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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.

Let X be exponential with mean 1/λ; that is, f_X(x) = λe^{-λx}, 0 < x < ∞. Find E[X | X > T].

This problem focuses on calculating the conditional expectation E[X | X > T] for an exponential distribution with mean 1/λ. The solution involves understanding conditional probability density functions and using integration to derive the result.

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