Question

3 points possible (graded)

Let X be an exponential random variable with parameter λ.

a) The probability that X > 5 is
b) The probability that X>5 given that X>2 is
c) Given that X>2 and for a small delta>0, the probability that 4<=X<=(4+2 delta) is approximately

Answer 1

Let \( X \) be an exponential random variable with parameter \( \lambda \). The probability density function (pdf) of \( X \) is given by:

\[
f_X(x) = \lambda e^{-\lambda x} \quad \text{for } x \geq 0
\]

The cumulative distribution function (CDF) is:

\[
F_X(x) = 1 - e^{-\lambda x} \quad \text{for } x \geq 0
\]

### a) The probability that \( X > 5 \)

The probability that \( X \) is greater than 5 is given by:

\[
P(X > 5) = 1 - F_X(5) = e^{-\lambda \cdot 5} = e^{-5\lambda}
\]

### b) The probability that \( X > 5 \) given that \( X > 2 \)

This can be calculated using the memoryless property of the exponential distribution. The memoryless property states that:

\[
P(X > s + t \mid X > s) = P(X > t)
\]

For \( s = 2 \) and \( t = 3 \) (since \( 5 - 2 = 3 \)):

\[
P(X > 5 \mid X > 2) = P(X > 3) = e^{-\lambda \cdot 3} = e^{-3\lambda}
\]

### c) Given that \( X > 2 \) and for a small \( \delta > 0 \), the probability that \( 4 \leq X \leq (4 + 2\delta) \) is approximately

For this case, we need the conditional probability \( P(4 \leq X \leq 4 + 2\delta \mid X > 2) \). We can use the definition of conditional probability:

\[
P(4 \leq X \leq 4 + 2\delta \mid X > 2) = \frac{P(4 \leq X \leq 4 + 2\delta)}{P(X > 2)}
\]

1. Calculate \( P(X > 2) \):

\[
P(X > 2) = e^{-2\lambda}
\]

2. Calculate \( P(4 \leq X \leq 4 + 2\delta) \):

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

For small \( \delta \), we can use the approximation \( e^{-\lambda(4 + 2\delta)} \approx e^{-4\lambda} (1 - 2\delta\lambda) \):

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

3. Now, substitute back into the conditional probability formula:

\[
P(4 \leq X \leq 4 + 2\delta \mid X > 2) = \frac{e^{-4\lambda} (2\delta\lambda)}{e^{-2\lambda}} = 2\delta\lambda e^{-2\lambda}
\]

Thus, given that \( X > 2 \) and for a small \( \delta > 0 \):

\[
P(4 \leq X \leq 4 + 2\delta \mid X > 2) \approx 2\delta\lambda e^{-2\lambda}
\]