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

To find \( \mathbb{E}[X \mid X > 1] \), we need to compute the conditional expectation of \( X \) given that \( X > 1 \). The random variable \( X \) follows an exponential distribution with mean \( 1/\lambda \), which has the probability density function (PDF) given by:

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

The conditional expectation \( \mathbb{E}[X \mid X > 1] \) is defined as:

\[
\mathbb{E}[X \mid X > 1] = \frac{\int_1^{\infty} x f_X(x) \, dx}{P(X > 1)}
\]

### Step 1: Calculate \( P(X > 1) \)
The probability \( P(X > 1) \) is given by the survival function of the exponential distribution:

\[
P(X > 1) = \int_1^{\infty} \lambda e^{-\lambda x} \, dx
\]

This integral can be computed as:

\[
P(X > 1) = \left[ -e^{-\lambda x} \right]_1^{\infty} = e^{-\lambda}
\]

### Step 2: Calculate the numerator \( \int_1^{\infty} x f_X(x) \, dx \)

The next step is to calculate the integral:

\[
\int_1^{\infty} x \lambda e^{-\lambda x} \, dx
\]

This integral can be solved using integration by parts. Let \( u = x \) and \( dv = \lambda e^{-\lambda x} dx \), then:

\[
\int_1^{\infty} x \lambda e^{-\lambda x} \, dx = \left[ -x e^{-\lambda x} \right]_1^{\infty} + \int_1^{\infty} e^{-\lambda x} \, dx
\]

Evaluating the boundary terms and the remaining integral:

\[
\left[ -x e^{-\lambda x} \right]_1^{\infty} = 0 - (-1 e^{-\lambda}) = e^{-\lambda}
\]
\[
\int_1^{\infty} e^{-\lambda x} \, dx = \frac{e^{-\lambda}}{\lambda}
\]

Thus, the total integral is:

\[
\int_1^{\infty} x \lambda e^{-\lambda x} \, dx = e^{-\lambda} + \frac{e^{-\lambda}}{\lambda}
\]

### Step 3: Calculate \( \mathbb{E}[X \mid X > 1] \)

Now, using the formula for conditional expectation:

\[
\mathbb{E}[X \mid X > 1] = \frac{e^{-\lambda} + \frac{e^{-\lambda}}{\lambda}}{e^{-\lambda}} = 1 + \frac{1}{\lambda}
\]

Thus, the conditional expectation is:

\[
\mathbb{E}[X \mid X > 1] = 1 + \frac{1}{\lambda}
\]

### Final Answer:
\[
\mathbb{E}[X \mid X > 1] = 1 + \frac{1}{\lambda}
\]

Do you have any questions or would you like further details on any of the steps?

Here are 5 related questions:
1. What is the general formula for the conditional expectation of an exponential random variable given a threshold \( X > a \)?
2. How is the survival function of an exponential distribution derived?
3. Can you explain the intuition behind the exponential distribution's memoryless property?
4. How does the mean of an exponential distribution change if the rate parameter \( \lambda \) is doubled?
5. What is the variance of the conditional distribution of \( X \) given \( X > 1 \)?

**Tip:** When solving problems involving exponential distributions, remember that the exponential distribution is memoryless, meaning past events do not influence future outcomes.

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

Explore how to calculate the conditional expectation of an exponential random variable given that X > 1. This detailed solution uses concepts from probability theory, including the exponential distribution's PDF and survival function. Suitable for undergraduate students studying probability.

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