[ F_X(x) = 2 \left[ -\frac{1}{2} e^{-2t} \right]_0^x = 2 \left( -\frac{1}{2} e^{-2x} + \frac{1}{2} e^0 \right) = 2 \left( -\frac{1}{2} e^{-2x} + \frac{1}{2} \right). \]
This simplifies to:
\[ F_X(x) = 1 - e^{-2x} \quad \text{for } x > 0. \]
Summary:
The cumulative distribution function (CDF) of the exponential random variable \( X \) with parameter \( 2 \) is:
\[ F_X(x) = \begin{cases} 0 & \text{for } x \leq 0 \ 1 - e^{-2x} & \text{for } x > 0 \end{cases}. \]