The question asks to find the expected value $\mathbb{E}[X | X > T]$, where $X$ follows an exponential distribution with the probability density function (PDF):

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

This is a conditional expectation for an exponential random variable. The solution follows these steps:

Step 1: Understanding the Conditional Expectation

We need to compute $\mathbb{E}[X | X > T]$, the expected value of $X$ given that $X > T$. This can be expressed as:

$\mathbb{E}[X | X > T] = \int_T^{\infty} x \cdot f_{X | X > T}(x) \, dx$

where $f_{X | X > T}(x)$ is the conditional probability density function of $X$, given that $X > T$.

Step 2: Deriving the Conditional Density $f_{X | X > T}(x)$

The conditional PDF is given by:

$f_{X | X > T}(x) = \frac{f_X(x)}{\mathbb{P}(X > T)}$

where $\mathbb{P}(X > T)$ is the probability that $X$ exceeds $T$. For an exponential distribution, the survival function $\mathbb{P}(X > T)$ is:

$\mathbb{P}(X > T) = \int_T^{\infty} \lambda e^{-\lambda x} \, dx = e^{-\lambda T}$

Thus, the conditional PDF becomes:

$f_{X | X > T}(x) = \frac{\lambda e^{-\lambda x}}{e^{-\lambda T}} = \lambda e^{-\lambda (x - T)}, \quad x \geq T$

Step 3: Compute the Conditional Expectation

The conditional expectation is now:

$\mathbb{E}[X | X > T] = \int_T^{\infty} x \cdot \lambda e^{-\lambda (x - T)} \, dx$

This can be simplified by making the substitution $u = x - T$, which gives $du = dx$, and shifting the limits accordingly:

$\mathbb{E}[X | X > T] = \int_0^{\infty} (u + T) \lambda e^{-\lambda u} \, du$

This separates into two integrals:

$\mathbb{E}[X | X > T] = \lambda \int_0^{\infty} u e^{-\lambda u} \, du + T \lambda \int_0^{\infty} e^{-\lambda u} \, du$

The first integral $\int_0^{\infty} u e^{-\lambda u} \, du$ is the mean of an exponential distribution, which equals $\frac{1}{\lambda}$. The second integral $\int_0^{\infty} e^{-\lambda u} \, du$ equals $\frac{1}{\lambda}$ as well. Therefore:

$\mathbb{E}[X | X > T] = \frac{1}{\lambda} + T$

Final Answer:

$\mathbb{E}[X | X > T] = \frac{1}{\lambda} + T$

This shows that the conditional expectation of $X$ given $X > T$ is shifted by $T$ plus the mean $\frac{1}{\lambda}$ of the exponential distribution.

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How is the exponential distribution used to model real-world processes?
2. What is the memoryless property of the exponential distribution, and how does it apply here?
3. How does the result $\mathbb{E}[X | X > T] = \frac{1}{\lambda} + T$ illustrate the shift in the expected value?
4. Can you compute $\mathbb{P}(X > T)$ for different values of $\lambda$ and $T$?
5. What is the interpretation of the expected value when $T = 0$ in this context?

Tip: The exponential distribution is often used to model waiting times between independent events occurring at a constant rate.