Given $X \sim \text{Exp}(1)$, meaning $X$ follows an exponential distribution with rate parameter $\lambda = 1$, we are tasked with finding the conditional expectation $E(X | X \geq 2)$.

For an exponential random variable with rate parameter $\lambda$, the probability density function (PDF) is given by:

$f_X(x) = \lambda e^{-\lambda x}, \quad x \geq 0$

In this case, since $\lambda = 1$, the PDF simplifies to:

$f_X(x) = e^{-x}, \quad x \geq 0$

Conditional Expectation for Exponential Distribution

For an exponential distribution, the conditional expectation $E(X | X \geq a)$ is given by the following formula:

$E(X | X \geq a) = a + \frac{1}{\lambda}$

In our case, since $\lambda = 1$ and $a = 2$, we substitute these values into the formula:

$E(X | X \geq 2) = 2 + \frac{1}{1} = 2 + 1 = 3$

Thus, the conditional expectation is:

$E(X | X \geq 2) = 3$

Would you like further details on the derivation, or any other questions?

Related Questions:

1. How is the exponential distribution derived from the Poisson process?
2. What is the memoryless property of the exponential distribution?
3. How does the conditional expectation formula generalize to other values of $\lambda$?
4. Can you derive the conditional probability $P(X \geq b | X \geq a)$ for exponential distributions?
5. What are some real-world examples of phenomena modeled by the exponential distribution?

Tip:

For any $X \sim \text{Exp}(\lambda)$, the distribution has the memoryless property: $P(X \geq t + s | X \geq t) = P(X \geq s)$.