Question

Let X be an exponential random variable with parameter lambda = 2 Find the values of the following. Use 'e' for the base of the natural logarithm

a) E[(3 X + 1) ^ 2] =

b) P(1 <= X <= 2) =

Answer 1

To solve for the values related to the exponential random variable \( X \) with parameter \( \lambda = 2 \), we can use the properties of the exponential distribution.

### a) \( E[(3X + 1)^2] \)

To find \( E[(3X + 1)^2] \), we first expand the expression:

\[
(3X + 1)^2 = 9X^2 + 6X + 1
\]

Now, we can use the linearity of expectation:

\[
E[(3X + 1)^2] = E[9X^2] + E[6X] + E[1]
\]

Since expectations can be factored out:

\[
E[(3X + 1)^2] = 9E[X^2] + 6E[X] + 1
\]

Next, we calculate \( E[X] \) and \( E[X^2] \):

- The mean of an exponential random variable \( E[X] \) is given by \( \frac{1}{\lambda} \):

\[
E[X] = \frac{1}{2} = 0.5
\]

- The second moment \( E[X^2] \) can be calculated using the formula \( E[X^2] = \frac{2}{\lambda^2} \):

\[
E[X^2] = \frac{2}{2^2} = \frac{2}{4} = 0.5
\]

Now substituting these back into the equation for \( E[(3X + 1)^2] \):

\[
E[(3X + 1)^2] = 9(0.5) + 6(0.5) + 1 = 4.5 + 3 + 1 = 8.5
\]

Thus, the answer for part a) is:

\[
\boxed{8.5}
\]

### b) \( P(1 \leq X \leq 2) \)

To find the probability \( P(1 \leq X \leq 2) \) for an exponential random variable, we can use the cumulative distribution function (CDF):

The CDF for an exponential random variable \( X \) is given by:

\[
F(x) = 1 - e^{-\lambda x}
\]

For our case, \( \lambda = 2 \):

1. First, calculate \( P(X \leq 2) \):

\[
P(X \leq 2) = F(2) = 1 - e^{-2 \times 2} = 1 - e^{-4}
\]

2. Next, calculate \( P(X \leq 1) \):

\[
P(X \leq 1) = F(1) = 1 - e^{-2 \times 1} = 1 - e^{-2}
\]

3. Therefore, the probability \( P(1 \leq X \leq 2) \) is:

\[
P(1 \leq X \leq 2) = P(X \leq 2) - P(X \leq 1) = (1 - e^{-4}) - (1 - e^{-2}) = e^{-2} - e^{-4}
\]

Thus, the answer for part b) is:

\[
\boxed{e^{-2} - e^{-4}}
\]