Question

Let X be an exponential random variable with parameter 2

Find the CDF of X Express your answer in terms of z using standard notation. Use e for the base of the natural logarithm

a) For x <= 0 ,F_{X}(x) =

b) For x > 0 ,F_{X}(x) =

Answer 1

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:

\[
F_X(x) =
\begin{cases}
0 & \text{for } x \leq 0 \\
1 - e^{-2x} & \text{for } x > 0
\end{cases}.
\]